Research Article On Degrees of Modular Common Divisors and...

Research ArticleOn Degrees of Modular Common Divisors andthe Big Prime gcd Algorithm

Vahagn Mikaelian12

1Yerevan State University Alex Manoogian 1 0025 Yerevan Armenia2American University of Armenia 40 Marshal Baghramyan Ave 0019 Yerevan Armenia

Correspondence should be addressed to Vahagn Mikaelian vmikaelianysuam

Received 6 July 2016 Accepted 28 August 2016

Academic Editor Nawab Hussain

Copyright copy 2016 Vahagn Mikaelian This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We consider a few modifications of the Big prime modular gcd algorithm for polynomials in Z[119909] Our modifications are basedon bounds of degrees of modular common divisors of polynomials on estimates of the number of prime divisors of a resultantand on finding preliminary bounds on degrees of common divisors using auxiliary primesThese modifications are used to suggestimproved algorithms for gcd calculation and for coprime polynomials detection To illustrate the ideas we apply the constructedalgorithms on certain polynomials in particular on polynomials from Knuthrsquos example of intermediate expression swell

1 Introduction

This work is one of the articles in which we would like topresent parts from new introduction to computer algebra [1]that currently is under preparation In [1] we try to give aldquomore algebraicrdquo and detailed view on some of the areas ofcomputer algebra such as algorithms on the Euclidean ringsextensions of fields operators in spaces on finite fields andfactorization in UFDs

The Big prime modular gcd algorithm is one of the firstandmost popular algorithms of computer algebra In its clas-sical form it allows calculating the greatest common divisorgcd(119891(119909) 119892(119909)) for any nonzero polynomials 119891(119909) 119892(119909) isin

Z[119909]There are a fewmodifications of this algorithm for otherUFDs such asmultivariate polynomial rings Attention to thegcd calculation is partially explained by the first examples thatwere built to explain importance of application of algebraicmethods to computer science In particular Knuthrsquos well-known example of intermediate expression swell discussesthe polynomials

119891 (119909) = 1199098+ 1199096minus 31199094minus 31199093+ 81199092+ 2119909 minus 5

119892 (119909) = 31199096+ 51199094minus 41199092minus 9119909 + 21

(1)

and it shows that calculation of gcd(119891(119909) 119892(119909)) by traditionalEuclidean algorithm on rational numbers generates very

large integers to deal with whereas consideration of thesepolynomials modulo 119901 that is consideration of their imagesunder ring homomorphism 120593

119901 Z[119909] rarr Z

119901[119909] (where

Z119901[119909] is the polynomial ring over the residue ring Z

119901cong

Z119901Z) very easily shows that gcd(119891(119909) 119892(119909)) = 1 (see [2] andalso [3ndash6]) We are going to use polynomials (1) as examplesbelow to apply the algorithms below (see Examples 16 18 2225 27 and 29)

The main idea of the Big prime modular gcd algorithmis that for the given polynomials 119891(119909) 119892(119909) isin Z[119909] onemay first consider their images 119891

119901(119909) = 120593

119901(119891(119909)) 119892

119901(119909) =

120593119901(119892(119909)) isin Z

119901[119909] under 120593

119901 Unlike Z[119909] the ring Z

119901[119909] is

an Euclidean domain since it is a polynomial ring over a fieldso gcd(119891

119901(119909) 119892119901(119909)) can be computed in it by the well-known

Euclidean algorithm There remains ldquoto liftrdquo a certain fold119905 sdot gcd(119891

119901(119909) 119892119901(119909)) of it to the ring Z[119909] to reconstruct the

preimage gcd(119891(119909) 119892(119909)) The ldquoliftingrdquo procedure consistsof selecting the suitable value for prime 119901 then finding inZ[119909] an appropriate preimage for gcd(119891

119901(119909) 119892119901(119909)) and then

checking if that preimage divides both 119891(119909) and 119892(119909) If yesit is gcd(119891(119909) 119892(119909)) we are looking for If not then a new119901 need be selected to repeat the process Arguments basedon resultants and on Landau-Mignotte bounds show that wecan effectively choose 119901 such that the number of requiredrepetitions is ldquosmallrdquo

Hindawi Publishing CorporationInternational Journal of Mathematics and Mathematical SciencesVolume 2016 Article ID 3262450 13 pageshttpdxdoiorg10115520163262450

2 International Journal of Mathematics and Mathematical Sciences

The first aim of this work is to present in Sections 2ndash5 a slightly modified argumentation of the algorithm basedon comparison of the degrees of common divisors of 119891(119909)and 119892(119909) in Z[119909] and of 119891

119901(119909) and 119892

119901(119909) in Z

119901[119909] (see

Algorithm 2) This approach allows some simplification ofa step of the algorithm for some primes 119901 we need notreconstruct the preimage of 119905 sdot gcd(119891

119901(119909) 119892119901(119909)) but we

immediately get an indication that this prime is not suitableand we should proceed to a new 119901 (see Remark 15)

Then in Section 6 we discuss the problem if the Big primemodular gcd algorithm could output the correct answer usingjust one prime 119901 or not The answer is positive but for somereasons it should not be used to improve the algorithm (tomake it work with one 119901) because it evolves a very largeprime (see Remark 19) Instead we show that we can estimatethe maximal number of 119901 (repetitions of steps) that may beused in traditional Big prime modular gcd algorithm Forexample for polynomials (1) of Knuthrsquos example this numberis at most 31 Estimates of this type can be found in literatureelsewhere We just make the bound considerably smaller (seeRemark 23)

The obtained bounds on the number of primes 119901 areespecially effective when we are interested not in gcd but justin detection if the polynomials119891(119909) 119892(119909) isin Z[119909] are coprimeor not We consider this in Section 7 (see Algorithm 3)

In Section 8 we consider four other ideas to modify theBig prime modular gcd algorithm Two first ideas are basedon checking the number of primes 119901 The third idea is basedon using an auxiliary prime 119902 to estimate the degree ofgcd(119891(119909) 119892(119909)) by means of the degree of gcd(119891

119902(119909) 119892119902(119909))

(see Algorithm4) Example 27 shows howmuch better resultswe may get by this modification The fourth idea combinesboth approaches it uses a set of auxiliary primes 119902

1 119902

119896+1

to correctly find the degree of gcd(119891(119909) 119892(119909)) and thenwe use a modified version of Landau-Mignotte bound tofind a single big prime 119901 by which we can calculate thegcd(119891

119901(119909) 119892119901(119909))

The arguments used here can be generalized for thecase of polynomials on general UFDs From the uniquefactorization in a UFD it easily follows that gcd alwaysexists and it is easy to detect if the given common divisorof maximal degree is gcd or not The less simple part is tofind ways to compute gcd (without having the prime-powerfactorization) That can be done for some classes of UFDssuch as multivariate polynomials on fields

2 gcd in Polynomial Rings and the Degrees ofCommon Divisors

Theproblem of finding the greatest common divisor gcd(119886 119887)of any nonzero elements 119886 119887 in a ring 119877 can be separated totwo tasks

(1) Finding out if gcd(119886 119887) in general exists for 119886 119887 isin 119877(2) Finding an effective way to calculate gcd(119886 119887)

The Euclidean algorithm gives an easy answer to bothof these tasks in any Euclidean domain that is an integritydomain119877 possessingEuclidean norm 120575 1198770 rarr Ncup0 such

that 120575(119886119887) ge 120575(119886) hold for any nonzero elements 119886 119887 isin 119877 andfor any 119886 119887 isin 119877 where 119887 = 0 there exist elements 119902 119903 isin 119877such that 119886 = 119902119887 + 119903 where either 119903 = 0 or 119903 = 0 and120575(119903) lt 120575(119887) [4 5 7ndash10] The Euclidean algorithm works forany polynomial ring119870[119909] over a field119870 such asQ[119909]R[119909]C[119909] and Z

119901[119909] because these rings can easily be turned to

an Euclidean domain by defining 120575(119891(119909)) = deg119891(119909) for anynonzero 119891(119909) isin 119870[119909]

The situation is less simple in non-Euclidean domainseven in such a widely used ring as the ring Z[119909] of polyno-mials with integer coefficients That Z[119909] is not an Euclideandomain easy to show by elements 119909 2 isin Z[119909] IfZ[119909]were anEuclidean domain it would contain elements 119906(119909) V(119909) suchthat 119909 sdot 119906(119909) + 2 sdot V(119909) = gcd(119909 2) = plusmn1 which is not possible

The first of two tasks mentioned above namely existenceof gcd can be accomplished for Z[119909] by proving that Z[119909] isa UFD that is an integrity domain in which every nonzeroelement 119886 has a factorization 119886 = 120598 119901

1sdot sdot sdot 119901119896 where 120598 isin 119877

lowast

is a unit (invertible) element in 119877 the elements 119901119894are prime

for all 119894 = 1 119896 and the factorization above is uniquein the sense that if 119886 has another factorization of that type120579 1199021sdot sdot sdot 119902119904 where 120579 isin 119877

lowast and the elements 119902119894are prime then

119896 = 119904 and (perhaps after some reordering of the prime factors)the respective prime elements are associated 119901

119894asymp 119902119894for all

119894 = 1 119896 For briefness in the sequel we will often omit thephrase ldquoperhaps after some reordering of the prime factorsrdquoand this will cause no confusion

After merging the associated prime elements together weget a unique factorization into prime-power elements

119886 = ] 1199011205721

1sdot sdot sdot 119901120572119899

119899

] isin 119877lowast 120572119894isin N 119901

119894asymp 119901119895for any 119894 = 119895 119894 119895 = 1 119899

(2)

(in some arguments below we may admit that some of thefactors 119901120572119894

119894participate with degrees 120572

119894= 0 this makes some

notations simpler) From this it is easy to see that in a UFD119877 gcd(119886 119887) exists for any nonzero elements 119886 119887 isin 119877 Assume119887 isin 119877 has the factorization

119887 = 120581 1199011205721015840

1

1sdot sdot sdot 1199011205721015840

119899

119899 120581 isin 119877

lowast (3)

(we use the same primes 119901119894in both factorizations because

if eg 119901119894is not actually participating in one of those

factorizations we can add it as 119901120572119894119894with 120572

119894= 0) Then

gcd (119886 119887) = 119889 = 1199011205741

1sdot sdot sdot 119901120574119899

119899 (4)

where 120574119894= min120572

119894 1205721015840

119894 This follows from uniqueness of

factorization in UFD For if ℎ is a common divisor of 119886 119887 andif 119901119894is a prime divisor of ℎ then it also is a prime divisor of 119886

and of 119887 The elements 119901119894cannot participate in factorization

of ℎ by a power greater than min120572119894 1205721015840

119894 because then 119886 (or

119887) would have an alternative factorization in which 119901119894occurs

more than 120572119894(or 1205721015840119894) times

The shortest way to see that Z[119909] is a UFD is to applyGaussrsquos theorem if the ring 119877 is a UFD then the polynomialring 119877[119909] also is a UFD [4 7 8 10 11] SinceZ is a UFD (thatfact is known as ldquothe fundamental theorem of arithmeticrdquo)Z[119909] also is a UFD

International Journal of Mathematics and Mathematical Sciences 3

Clearly gcd(119886 119887) is defined up to a unit multiplier from119877lowast For integers from 119877 = Z or for polynomials from

119877 = Z[119909] this unit multiplier can be just minus1 or 1 So tosay gcd(119886 119887) is defined ldquoup to the sign plusmn1rdquo because Zlowast =

Z[119909]lowast

= minus1 1 And for polynomials from 119877 = Z119901[119909]

gcd(119886 119887) is defined up to any nonzero multiplier 119905 isin Zlowast119901=

1 119901minus1 Taking this into account we can use gcd(119886 119887) =1 and gcd(119886 119887) asymp 1 as equivalent notations since associatedelements are defined up to a unit multiplier Notice that insome sources they prefer to additionally introduce a normalform of gcd to distinguish one fixed instance of gcd Insteadof using that extra term we will just in a few places refer tothe ldquopositive gcdrdquo meaning that we take for example 2 =

gcd(6 8) and not minus2Furthermore since the content cont(119891(119909)) of a poly-

nomial 119891(119909) is gcd for some elements (coefficients of thepolynomials) the constant and the primitive part pp(119891(119909)) =119891(119909)cont(119891(119909)) can also be considered up to a unit multi-plier For a nonzero polynomial 119891(119909) isin Z[119909] we can choosecont(119891(119909)) so that sgn cont(119891(119909)) = sgn lc(119891(119909)) that isthe cont(119891(119909)) has the same sign as the leading coefficientof 119891(119909) Then the leading coefficient lc(pp(119891(119909))) of theprimitive part pp(119891(119909)) = 119891(119909)cont(119891(119909)) will be positiveWe will use this below without special notification

Now we would like to little restrict the algebraic back-ground we use Two main algebraic systems used in the Bigprimemodular gcd algorithm are the Euclidean domains andthe UFDs However their usage is ldquoasymmetricrdquo in the sensethat the Euclidean domains and Euclidean algorithm areused in many parts of the Big prime modular gcd algorithmwhereas the UFDs are used just to prove that gcd does existMoreover it is easy to understand that (2) and (4) mayhardly be effective tools to calculate gcd since they are usingfactorization of elements to primes while finding such afactorization is amore complicated task than finding just gcdThus it is reasonable to drop the UFDs from considerationand to obtain (2) directly using Gaussrsquos lemma on primitivepolynomials in Z[119909] (a polynomial 119891(119909) isin Z[119909] is primitiveif cont(119891(119909)) asymp 1 that is pp(119891(119909)) = 119891(119909)cont(119891(119909)) asymp

119891(119909))ByGaussrsquos lemma a product of twoprimitive polynomials

is primitive in Z[119909] [4 7 8 10 11] So if

119891 (119909) = cont (119891 (119909)) sdot pp (119891 (119909))

119892 (119909) = cont (119892 (119909)) sdot pp (119892 (119909)) (5)

thencont (119891 (119909) sdot 119892 (119909)) = cont (119891 (119909)) sdot cont (119892 (119909))

pp (119891 (119909) sdot 119892 (119909)) = pp (119891 (119909)) sdot pp (119892 (119909)) (6)

The following is easy to deduce from Gaussrsquos lemma

Lemma 1 If 119891(119909) 119905(119909) isin Z[119909] and 119905(119909) is primitive then if119905(119909) divides 119891(119909) in the ring Q[119909] and then 119905(119909) also divides119891(119909) in Z[119909]

The unique factorization of any nonzero 119891(119909) isin Z[119909]

is easy to obtain from decompositions (6) above and from

Lemma 1 Let us just outline it the details can be found in [1 45 7 8] By the fundamental theoremof arithmetic cont(119891(119909))can in a unique way be presented as a product of powers ofprimes cont(119891(119909)) = ] 119901

1205721

1sdot sdot sdot 119901120572119899

119899 So if deg119891(119909) = 0 then

we are doneAssume deg119891(119909) gt 0 If 119891(119909) is not prime then by

repeatedly splitting it to products of factors of lower degree asmany times as needed we will eventually get a presentationof 119891(119909) as a product of cont(119891(119909)) and of some finitely manyprimitive prime polynomials 119902

119894(119909) of degrees greater than 0

We do not yet have the uniqueness of this decomposition butwe can still group the associated elements together to get thepresentation

119891 (119909) = cont (119891 (119909)) sdot pp (119891 (119909))

= ] 1199011205721

1sdot sdot sdot 119901120572119899

119899sdot 1199021205731

1(119909) sdot sdot sdot 119902

120573119898

119898(119909)

(7)

If119891(119909) has another alternative presentation of this sort and if119905(119909) is one of the primitive prime factors (of degree greaterthan 0) of that presentation then product (7) is divisibleby 119905(119909) By Lemma 1 119905(119909) divides 119891(119909) also in Z[119909] Since119905(119909) is prime it is associated with one of 119902

119894(119909) Eliminate

one instance of this 119902119894(119909) in (7) and consider 119891(119909)119902

119894(119909) If

119891(119909)119902119894(119909) also is divisible by 119902

119894(119909) we repeat the process

If not we turn to other primitive prime polynomials (ofdegree greater than 0) dividing what remains from (7)after elimination After finitely many steps (7) will become] 1199011205721

1sdot sdot sdot 119901120572119899

119899 and also from the other alternative presentation

a constant should be left only So we apply the fundamentaltheorem of arithmetic one more time to get that (7) is theunique factorization

We see that (7) is a particular case of (2) The proof aboveavoided usage of Gaussrsquos theorem and the formal definitionsof the UFDs And we see that the prime elements ofZ[119909] areof two types prime numbers and primitive prime polynomialsof degrees greater than 0

Existence of gcd(119891(119909) 119892(119909)) for any two nonzero poly-nomials in 119891(119909) 119892(119909) isin Z[119909] can be deduced from (7) inanalogy with (4) If

119892 (119909) = ]1015840 1199011205721015840

1

1sdot sdot sdot 1199011205721015840

119899

119899sdot 1199021205731015840

1

1(119909) sdot sdot sdot 119902

1205731015840

119898

119898(119909) (8)

then

gcd (119891 (119909) 119892 (119909)) = 120581 1199011205741

1sdot sdot sdot 119901120574119899

119899sdot 1199021205751

1(119909) sdot sdot sdot 119902

120575119898

119898(119909) (9)

where 120581 = plusmn1 120574119894= min120572

119894 1205721015840

119894 and 120575

119895= min120573

119895 1205731015840

119895

(119894 = 1 119899 119895 = 1 119898) However like we admittedearlier (4) and (9) are no effective tools to calculate gcd Wewill turn to gcd calculation algorithm in the next sections (seeAlgorithms 4 and 5)

Equations (4) and (9) allow us to get some informationthat will be essential later Observe that the following defi-nition of gcd often used in elementary mathematics is nolonger true for general polynomial rings ldquo119889(119909) is the greatestcommon divisor of 119891(119909) and 119892(119909) if it is their common divisorof maximal degreerdquo For example for 119891(119909) = 12119909

2+ 24119909 + 12

and 119892(119909) = 8119909 + 8 the maximum of degree of their common


divisors is 1 Nevertheless ℎ(119909) = 119909+1 is not gcd(119891(119909) 119892(119909))although ℎ(119909) | 119891(119909) ℎ(119909) | 119892(119909) and deg ℎ(119909) = 1 For 119909+1is not divisible by the common divisor 2119909 + 2 We can detectthe cases when the divisor of highest degree is gcd

Lemma 2 For polynomials 119891(119909) 119892(119909) isin Z[119909] their commondivisor of maximal degree ℎ(119909) is their gcd if and only ifcont(ℎ(119909)) asymp gcd(cont(119891(119909)) cont(119892(119909)))

The lemma easily follows from (7) (8) and (9) We seethat in example above the condition was missing cont(119909 +

1) = 1 but gcd(cont(119891(119909)) cont(119892(119909))) = gcd(12 8) asymp 4 asymp 1In fact gcd(119891(119909) 119892(119909)) asymp 4119909 + 4

Corollary 3 For primitive polynomials 119891(119909) 119892(119909) isin Z[119909]their common divisor of maximal degree ℎ(119909) is their gcd ifand only if ℎ(119909) is primitive

In the case if polynomials are over a field the situation issimpler For any field 119870 the polynomial ring 119870[119909] is a UFD(and even an Euclidean domain) Any nonzero 119891(119909) isin 119870[119909]

has a factorization

119891 (119909) = 120579 sdot 1199021205731

1(119909) sdot sdot sdot 119902

120573119898

119898(119909)

120579 isin 119870lowast deg 119902

119894(119909) gt 0 119894 = 1 119898

(10)

which is unique in the sense mentioned above Since allnonzero scalars in119870 are units what we in (7) above had as aproduct of some prime numbers actually ldquomergesrdquo in 119870 intoa unit

] sdot 11990112057211sdot sdot sdot 119901120572119899

119899= 120579 isin 119870

lowast= 119870 0 (11)

Comparing factorizations of type (10) for any nonzeropolynomials 119891(119909) 119892(119909) isin 119870[119909] we easily get the following

Lemma 4 For any nonzero polynomials 119891(119909) 119892(119909) isin 119870[119909]

over a field119870 their common divisor of maximal degree ℎ(119909) istheir gcd

This in particular is true for rings mentioned aboveQ[119909] R[119909] C[119909] and Z

119901[119909] We will use this fact later

to construct the Big prime modular gcd algorithm and itsmodifications

The analog of Lemma 4 was not true for Z[119909] becausein factorization (9) we have the nonunit prime-power fac-tors 119901

120574119894

119894which do participate in factorization of 119889(119909) =

gcd(119891(119909) 119892(119909)) but which add nothing to the degree of 119889(119909)This is why maximality of the degree is no longer the onlycriterion in Z[119909] to detect if the given ℎ(119909) is gcd or not

3 Some Notations for Modular Reductions

The following notations adopted from [1] are to make ourarguments shorter and more uniform when we deal withnumerals polynomials and matrices As above letZ

119901be the

residue ring (finite Galois field Z119901= F119901= 0 119901 minus 1) and

let

120593119901 Z 997888rarr Z

119901 (12)

be the rings homomorphism mapping each 119911 isin Z to theremainder after division of 119911 by 119901 That is 120593

119901(119911) equiv 119911(mod

119901) and 120593119901(119911) isin Z

119901

Weuse the same symbol120593119901to denote the homomorphism

120593119901 Z [119909] 997888rarr Z

119901 [119909] (13)

where Z119901[119909] is the ring of polynomials over Z

119901and 120593

119901is

mapping each of the coefficients 119886119894of 119891(119909) isin Z[119909] to the

remainder after division of 119886119894by 119901

Similarly we define the homomorphism of matrix rings

120593119901 119872119898119899 (Z) 997888rarr 119872

119898119899(Z119901) (14)

which maps each of the elements 119886119894119895of a matrix119860 isin 119872

119898119899(Z)

to the remainder after division of 119886119894119895by 119901

Using the same symbol 120593119901for numeric polynomial

and matrix homomorphisms causes no misunderstandingbelow and it is more comfortable for some reasons Thesehomomorphisms are called ldquomodular reductionsrdquo or justldquoreductionsrdquo We can also specify these homomorphisms asldquonumeric modular reductionrdquo ldquopolynomial modular reduc-tionrdquo or ldquomatrix modular reductionrdquo where needed [1]

For 119886 isin Z denote 120593119901(119886) = 119886

119901 For 119891(119909) isin Z[119909] denote

120593119901(119891(119909)) = 119891

119901(119909) isin Z

119901[119909] So if

119891 (119909) = 1198860119909119899+ sdot sdot sdot + 119886

119899 (15)

then119891119901(119909) = 120593

119901(119891 (119909)) = 120593

119901(1198860) 119909119899+ sdot sdot sdot + 120593

119901(119886119899)

= 1198860119901119909119899+ sdot sdot sdot + 119886

119899119901isin Z119901 [119909]

(16)

And for a matrix 119860 isin 119872119898119899

(Z) denote 120593119901(119860) = 119860

119901isin

119872119898119899

(Z119901) If 119860 = 119886

119894119895119898times119899

then 119860119901

= 120593119901(119886119894119895)119898times119899

=

119886119894119895119901

119898times119899

4 Problems at Lifting the Modular gcd to Z[119909]

Now we turn to the second task mentioned earlier effectivecalculation of the actual gcd(119891(119909) 119892(119909)) for the given nonzeropolynomials 119891(119909) 119892(119909) isin Z[119909]

The ring Z119901[119909] is an Euclidean domain unlike the ring

Z[119909] So we can use the Euclidean algorithm to calcu-late gcd for any nonzero polynomials in Z

119901[119909] including

the modular images 119891119901(119909) and 119892

119901(119909) Since the notation

gcd(119891119901(119909) 119892119901(119909)) is going to be used repeatedly for brief-

ness denote by 119890119901(119909) gcd calculated by Euclidean algorithm

for 119891119901(119909) 119892119901(119909) Let us stress that gcd(119891

119901(119909) 119892119901(119909)) is not

determined uniquely since for any nonzero 119905 isin Z119901the

product 119905 sdot gcd(119891119901(119909) 119892119901(119909)) also is gcd for 119891

119901(119909) 119892119901(119909) We

are denoting just one of these gcdrsquos (namely that computedby the Euclidean algorithm) by 119890

119901(119909) This 119890

119901(119909) is unique

since at each step of the Euclidean algorithmwe have a uniqueaction to take (to see this just consider the steps of ldquolongdivisionrdquo used to divide 119891

119901(119909) by 119892

119901(119909) on field Z

119901)

The main idea of the algorithm is to calculate 119890119901(119909) asymp

gcd(119891119901(119909) 119892119901(119909)) for some suitable 119901 and to reconstruct

119889(119909) = gcd(119891(119909) 119892(119909)) by it We separate the process to fourmain problems that may occur and show how to overcomeeach one to arrive to a correctly working algorithm


41 Problem 1 Avoiding the Eliminating Coefficients Afterreduction 120593

119901 some of the coefficients of 119891(119909) and 119892(119909) may

change or even be eliminated So their images 119891119901(119909) =

120593119901(119891(119909)) and 119892

119901(119909) = 120593

119901(119892(119909)) may keep very little

information to reconstruct 119889(119909) based on 119890119901(119909)

Example 5 If 119891(119909) = 71199092+ 22 and 119892(119909) = 49119909

3+ 154119909

then for 119901 = 7 we get 119891119901(119909) = 1 and 119892

119901(119909) = 0 So

these values contain no reliable information to reconstructgcd(119891(119909) 119892(119909))

The first simple idea to avoid such elimination is to take119901 larger than the absolute value of all coefficients of 119891(119909) and119892(119909) This however is not enough since a divisor ℎ(119909) of apolynomial 119891(119909) may have coefficients larger than those of119891(119909) Moreover using the cyclotomic polynomials for largeenough 119899

120601119899(119909) = prod

119896=1119899(119896119899)=1

(119909 minus 1198902119894120587119896119899

) (17)

one can get divisors of 119891(119909) = 119909119899minus 1 which have a coefficient

larger than any pregiven number [1 4 8] Since we do notknow the divisors of 119891(119909) and 119892(119909) we cannot be sure ifthe abovementioned large 119901 will be large enough to preventelimination of coefficients of ℎ(119909) To overcome this one canuse the Landau-Mignotte bounds (in different sources thebounds on coefficients of the divisors are called differentlyassociating them with names of L Landau or M Mignotteor with both of them these authors have different roles indevelopment of the formulas which in turn are consequenceof a formula by A L Cauchy) as done in [4ndash6] For apolynomial 119891(119909) given by (15) denote its norm by 119891(119909) =

radicsum119899

119894=01198862119894

Theorem6 (L LandauMMignotte) Let119891(119909) = 1198860119909119899+sdot sdot sdot+

119886119899and ℎ(119909) = 119888

0119909119896+ sdot sdot sdot + 119888

119896be nonzero polynomials in Z[119909]

If ℎ(119909) is a divisor of 119891(119909) then

119899

sum119894=0

10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot10038161003816100381610038161003816100381610038161003816

1198880

1198860

10038161003816100381610038161003816100381610038161003816sdot1003817100381710038171003817119891 (119909)

1003817100381710038171003817 (18)

The proof is based on calculations on complex numbersand it can be found for example in [1 4] We are going to usethe Landau-Mignotte bounds in the following two shapes

Corollary 7 In notations of Theorem 6 there is the followingupper bound for the coefficients of ℎ(119909)

10038161003816100381610038161198881198941003816100381610038161003816 le 119873119891= 2119899minus1 1003817100381710038171003817119891 (119909)

1003817100381710038171003817 (19)

Proof To obtain this from (18) first notice that |11988801198860| le 1

Next if 119896 = deg ℎ(119909) = deg119891(119909) = 119899 then 119891(119909) = 119903 sdot

ℎ(119909) where 119903 is a nonzero integer Then |119888119894| le max|119888

119894| | 119894 =

0 119899 le max|119886119894| | 119894 = 0 119899 le 119891(119909)

Finally if 119896 = deg ℎ(119909) le 119899 minus 1 (119896 is unknown to us) thenwe can simply replace in (18) the value 2119896 by 2119899minus1

Remark 8 In literature they use the rather less accuratebound |119888

119894| le 2119899119891(119909) but the second paragraph of our proof

above allows replacing 2119899 by 2119899minus1 See also Remark 23

Corollary 9 In notations ofTheorem 6 if ℎ(119909) also is a divisorof the polynomial 119892(119909) = 119887

0119909119898+ sdot sdot sdot + 119887

119898 then there is the

following upper bound for the coefficients of ℎ(119909)

10038161003816100381610038161198881198941003816100381610038161003816 le 119873119891119892

= 2min119899119898

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(20)

Proof To obtain this from (18) just notice that if ℎ(119909)

is a common divisor for 119891(119909) and 119892(119909) then its leadingcoefficient 119888

0divides both 119886

0and 1198870

Formula (20) provides the hint to overcome Problem1 about eliminating coefficients mentioned at the start ofthis subsection Although the divisors ℎ(119909) of 119891(119909) and119892(119909) are yet unknown we can compute 119873

119891119892and take

119901 gt 119873119891119892 If we apply the reduction 120593

119901for this 119901 we can

be sure that none of the coefficients of ℎ(119909) has changedldquomuchrdquo under that homomorphism for 120593

119901does not alter

the nonnegative coefficients of ℎ(119909) and it just adds 119901 to allnegative coefficients of ℎ(119909) The same holds true for 119889(119909) =gcd(119891(119909) 119892(119909))

42 Problem 2 Negative Coefficients and Reconstruction ofthe Preimage The reduction 120593

119901is not a bijection and 119889

119901(119909)

has infinitely many preimages in Z[119909] But the relativelyuncomplicated relationship between coefficients of 119889(119909) and119889119901(119909) obtained in previous subsection may allow us to

reconstruct 119889(119909) if we know 119889119901(119909) The condition 119901 gt 119873

119891119892

puts a restriction on the preimage 119889(119909) the coefficients of119889(119909) either are equal to respective coefficients of119889

119901(119909) (if they

are nonnegative) or are the respective coefficients of 119889119901(119909)

minus 119901 (if they are negative) Reconstruction may causeproblems connected with negative coefficients

Example 10 If for some polynomials 119891(119909) 119892(119909) we have119873119891119892

= 15 we can take the prime say 119901 = 17 gt 119873119891119892 Assume

we have somehow calculated 11988917(119909) = 12119909

3+ 3119909 + 10 we

can be sure that 119889(119909) is not the preimage 291199093 minus 171199092 + 20119909+27 because 119889(119909) cannot have coefficients greater than 15 byabsolute value But we still cannot be sure if the preimage119889(119909)is 121199093 +3119909+10 or minus51199093 +3119909+10 or maybe minus51199093 minus14119909minus7

It is easy to overcome this by just taking a larger value

119901 gt 2 sdot 119873119891119892 (21)

If the coefficient 119888119894of 119889(119909) is nonnegative then 120593

119901(119888119894) =

119888119894lt 1199012 and if it is negative then 120593

119901(119888119894) = 119888119894+ 119901 gt 1199012

This provides us with the very simple method as shown inAlgorithm 1 to reconstruct 119889(119909) if we have already computed119889119901(119909) for sufficiently large prime 119901


Input For an unknown polynomial 119889(119909) isin Z[119909] we knowthe upper bound119873 of absolute values of its coefficientsand for arbitrarily large prime number 119901 we have themodular image 119889

119901(119909) = 119887

0119901119909119896+ sdot sdot sdot + 119887

119896119901isin Z119901[119909]

Reconstruct the polynomial 119889(119909)(01) Choose any prime 119901 gt 2 sdot 119873(02) Set 119896 = deg 119889

119901(119909)

(03) Set 119894 = 0(04) While 119894 le 119896

(05) if 119887119894119901

lt 1199012

(06) set 119887119894= 119887119894119901

(07) else(08) set 119887

119894= 119887119894119901

minus 119901(10) set 119894 = 119894 + 1(11) Output 119889(119909) = 119887

0119909119896+ sdot sdot sdot + 119887

119896

Algorithm 1 The polynomial reconstruction by modular image

43 Problem 3 Finding the Correct Fold of the Modular gcdof Right Degree Now additionally assume the polynomials119891(119909) 119892(119909) isin Z[119909] to be primitive Since cont(119891(119909)) andcont(119892(119909)) are defined up to the sign plusmn1 we can without lossof generality admit the leading coefficients of119891(119909) 119892(119909) to bepositive

Below in Problem 4 we will see that for some 119901 thepolynomial 119890

119901(119909) computed by the Euclidean algorithm in

Z119901[119909] may not be the image of 119889(119909) and moreover its

degree may be different from that of 119889(119909) This means thatby applying Algorithm 1 to 119890

119901(119909) we may not obtain 119889(119909)

Assume however we have 119901 which meets the condition 119901 gt

2 sdot 119873119891119892

and for which

deg 119889 (119909) = deg 119890119901(119909) (22)

By Corollary 3 a common divisor of 119891(119909) 119892(119909) is 119889(119909) =

gcd(119891(119909) 119892(119909)) if and only if it is primitive and if its degreeis the maximum of degrees of all common divisors Since120593119901does not change the degree of 119889(119909) we get by Lemma 4

(applied for the field119870 = Z119901) that119889

119901(119909) is gcd of119891

119901(119909) 119892119901(119909)

in Z119901[119909] This correspondence surely is not one-to-one

because in Z[119909] gcd is calculated up to the unit element ofZ[119909] which is plusmn1 whereas in Z

119901[119909] gcd is calculated up to

the unit element ofZ119901[119909] which can be any nonzero number

119905 isin Zlowast119901= 1 119901minus1 So the polynomial 119890

119901(119909) calculated by

the Euclidean algorithmmay not be the image 119889119901(119909) of 119889(119909)

Example 11 For 119891(119909) = 1199092+ 4119909 + 3 and 119892(119909) = 119909

2+ 2119909 + 1

whichever prime 119901 gt 4 we take we will get by the Euclideanalgorithm

119890119901(119909) = gcd (119891

119901(119909) 119892

119901(119909)) = 2119909 + 2 isin Z

119901 [119909] (23)

But in Z[119909] we have 119889(119909) = 119909 + 1 So regardless how large 119901we choose we will never get 120593

119901(119909 + 1) = 2119909 + 2

In other words we are aware that the image 119889119901(119909) is one

of the folds 119905 sdot 119890119901(119909) of 119890

119901(119909) for some 119905 isin 1 119901 minus 1 but

we are not aware which 119905 is that

The leading coefficient 1198880= lc(119889(119909)) of 119889(119909) can also be

assumed to be positive Denote by 119908 the positive gcd(1198860 1198870)

Since both 1198880and 119908 are not altered by 120593

119901 their fraction 119908119888

0

also is not altered Take such 119905 that

lc (119905 sdot 119890119901(119909)) = 119908 (24)

Even if 119905 sdot 119890119901(119909) is not the image 119889

119901(119909) it is the image of 119897 sdot

119889(119909) where 119897 divides1199081198880 If we calculate the preimage 119896(119909) isin

Z[119909] of 119905 sdot 119890119901(119909) by Algorithm 1 we will get a polynomial

which is either 119889(119909) or some fold of 119889(119909) Since 119891(119909) 119892(119909)are primitive it remains to go to the primitive part 119889(119909) =

pp(119896(119909))The general case when 119891(119909) 119892(119909) may not be primi-

tive can easily be reduced to the following for arbitrary119891(119909) 119892(119909) take their decompositions by formula (5) and set

119903 = gcd (cont (119891 (119909)) cont (119892 (119909))) isin Z (25)

Then assign 119891(119909) = pp(119891(119909)) 119892(119909) = pp(119892(119909)) and do thesteps above for these new polynomials After 119889(119909) = pp(119896(119909))is computed we get the final answer as 119903 sdot 119889(119909) = 119903 sdot pp(119896(119909))

Notice that for Algorithm 1 we need 119901 to be greater thanany coefficient |119888

119894| of the polynomial we reconstruct The

bound 119901 gt 2 sdot 119873119891119892

assures that 119901 meets this condition for119889(119909) We however reconstruct not 119889(119909) but 119897 sdot 119889(119909) whichmay have larger coefficients One could overcome this pointby taking 119901 gt 119908 sdot 2 sdot 119873

119891119892 but this is not necessary because

as we see later while the Big prime modular gcd algorithmworks the value of 119901 will grow and this issue will be covered

44 Problem 4 Finding the Right Degree for the Modular gcdAs we saw one can reconstruct 119889(119909) if we find 119901 gt 2 sdot 119873

119891119892

such that condition (22) holds Consider an example to seethat (22) may actually not hold for some 119901 even if 120593

119901is not

altering the coefficients of 119891(119909) and 119892(119909)

Example 12 For 119891(119909) = 1199092+ 1 and 119892(119909) = 119909 + 1 we have

119889(119909) = gcd(119891(119909) 119892(119909)) = 1 Taking 119901 = 2 we get 1198912(119909) =

1199092+ 1 and 119892

2(119909) = 119909 + 1 In Z

2[119909] we have 119891

2(119909) = 119909

2+ 1 =

1199092+12= (119909+1)(119909+1) thus 119890

2(119909) = gcd(119891

2(119909) 1198922(119909)) = 119909+1

We get that 1 = deg(119909 + 1) gt deg(119909) = 0 In particularwhatever 119905 we take 119905 sdot (119909 + 1) is not the image of 119889(119909) = 1

under 1205932

The idea to overcome this problem is to show that thenumber of primes 119901 for which (22) falsifies is ldquosmallrdquo Soif the selected 119901 is not suitable we take another 119901 and do thecalculation again by the new prime And we will not have torepeat these steps for many times (we will turn to this pointin Section 6)

The proof of the following theorem and the definition ofthe resultant res(119891(119909) 119892(119909)) (ie of the determinant of theSylvestermatrix 119878

119891119892of polynomials119891(119909) 119892(119909)) can be found

for example in [1 4 7 10]The resultant is a comfortable toolto detect if the given polynomials are coprime

Theorem 13 Let 119877 be an integrity domain The polynomials119891(119909) 119892(119909) isin 119877[119909] are coprime if and only if res(119891(119909) 119892(119909)) =

0


Input non-zero polynomials 119891(119909) 119892(119909) isin Z[119909]Calculate their greatest common divisor gcd(119891(119909) 119892(119909)) isin Z[119909](01) Calculate cont(119891(119909)) cont(119892(119909)) in the Euclidean domain Z choose their signs so that sgn cont(119891(119909)) = sgn lc(119891(119909))

and sgn cont(119892(119909)) = sgn lc(119892(119909))(02) Set 119891(119909) = pp(119891(119909)) and 119892(119909) = pp(119892(119909))(03) Calculate 119903 in the Euclidean domain Z by (25)(04) Set 119886

0= lc(119891(119909)) and 119887

0= lc(119892(119909)) (they are positive by our selection of signs for cont(119891(119909)) and cont(119892(119909)))

(05) Calculate the positive 119908 = gcd(1198860 1198870) in the Euclidean domain Z

(06) Set119863 = mindeg119891(119909) deg119892(119909) + 1(07) Compute the Landau-Mignotte bound119873

119891119892by (20)

(08) Choose a new prime number 119901 gt 2 sdot 119873119891119892

(09) Apply the reduction 120593119901to calculate the modular images 119891

119901(119909) 119892

119901(119909) isin Z

119901[119909]

(10) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892

119901(119909)) in the Euclidean domain Z

119901[119909]

(11) If119863 le deg 119890119901(119909)

(12) go to step (08)(13) else(14) choose a 119905 such that the lc(119905 sdot 119890

119901(119909)) = 119908

(15) call Algorithm 1 to calculate the preimage 119896(119909) of 119905 sdot 119890119901(119909)

(16) calculate cont(119896(119909)) in the Euclidean domain Z(17) set 119889(119909) = pp(119896(119909)) = 119896(119909)cont(119896(119909))(18) if 119889(119909) | 119891(119909) and 119889(119909) | 119892(119909)(19) go to step (23)(20) else(21) set119863 = deg 119890

119901(119909)

(22) go to step (08)(23) Output the result gcd(119891(119909) 119892(119909)) = 119903 sdot 119889(119909)

Algorithm 2 Big prime modular gcd algorithm

Input non-zero polynomials 119891(119909) 119892(119909) isin Z[119909]Detect if 119891(119909) and 119892(119909) are coprime(01) Calculate cont(119891(119909)) cont(119892(119909)) in the Euclidean domain Z(02) Calculate 119903 in the Euclidean domain Z by (25)(03) If 119903 asymp 1

(04) output the result 119891(119909) and 119892(119909) are not coprime and stop(05) Set 119886

0= lc(119891(119909)) and 119887

0= lc(119892(119909))

(06) Calculate 119908 = gcd(1198860 1198870) in the Euclidean domain Z

(07) Set 119891(119909) = pp(119891(119909)) and 119892(119909) = pp(119892(119909))(08) Compute the bound 119860

119891119892for polynomials 119891(119909) 119892(119909) by (32)

(09) Find the maximal 119896 for which 119901119896 le 119860

119891119892

(10) Set 119894 = 1(11) While 119894 = 119896 + 1

(12) choose a new prime 119901 ∤ 119908(13) apply the reduction 120593

119901to calculate the modular images 119891

119901(119909) 119892

119901(119909) isin Z

119901[119909]

(14) calculate 119890119901= gcd(119891

119901(119909) 119892


119901[119909]

(15) if deg 119890119902119894= 0

(16) output the result 119891(119909) and 119892(119909) are coprime and stop(17) set 119894 = 119894 + 1(18) If 119894 lt 119896 + 1

(19) go to step (12)(20) else(21) output the result 119891(119909) and 119892(119909) are not coprime

Algorithm 3 Coprime polynomials detection modular algorithm


Input non-zero polynomials 119891(119909) 119892(119909) isin Z[119909]Calculate their greatest common divisor gcd(119891(119909) 119892(119909)) isin Z[119909](01) Calculate cont(119891(119909)) cont(119892(x)) in the Euclidean domain Z choose their signs

so that sgn cont(119891(119909)) = sgn lc(119891(119909)) and sgn cont(119892(119909)) = sgn lc(119892(119909))(02) Set 119891(119909) = pp(119891(119909)) and 119892(119909) = pp(119892(119909))(03) Calculate 119903 in the Euclidean domain Z by (25)(04) Set 119886

0= lc(119891(119909)) and 119887

0= lc(119892(119909)) (they are positive by our selection of signs

for cont(119891(119909)) and cont(119892(119909)))(05) Calculate the positive 119908 = gcd(119886

0 1198870) in the Euclidean domain Z

(06) Set119863 = mindeg119891(119909) deg119892(119909) + 1(07) Choose a prime number 119902 ∤ 119908(08) Apply the reduction 120593


119902(119909) 119892

119902(119909) isin Z

119902[119909]

(09) Calculate 119890119902(119909) = gcd(119891

119902(119909) 119892


119902[119909]

(10) Set 119904(119902 119891 119892) = deg 119890119902(119909)

(11) Calculate119872119902119891119892

by (40) using the value of 119904(119902 119891 119892)(12) Choose a new prime number 119901 gt 2 sdot 119872

119902119891119892


119901(119909) 119892

119901(119909) isin Z

119901[119909]

(14) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(15) If119863 le deg 119890119901(119909)


119901(119909)) = 119908



119901(119909)


Algorithm 4 Big prime modular gcd algorithm with a preliminary estimate on divisor degree

The following fact in a little different shape can be foundin [4] or [5]

Corollary 14 If the prime 119901 does not divide at least oneof the leading coefficients 119886

0 1198870of polynomials respectively

119891(119909) 119892(119909) isin Z[119909] then deg 119889(119909) le deg 119890119901(119909) If 119901 also

does not divide 119877 = res(119891(119909)119889(119909) 119892(119909)119889(119909)) where 119889(119909) =gcd(119891(119909) 119892(119909)) then

deg 119889 (119909) = deg 119889119901(119909) = deg gcd (119891

119901(119909) 119892

119901(119909))

= deg 119890119901(119909)

(26)

Proof Since 1198880

= lc(119889(119909)) divides 119908 = gcd(1198860 1198870) then

120593119901(1198880) = 0 by the choice of 119901 Thus deg 119889(119909) = deg 119889

119901(119909) le

deg gcd(119891119901(119909) 119892119901(119909))

Since 119889119901(119909) = 0 we can consider the fractions

119891119901(119909)119889119901(119909) and 119892

119901(119909)119889119901(119909) in Z

119901[119909] From unique factor-

izations of 119891119901(119909) and 119892

119901(119909) in UFD Z

119901[119909] it is very easy to

deduce that

119890119901 (119909) asymp gcd (119891

119901 (119909) 119892119901 (119909))

asymp 119889119901(119909) sdot gcd(

119891119901(119909)

119889119901(119909)

119892119901(119909)

119889119901(119909)

)

(27)

In particular deg 119889(119909) = deg 119889119901(119909) le deg 119890

119901(119909) And the

inequality deg119889119901(119909) = deg 119890

119901(119909)may occur only if

deg gcd(119891119901(119909)

119889119901 (119909)

119892119901(119909)

119889119901 (119909)

) gt 0 (28)

that is when 119891119901(119909)119889119901(119909) and 119892

119901(119909)119889119901(119909) are not

coprime in Z119901[119909] or by Theorem 13 when res(119891

119901(119909)

119889119901(119909) 119892119901(119909)119889119901(119909)) = 0 The latter is the determinant of

Sylvester matrix 119878119891119901119889119901119892119901119889119901

Consider the matrix rings ho-momorphism (matrix modular reduction)

120593119901 119872119898+119899

(Z) 997888rarr 119872119898+119899

(Z119901) (29)

where 119899 = deg119891(119909) 119898 = deg119892(119909) (as mentioned earlier weuse the same symbol 120593

119901for numeric polynomial and matrix

reductions) Since 120593119901(119878119891119889119892119889

) = 119878119891119901119889119901119892119901119889119901

and since thedeterminant of a matrix is a sum of products of its elementswe get

119877119901= 120593119901 (119877) = 120593

119901(res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909)))

= res(119891119901(119909)

119889119901(119909)

119892119901(119909)

119889119901(119909)

)

(30)


Input non-zero polynomials 119891(119909) 119892(119909) isin Z[119909]Calculate their greatest common divisor gcd(119891(119909) 119892(119909)) isin Z[119909](01) Calculate cont(119891(119909)) cont(119892(119909)) in the Euclidean domain Z choose their signs

so that sgn cont(119891(119909)) = sgn lc(119891(119909)) and sgn cont(119892(119909)) = sgn lc(119892(119909))(02) Set 119891(119909) = pp(119891(119909)) and 119892(119909) = pp(119892(119909))(03) Compute the bound 119860



119891119892

(05) Calculate 119903 in the Euclidean domain Z by (25)(06) Set 119886

0= lc(119891(119909)) and 119887




(08) Set 119904(119891 119892) = mindeg119891(119909) deg119892(119909)(09) Set 119894 = 1(10) While 119894 = 119896 + 1

(11) choose a new prime 119902119894∤ 119908

(12) apply the reduction 120593119902119894to calculate the modular images 119891

119902119894(119909) 119892

119902119894(119909) isin Z

119902119894[119909]

(13) calculate 119890119902119894= gcd(119891

119902119894(119909) 119892


119901[119909]

(14) if deg 119890119902119894le 119904(119891 119892)

(15) set 119904(119891 119892) = deg 119890119902119894

(16) if deg 119890119902119894= 0

(17) set 119889(119909) = 1(18) go to step (32)(19) set 119894 = 119894 + 1(20) Calculate119872

119891119892by (43) using the value of 119904(119891 119892)



119901(119909) 119892

119901(119909) isin Z

119901[119909]

(23) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(24) If deg 119890119901= 119904(119891 119892)

(25) choose a 119905 such that the lc(119905 sdot 119890119901(119909)) = 119908


(27) calculate cont(119896(119909)) in the Euclidean domain Z(28) set 119889(119909) = pp(119896(119909)) = 119896(119909)cont(119896(119909))(29) go to step (32)(30) else(31) go to step (21)(32) Output the result gcd(119891(119909) 119892(119909)) = 119903 sdot 119889(119909)

Algorithm 5 Big prime modular gcd algorithm with preliminary estimates on divisor degrees by multiple primes

So 119877119901can be zero if and only if 119877 is divisible by 119901 The

polynomials 119891(119909)119889(119909) and 119892(119909)119889(119909) are coprime in Z[119909]

and their resultant is not zero by Theorem 13 And 119877 cannotbe a positive integer divisible by 119901 since that contradicts thecondition of this corollary

Corollary 14 shows that if for some 119901 equality (22) doesnot hold for polynomials 119891(119909) 119892(119909) isin Z[119909] then 119901 eitherdivides 119886

0and 1198870or divides the resultant 119877 We do not know

119877 since we do not yet know 119889(119909) to calculate the resultant119877 = res(119891(119909)119889(119909) 119892(119909)119889(119909)) But since the number of suchprimes 119901 is just finite we can arrive to the right 119901 after tryingthe process for a few primes We will turn to this again inSection 6

5 The Big Prime Modular gcd Algorithm

Four steps of the previous section provide us with the follow-ing procedureWe keep all the notations from Section 4 Takethe primitive polynomials 119891(119909) 119892(119909) isin Z[119909] Without loss of

generality we may assume 1198860 1198870gt 0 Take any 119901 gt 2 sdot 119873

119891119892

Then 119901 ∤ 119908 = gcd(1198860 1198870) since 119886

0 1198870le 119873119891119892 Calculate

119890119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) in Z

119901[119909] by Euclidean algorithm

Then choose 119905 so that (24) holds Construct 119896(119909) applyingAlgorithm 1 to 119905 sdot 119890

119901(119909) If the primitive part 119889(119909) = pp(119896(119909))

divides both 119891(119909) and 119892(119909) then gcd for these primitivepolynomials is found 119889(119909) = gcd(119891(119909) 119892(119909)) That followsfrom consideration about divisor degrees above if 119891(119909) 119892(119909)had a common divisor ℎ(119909) of degree greater than deg 119889(119909)then since the degree of ℎ(119909) is not altered by 120593

119901 we would

get deg ℎ119901(119909) gt deg 119889

119901(119909) = deg 119889(119909) = deg 119890

119901(119909) which

contradicts the maximality of deg 119889119901(119909) by Lemma 4

This means that if for 119901 gt 2 sdot 119873119891119892

we get 119889(119909) ∤ 119891(119909) or119889(119909) ∤ 119892(119909) we have the case when 119901 divides the resultant119877 Then we just ignore the calculated polynomial chooseanother 119901 gt 2 sdot119873

119891119892 and redo the steps for it Repeating these

steps for finitely many times we will eventually arrive to thecorrect 119889(119909) for the primitive polynomials 119891(119909) 119892(119909)

The case of arbitrary nonzero polynomials can easilybe reduced to this By arguments mentioned earlier we


should calculate 119889(119909) for primitive polynomials pp(119891(119909)) andpp(119892(119909)) and then output the final answer as 119903 sdot 119889(119909) where 119903is defined by (25) The process we described is the traditionalform of the Big prime modular gcd algorithm

Remark 15 Since our approach in Section 4 evolved themaximality of degrees of the common divisors we canshorten some of the steps of our algorithm Let us store ina variable say 119863 the minimal value for which we alreadyknow it is not deg gcd(119891(119909) 119892(119909)) As an initial 119863 we maytake for example119863 = mindeg119891(119909) deg119892(119909)+1 Each timewe calculate 119890

119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) check if deg 119890

119901(119909) is

equal to or larger than the current119863 If yes we already knowthat we have an ldquoinappropriaterdquo 119901 Then we no longer needto use Algorithm 1 to reconstruct 119896(119909) and to get 119889(119909) =

pp(119896(119909)) We just skip these steps and proceed to the next119901 Reconstruct 119889(119909) and check if 119889(119909) | 119891(119909) and 119889(119909) |

119892(119909) only when deg 119890119901(119909) lt 119863 Then if 119889(119909) does not

divide 119891(119909) or 119892(119909) we have discovered a new bound 119863 fordeg(gcd(119891(119909) 119892(119909))) So set 119863 = deg 119890

119901(119909) and proceed to

the next 119901 If in next step we get deg 119890119901(119909) ge 119863 we will

again be aware that the steps of reconstruction of 119889(119909) needbe skipped

We constructed Algorithm 2Turning back to Remark 15 notice that for some prime

numbers 119901 we skip steps (14)ndash(18) of Algorithm 2 anddirectly jump to step (08) In fact Remark 15 has mainlytheoretical purpose to display how usage of UFD propertiesand comparison of divisor degrees may reduce some of thesteps of the Big prime modular gcd algorithm In practicalexamples the set of primes we use contains few primesdividing 119877 = res(119891(119909)119889(119909) 119892(119909)119889(119909)) so we may notfrequently get examples where steps (14)ndash(18) are skipped

Example 16 Let us apply Algorithm 2 to polynomials (1)mentioned in Knuthrsquos example above Since 119891(119909) = radic113

and 119892(119909) = radic570

119873119891119892

= 2min86

sdot gcd (1 3) sdotminradic113

1radic570

3

lt 512

(31)

And we can take the prime 119901 = 1031 gt 2 sdot 119873119891119892 It is not

hard to compute that gcd(1198911031

(119909) 1198921031

(119909)) asymp 1 So119891(119909) and119892(119909) are coprime It is worth to compare 119901 = 1031withmuchsmaller values119901 = 67 and119901 = 37 obtained below for the samepolynomials in (1) in Example 27 usingmodifiedAlgorithm4

In [1]we also applyAlgorithm2 to other polynomialswithcases when the polynomials are not coprime

6 Estimating the Prime Divisorsof the Resultant

Although at the start of the Big primemodular gcd algorithmwe cannot compute the resultant 119877 = res(119891(119909)119889(119909) 119892(119909)119889(119909)) for the given119891(119909) 119892(119909) isin Z[119909] (we do not know 119889(119909))

we can nevertheless estimate the value of 119877 and the numberof its prime divisors Denote

119860119891119892

= radic(119899 + 1)119898(119898 + 1)

119899sdot 119873119898

119891119873119899

119892

= 22119899119898minus119899minus119898radic(119899 + 1)

119898(119898 + 1)

119899

sdot1003817100381710038171003817119891 (119909)

1003817100381710038171003817119898 1003817100381710038171003817119892 (119909)

1003817100381710038171003817119899

(32)

Lemma 17 For any polynomials 119891(119909) 119892(119909) isin Z[119909] and forany of their common divisors 119889(119909) the following holds

10038161003816100381610038161003816100381610038161003816res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909))10038161003816100381610038161003816100381610038161003816=10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le 119860119891119892 (33)

Proof By Corollary 7 the coefficients of fractions 119891(119909)119889(119909)and119892(119909)119889(119909) are bounded respectively by119873

119891= 2119899minus1

119891(119909)

and 119873119892= 2119898minus1

119892(119909) where 119899 = deg119891(119909) 119898 = deg119892(119909)Since the numbers of summands in these fractions are atmost119899 + 1 and119898 + 1 respectively we get

10038171003817100381710038171003817100381710038171003817

119891 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119899 + 1)1198732

119891

10038171003817100381710038171003817100381710038171003817

119892 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119898 + 1)1198732

119892

(34)

Applying Hadamardrsquos maximal determinant bound [4] to theSylvester matrix 119878

119891119889119892119889 we get that

|119877| =10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le (radic(119899 + 1)119873

119891)119898

sdot (radic(119898 + 1)119873119892)119899

(35)

The bound of (32) is very rough To see this apply it topolynomials (1) of Knuthrsquos example

Example 18 For polynomials (1) we have 119891(119909) = radic113 and119892(119909) = radic570 So we can estimate 119873

119891lt 1408 119873

119892lt 768

and119873119891119892

lt 512 Thus

|119877| le radic(8 + 1)6(6 + 1)

8sdot 1408

6sdot 5708= 120596

= 16505374299582118582810249858265119890 + 48

(36)

which is a too large number to comfortably operate with

Remark 19 If in Algorithm 2 we use a prime

119901 gt 2 sdot 119860119891119892 (37)

then we will get that 119901 ∤ 119877 = res(119891(119909)119889(119909) 119892(119909)119889(119909))whatever the greatest common divisor 119889(119909) is And clearly119901 ∤ 119908 holds for 119908 = gcd(119886

0 1198870) So in this case Algorithm 2

will output the correct pp(119896(119909)) using just one 119901 and we willnot have to take another 119901 ∤ 119908 after step (18) HoweverExample 18 shows why it is not reasonable to choose 119901 by rule(37) to have in Algorithm 2 one cycle only it is easier to govia a few cycles for smaller 119901 rather than to operate with ahuge 119901 which is two times larger than the bound 120596 obtainedin Example 18


Nevertheless the bound119860119891119892

may be useful if we remem-ber that the process in Algorithm 2 concerned not the valueof res(119891(119909)119889(119909) 119892(119909)119889(119909)) but the number of its distinctprime divisors Let us denote by 119901

119896 the product of the first

119896 primes 119901119896 = 119901

1sdot 1199012sdot sdot sdot 119901119896(where 119901

1= 2 119901

2= 3 etc)

They sometimes call 119901119896 the ldquo119896th primorialrdquoThe following is

essential

Lemma 20 The number of pairwise distinct prime divisors ofa positive integer 119899 is less than or equal tomax119896 | 119901

119896 le 119899

From Lemmas 17 and 20 we easily get the following

Corollary 21 For any polynomials 119891(119909) 119892(119909) isin Z[119909] andfor any of their common divisors 119889(119909) the number of pairwisedistinct prime divisors of res(119891(119909)119889(119909) 119892(119909)119889(119909)) is at most119896 where 119896 is the largest number for which 119901

119896 le 119860

119891119892

Primorial (as a function on 119896) grows very rapidly Say for119896 = 10 it is more than six billions 119901

10 = 6 469 693 230

This observation allows using the bound119860119891119892

in the followingway although the value of119860

119891119892as a function on 119899 = deg119891(119909)

and 119898 = deg119892(119909) and on the coefficients of 119891(119909) and119892(119909) grows rapidly the number of its distinct prime divisorsmay not be ldquovery largerdquo thanks to the fact that 119901

119896 also

grows rapidly Consider this on polynomials and values fromExample 18

Example 22 It is easy to compute that

11990130 = 31610054640417607788145206291544119890 + 46

lt 120596

11990131 = 4014476939333036189094441199026119890 + 48

gt 120596

(38)

where 120596 is the large number from Example 18 Thismeans that the number of prime divisors of 119877 =

res(119891(119909)119889(119909) 119892(119909)119889(119909)) whatever the divisor 119889(119909) is isnot greater than 30 And whichever 30 + 1 = 31 distinctprimes we take at least one of them will not be a divisorof 119877 That is Algorithm 2 for the polynomials of Knuthrsquosexample will output the correct answer in not more than 31

cycles We cannot find 31 primes 119901 ∤ 119908 so that Algorithm 2arrives to a wrong 119889(119909) = pp(119896(119909)) on step (18) for all ofthem

Remark 23 Let us stress that estimates on the number ofprime divisors of the resultant and the analog of Algorithm 3can be found elsewhere for example in [4] So the only newswe have is that here we use a slightly better value for 119873

119891

and119873119892to get 2119899+119898 times smaller bound for 119860

119891119892 Namely in

Corollary 7 we estimate |119888119894| not by 2119899119891(119909) but by 2119899minus1119891(119909)

(see (19) and Remark 8) This makes the bound 119860119891119892

informula (32) 2119899+119898 times lower since 119873

119891and 119873

119892appear 119898

and 119899 times respectively

7 An Algorithm to CheckCoprime Polynomials

The first application of the bounds found in previous sec-tion is an algorithm checking if the given polynomials119891(119909) 119892(119909) isin Z[119909] are coprime Present the polynomials as119891(119909) = cont(119891(119909)) sdot pp(119891(119909)) and 119892(119909) = cont(119892(119909)) sdotpp(119892(119909)) If 119903 = gcd(cont(119891(119909)) cont(119892(119909))) asymp 1 then119891(119909) 119892(119909) are not coprime and we do not have to check theprimitive parts at all

If 119903 asymp 1 then switch to the polynomials 119891(119909) = pp(119891(119909))and 119892(119909) = pp(119892(119909)) By Corollary 21 the number of distinctprime divisors of res(119891(119909)119889(119909) 119892(119909)119889(119909)) is less than orequal to 119896 where 119896 is the largest number for which 119901

119896 le

119860119891119892Consider any 119896 + 1 primes 119901

1 119901

119896+1 each not dividing

119908 = gcd(1198860 1198870) where 119886

0= lc(119891(119909)) and 119887

0= lc(119892(119909))

If gcd(119891119901119894

(119909) 119892119901119894

(119909)) = 1 for at least one 119901119894 then 119891(119909) and

119892(119909) are coprime because 0 = deg gcd(119891119901119894

(119909) 119892119901119894

(119909)) ge

deg gcd(119891(119909) 119892(119909)) and 119903 asymp 1And if gcd(119891

119901119894

(119909) 119892119901119894

(119909)) = 1 for all 119894 = 1 119896 + 1 then119891119901119894

(119909) and 119892119901119894

(119909) are not coprime for at least one 119901119894 which is

not dividing res(119891(119909)119889(119909) 119892(119909)119889(119909))Thismeans that119891(119909)and 119892(119909) are not coprime We got Algorithm 3

Two important advantages of this algorithm are that herewe use much smaller primes 119901 (we just require 119901 ∤ 119908 not119901 gt 2 sdot 119873

119891119892) and in Algorithm 3 unlike in Algorithm 2

we never need to find 119905 to compute the preimage 119896(119909) of 119905 sdotgcd(119891

119901(119909) 119892119901(119909)) and the primitive part pp(119896(119909))

Remark 24 As is mentioned by Knuth in [2] in a probabilis-tic sense the polynomials aremuchmore likely to be coprimethan the integer numbers So it is reasonable to first test byAlgorithm 3 if the given polynomials 119891(119909) 119892(119909) are coprimeand only after that apply Algorithm 2 to find their gcd incase if they are not coprime See also Algorithm 5 where wecombine both of these approaches with a better bound forprime 119901

Example 25 Apply Algorithm 3 to polynomials (1) fromKnuthrsquos example As we saw in Example 22 119896 = 30 For 119901 = 2we get119891

2(119909) = 119909

8+1199096+1199094+1199093+1 1198922(119909) = 119909

6+1199094+119909+1 which

are not coprime since gcd(1198912(119909) 1198922(119909)) = 119909

2+ 119909 + 1 = 1

And for 119901 = 3 we get 1198913(119909) = 119909

8+ 1199096+ 21199092+ 2119909 + 1

1198923(119909) = 2119909

4+21199092 which are coprime So gcd(119891(119909) 119892(119909)) = 1

Example 26 If119891(119909) = 1199092+2119909+1 and119892(119909) = 119909+1Then119873

119891=

2radic6119873119892= radic2 and119860

119891119892lt 39 Since 2 sdot 3 sdot 5 sdot 7 = 210 gt 39 we

get that 119896 = 3 and gcd(119891(119909) 119892(119909)) = 1 if gcd(119891119901(119909) 119892119901(119909)) =

1 for any four primes (not dividing119908) It is easy to check thatgcd(119891

119901(119909) 119892119901(119909)) = 1 for 119901 = 2 3 5 7

8 Other Modifications of Algorithms

The bounds mentioned in Section 6 can be applied to obtainmodifications of Algorithm 2 Let us outline four ideas ofwhich only the last two will be written down as algorithms

For the nonzero polynomials 119891(119909) 119892(119909) isin Z[119909] let usagain start by computing 119903 = gcd(cont(119891(119909)) cont(119892(119909)))


and switching to the primitive parts 119891(119909) = pp(119891(119909)) and119892(119909) = pp(119892(119909)) assuming that their leading coefficients1198860and 119887

0are positive Calculate 119873

119891 119873119892by Corollary 7

119873119891119892

by Corollary 9 and 119860119891119892

by (32) Find the maximal119896 for which 119901

119896 le 119860

119891119892 Then take any 119896 + 1 primes

1199011 119901

119896+1 each greater than 2 sdot119873

119891119892 We do not know 119889(119909)

but we are aware that the number of prime divisors of 119877 =

res(119891(119909)119889(119909) 119892(119909)119889(119909)) is less than or equal to 119896 So at leastone of the primes 119901

1 119901

119896+1is not dividing 119877 To find it

compute the degrees of 119890119901119894

(119909) for all 119894 = 1 119896 + 1 Take any119901119894 for which deg 119890

119901119894

(119909) is the minimal (in case there is morethan one 119901

119894with this property take one of them preferably

the smallest of all)By our construction deg 119890

119901119894

(119909) = deg gcd(119891(119909) 119892(119909))holds So we can proceed to the next steps choose 119905 such thatlc(119905 sdot 119890

119901119894

(119909)) = 119908 = gcd(1198860 1198870) then find by Algorithm 1 the

preimage 119896(119909) of 119905 sdot 119890119901119894

(119909) then proceed to its primitive part119889(119909) = pp(119896(119909)) and then output the final answer as 119903 sdot 119889(119909)

The advantage of this approach is thatwe donot have to govia steps (14)ndash(18) of Algorithm 2 for more than one prime119901 Also we do not have to take care of the variable 119863 Butthe disadvantage is that we have to compute 119890

119901119894

(119909) for largeprimes for 119896 + 1 times (whereas in Algorithm 2 the correctanswer could be discovered after consideration of fewerprimes) Clearly the disadvantage is a serious obstacle sincerepetitions for 119896 + 1 large primes consume more labour thansteps (14)ndash(18) of Algorithm 2 So this is just a theoreticalidea not an approach for an effective algorithm

The disadvantage can be reduced in the following way inprevious arguments after we find 119901

119896 and 119896 select the prime

numbers 1199011 119901

119896+1each satisfying the condition 119901

119894∤ 119908

This is a much weaker condition than the condition 119901119894gt

2 sdot 119873119891119892

used above so we will surely get smaller primesTake 119872 to be the minimum of all degrees deg 119890

119901119894

(119909) forall 119894 = 1 119896 + 1 Since none of the primes 119901

1 119901

119896+1

divides 119908 for any 119894 = 1 119896 + 1 we have deg 119890119901119894

(119909) ge

deg gcd(119891(119909) 119892(119909)) On the other hand since at least one ofthe primes 119901

1 119901

119896+1does not divide 119877 we know that for

that 119901119894 the degree of 119890

119901119894

(119909) is equal to deg gcd(119891(119909) 119892(119909))Combining these we get that deg gcd(119891(119909) 119892(119909)) = 119872Since we know 119872 we can take a prime 119901 gt 2 sdot 119873

119891119892and

compute 119890119901(119909) and check its degree if deg 119890

119901(119909) = 119872 then

we have a wrong 119901 (we no longer need go to steps (14)ndash(18)of Algorithm 2 to discover that)Then choose a new value for119901 and repeat the step And if deg 119890

119901(119909) = 119872 then we have

the right 119901 We calculate 119905 the preimage 119896(119909) of 119905 sdot 119890119901119894

(119909) andthen 119889(119909) = pp(119896(119909)) and output the answer 119903 sdot 119889(119909) (seeAlgorithm 5 for a better version of this idea)

The third modification not depending on 119860119891119892 can be

constructed by estimating deg gcd(119891(119909) 119892(119909)) = deg 119889(119909) bymeans of an auxiliary prime number 119902 By Landau-MignotteTheorem 6 if ℎ(119909) = 119888

0119909119896+ sdot sdot sdot + 119888

119896is any divisor of the

polynomials119891(119909) = 1198860119909119899+sdot sdot sdot+119886

119899and 119892(119909) = 119887

0119909119898+sdot sdot sdot+119887

119898

then |119888119894| le 2119896|11988801198860|119891(119909) and |119888

119894| le 2119896|11988801198870|119892(119909) Since

|11988801198860| is bounded by |gcd(119886

0 1198870)1198860| and |119888

01198870| is bounded

by |gcd(1198860 1198870)1198870| we get the following analog of (20)

10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot gcd (119886

0 1198870) sdotmin

1003817100381710038171003817119891 (119909)1003817100381710038171003817

100381610038161003816100381611988601003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(39)

Now assume 119902 is a prime not dividing 119908 and denote119904(119902 119891 119892) = deg gcd(119891

119902(119909) 119892119902(119909)) ByCorollary 14 deg 119889(119909) le

119904(119902 119891 119892) We get for the coefficients of 119889(119909) the followingbound |119888

119894| le 119872

119902119891119892 where

119872119902119891119892

= 2119904(119902119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(40)

119872119902119891119892

is a better bound for the coefficients of 119889(119909) because2119904(119902119891119892) may be considerably less than 2

min119899119898We can improve Algorithm 2 if we preliminarily find

119904(119902 119891 119892) by calculating gcd(119891119902(119909) 119892119902(119909)) for an ldquoauxiliaryrdquo

prime 119902 ∤ 119908 and then choose the ldquomainrdquo prime 119901 by therule 119901 gt 2 sdot 119872

119902119891119892(instead of 119901 gt 2 sdot 119873

119891119892) Observe that

if 119901 gt 2 sdot 119872119902119891119892

then also 119902 ∤ 119908 = gcd(1198860 1198870) because

min119891(119909)|1198860| 119892(119909)|119887

0| ge 1 Additionally we can intro-

duce the variable 119863 to store the values deg gcd(119891119901(119909) 119892119901(119909))

that we know are greater than deg 119889(119909) We get Algorithm 4

Example 27 Let us applyAlgorithm4 to polynomials (1) fromKnuthrsquos example Since 119908 = 1 take 119902 = 2 We have alreadycomputed in Example 25 that 119890

2(119909) = gcd(119891

2(119909) 1198922(119909)) =

1199092+ 119909 + 1 Then 119904(2 119891 119892) = deg 119890

2(119909) = 2 and

1198722119891119892

= 22sdot 1 sdotmin

radic113

1radic570

3 lt 3184 (41)

Take 119901 = 67 gt 2 sdot 1198722119891119892

It is easy to calculate thatgcd(119891

67(119909) 11989267(119909)) asymp 1 Compare this with Example 16

where we had to use much larger prime 119901 = 1031 Moreoverif we take as an auxiliary 119902 say 119902 = 3 then 119904(3 119891 119892) =

deg 1198903(119909) = 0 and 119872

3119891119892le 1592 So we can take an even

smaller prime 119901 = 37 gt 2 sdot 1198723119891119892

The ideas of Algorithms 3 and 4 can be combined to workwith more than one auxiliary prime 119902 Like we mentioned inRemark 24 Knuth in [2] recommends checking first if thepolynomials 119891(119909) 119892(119909) isin Z[119909] are coprime and proceedingto their gcd calculation only after we get that they arenot coprime (this is motivated by probabilistic arguments)Compute 119860

119891119892by formula (32) and find 119896 like we did in step

(09) ofAlgorithm3 119896 is themaximal number forwhich119901119896 le

119860119891119892 Then choose any 119896 + 1 primes 119902

1 119902

119896+1not dividing

119908 and start computing the modular gcdrsquos 1198901199021

(119909) 1198901199022

(119909)

(119896 + 1 times) If at some step we find deg 119890119902119894

(119909) = 0 then weare done the polynomials 119891(119909) 119892(119909) are coprime if 119903 asymp 1 ortheir gcd is the nontrivial scalar 119903 asymp 1 And if deg 119890

119902119894

(119909) gt 0

for all 119902119894 then we know that

(1) these polynomials are not coprime(2) the positive degree of gcd(119891(119909) 119892(119909)) is theminimum

119904 (119891 119892) = min deg 1198901199021(119909) deg 119890

119902119896+1

(119909) gt 0 (42)

This exact value of 119904(119891 119892) = deg gcd(119891(119909) 119892(119909)) is a betterresult than the estimate 119904(119902 119891 119892) obtained earlier by just one119902


Like above we can assume 119891(119909) 119892(119909) to be primitive (ifnot we can again denote 119903 = gcd(cont(119891(119909)) cont(119892(119909))) andswitch to the primitive parts 119891(119909) = pp(119891(119909)) and 119892(119909) =

pp(119892(119909))) Applying the Landau-MignotteTheorem 6 for thecoefficients 119888

119894of gcd(119891(119909) 119892(119909)) we get that |119888

119894| le 119872

119891119892

where

119872119891119892

= 2119904(119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(43)

Now we can take 119901 gt 2 sdot 119872119891119892

and by the Euclideanalgorithm calculate 119890

119901(119909) in Z

119901[119909] If deg 119890

119901(119909) gt 119904(119891 119892) we

drop this 119901 and choose another prime 119901 gt 2 sdot 119872119891119892 And if

deg 119890119901(119909) = 119904(119891 119892) then we proceed to the final steps we

choose 119905 then get the preimage 119896(119909) of 119905 sdot 119890119901(119909) then go to the

primitive part 119889(119909) = pp(119896(119909)) and output the final answeras gcd(119891(119909) 119892(119909)) = 119903 sdot pp(119896(119909))

Remark 28 This approach has the following advantagesFirstly the bound on primes 119901 is better than formula (40)since here we have not 2119904(119902119891119892) but 2119904(119891119892) Secondly we nolonger need to calculate the number 119905 the preimage 119896(119909) andthe primitive part 119889(119909) for more than one prime 119901 This isbecause if the selected 119901 gt 2 sdot 119872

119891119892is not appropriate we

already have an indicator of that deg 119890119901(119909) gt 119904(119891 119892)

We built Algorithm 5

Example 29 Let us apply Algorithm 5 again on polynomialsof Knuthrsquos example (1) As we saw in Example 22 119896 = 30 Sowemayhave to consider atmost 31 auxiliary primes 119902

119894 Butwe

in fact need just two of them because deg gcd(1198912(119909) 1198922(119909)) =

deg(1199092 + 119909 + 1) = 2 and deg gcd(1198913(119909) 1198923(119909)) = deg(1) = 0

(see Example 25) So in Algorithm 5 we jump from step (16)

to step (32) directly

Competing Interests

The author declares that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The author was supported in part by joint grant 15RF-054 ofRFBR and SCS MES RA (in frames of joint research projectsSCS and RFBR) and by 15T-1A258 grant of SCS MES RA

References

[1] V H Mikaelian Algorithmic Algebra Commutative Rings andFields Yerevan University Press Yerevan Armenia OfficiallyApproved by the Ministry of Higher Education and Science ofRA As Textbook for Universities 2015

[2] D Knuth ldquoThe art of computer programmingrdquo in Vol 2Seminumerical Algorithms Addison-Wesley Series in ComputerScience and Information Processing Addison-Wesley 2ndedition 1969

[3] W S Brown ldquoOn Euclidrsquos algorithm and the computation ofpolynomial greatest common divisorsrdquo Journal of the ACM vol18 pp 478ndash504 1971

[4] J von zur Gathen and J Gerhard Modern Computer AlgebraCambridge University Press Cambridge UK 3rd edition 2013

[5] J H Davenport Y Siret and E Tournier Computer AlgebraSystems and Algorithms for Algebraic Computation AcademicPress London UK 2nd edition 1993

[6] E V Pankratev Elements of Computer Algebra Study GuideBinom LZ Moscow Russia 2007

[7] S LangAlgebra RevisedThirdEdition vol 211 ofGraduate Textsin Mathematics Springer New York NY USA 2002

[8] P B Garrett Abstract Algebra Chapman amp HallCRC BocaRaton Fla USA 2008

[9] P M Cohn Introduction to RingTheory Springer London UK2000

[10] A I Kostrikin An Introduction to Algebra vol 2 of MainStructures FizMatLit Moscow Russia 2004 (Russian)

[11] P M Cohn Basic Algebra Groups Rings and Fields SpringerLondon UK 2003

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Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

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Stochastic AnalysisInternational Journal of


The first aim of this work is to present in Sections 2ndash5 a slightly modified argumentation of the algorithm basedon comparison of the degrees of common divisors of 119891(119909)and 119892(119909) in Z[119909] and of 119891

119901(119909) and 119892

119901(119909) in Z

119901[119909] (see

Algorithm 2) This approach allows some simplification ofa step of the algorithm for some primes 119901 we need notreconstruct the preimage of 119905 sdot gcd(119891

119901(119909) 119892119901(119909)) but we

immediately get an indication that this prime is not suitableand we should proceed to a new 119901 (see Remark 15)

Then in Section 6 we discuss the problem if the Big primemodular gcd algorithm could output the correct answer usingjust one prime 119901 or not The answer is positive but for somereasons it should not be used to improve the algorithm (tomake it work with one 119901) because it evolves a very largeprime (see Remark 19) Instead we show that we can estimatethe maximal number of 119901 (repetitions of steps) that may beused in traditional Big prime modular gcd algorithm Forexample for polynomials (1) of Knuthrsquos example this numberis at most 31 Estimates of this type can be found in literatureelsewhere We just make the bound considerably smaller (seeRemark 23)

The obtained bounds on the number of primes 119901 areespecially effective when we are interested not in gcd but justin detection if the polynomials119891(119909) 119892(119909) isin Z[119909] are coprimeor not We consider this in Section 7 (see Algorithm 3)

In Section 8 we consider four other ideas to modify theBig prime modular gcd algorithm Two first ideas are basedon checking the number of primes 119901 The third idea is basedon using an auxiliary prime 119902 to estimate the degree ofgcd(119891(119909) 119892(119909)) by means of the degree of gcd(119891

119902(119909) 119892119902(119909))

(see Algorithm4) Example 27 shows howmuch better resultswe may get by this modification The fourth idea combinesboth approaches it uses a set of auxiliary primes 119902

1 119902

119896+1

to correctly find the degree of gcd(119891(119909) 119892(119909)) and thenwe use a modified version of Landau-Mignotte bound tofind a single big prime 119901 by which we can calculate thegcd(119891

119901(119909) 119892119901(119909))

The arguments used here can be generalized for thecase of polynomials on general UFDs From the uniquefactorization in a UFD it easily follows that gcd alwaysexists and it is easy to detect if the given common divisorof maximal degree is gcd or not The less simple part is tofind ways to compute gcd (without having the prime-powerfactorization) That can be done for some classes of UFDssuch as multivariate polynomials on fields

2 gcd in Polynomial Rings and the Degrees ofCommon Divisors

Theproblem of finding the greatest common divisor gcd(119886 119887)of any nonzero elements 119886 119887 in a ring 119877 can be separated totwo tasks

(1) Finding out if gcd(119886 119887) in general exists for 119886 119887 isin 119877(2) Finding an effective way to calculate gcd(119886 119887)

The Euclidean algorithm gives an easy answer to bothof these tasks in any Euclidean domain that is an integritydomain119877 possessingEuclidean norm 120575 1198770 rarr Ncup0 such

that 120575(119886119887) ge 120575(119886) hold for any nonzero elements 119886 119887 isin 119877 andfor any 119886 119887 isin 119877 where 119887 = 0 there exist elements 119902 119903 isin 119877such that 119886 = 119902119887 + 119903 where either 119903 = 0 or 119903 = 0 and120575(119903) lt 120575(119887) [4 5 7ndash10] The Euclidean algorithm works forany polynomial ring119870[119909] over a field119870 such asQ[119909]R[119909]C[119909] and Z

119901[119909] because these rings can easily be turned to

an Euclidean domain by defining 120575(119891(119909)) = deg119891(119909) for anynonzero 119891(119909) isin 119870[119909]

The situation is less simple in non-Euclidean domainseven in such a widely used ring as the ring Z[119909] of polyno-mials with integer coefficients That Z[119909] is not an Euclideandomain easy to show by elements 119909 2 isin Z[119909] IfZ[119909]were anEuclidean domain it would contain elements 119906(119909) V(119909) suchthat 119909 sdot 119906(119909) + 2 sdot V(119909) = gcd(119909 2) = plusmn1 which is not possible

The first of two tasks mentioned above namely existenceof gcd can be accomplished for Z[119909] by proving that Z[119909] isa UFD that is an integrity domain in which every nonzeroelement 119886 has a factorization 119886 = 120598 119901

1sdot sdot sdot 119901119896 where 120598 isin 119877

lowast

is a unit (invertible) element in 119877 the elements 119901119894are prime

for all 119894 = 1 119896 and the factorization above is uniquein the sense that if 119886 has another factorization of that type120579 1199021sdot sdot sdot 119902119904 where 120579 isin 119877

lowast and the elements 119902119894are prime then

119896 = 119904 and (perhaps after some reordering of the prime factors)the respective prime elements are associated 119901

119894asymp 119902119894for all

119894 = 1 119896 For briefness in the sequel we will often omit thephrase ldquoperhaps after some reordering of the prime factorsrdquoand this will cause no confusion

After merging the associated prime elements together weget a unique factorization into prime-power elements

119886 = ] 1199011205721

1sdot sdot sdot 119901120572119899

119899

] isin 119877lowast 120572119894isin N 119901

119894asymp 119901119895for any 119894 = 119895 119894 119895 = 1 119899

(2)

(in some arguments below we may admit that some of thefactors 119901120572119894

119894participate with degrees 120572

119894= 0 this makes some

notations simpler) From this it is easy to see that in a UFD119877 gcd(119886 119887) exists for any nonzero elements 119886 119887 isin 119877 Assume119887 isin 119877 has the factorization

119887 = 120581 1199011205721015840

1

1sdot sdot sdot 1199011205721015840

119899

119899 120581 isin 119877

lowast (3)

(we use the same primes 119901119894in both factorizations because

if eg 119901119894is not actually participating in one of those

factorizations we can add it as 119901120572119894119894with 120572

119894= 0) Then

gcd (119886 119887) = 119889 = 1199011205741

1sdot sdot sdot 119901120574119899

119899 (4)

where 120574119894= min120572

119894 1205721015840

119894 This follows from uniqueness of

factorization in UFD For if ℎ is a common divisor of 119886 119887 andif 119901119894is a prime divisor of ℎ then it also is a prime divisor of 119886

and of 119887 The elements 119901119894cannot participate in factorization

of ℎ by a power greater than min120572119894 1205721015840

119894 because then 119886 (or

119887) would have an alternative factorization in which 119901119894occurs

more than 120572119894(or 1205721015840119894) times

The shortest way to see that Z[119909] is a UFD is to applyGaussrsquos theorem if the ring 119877 is a UFD then the polynomialring 119877[119909] also is a UFD [4 7 8 10 11] SinceZ is a UFD (thatfact is known as ldquothe fundamental theorem of arithmeticrdquo)Z[119909] also is a UFD




Z[119909]lowast









119891 (119909) = cont (119891 (119909)) sdot pp (119891 (119909))

119892 (119909) = cont (119892 (119909)) sdot pp (119892 (119909)) (5)


pp (119891 (119909) sdot 119892 (119909)) = pp (119891 (119909)) sdot pp (119892 (119909)) (6)






1205721

1sdot sdot sdot 119901120572119899

119899 So if deg119891(119909) = 0 then





119891 (119909) = cont (119891 (119909)) sdot pp (119891 (119909))

= ] 1199011205721

1sdot sdot sdot 119901120572119899

119899sdot 1199021205731

1(119909) sdot sdot sdot 119902

120573119898

119898(119909)

(7)


119894(119909) Eliminate


119894(119909) If




1sdot sdot sdot 119901120572119899





119892 (119909) = ]1015840 1199011205721015840

1

1sdot sdot sdot 1199011205721015840

119899

119899sdot 1199021205731015840

1

1(119909) sdot sdot sdot 119902

1205731015840

119898

119898(119909) (8)

then

gcd (119891 (119909) 119892 (119909)) = 120581 1199011205741

1sdot sdot sdot 119901120574119899

119899sdot 1199021205751

1(119909) sdot sdot sdot 119902

120575119898

119898(119909) (9)

where 120581 = plusmn1 120574119894= min120572

119894 1205721015840

119894 and 120575

119895= min120573

119895 1205731015840

119895



2+ 24119909 + 12









has a factorization

119891 (119909) = 120579 sdot 1199021205731

1(119909) sdot sdot sdot 119902

120573119898

119898(119909)

120579 isin 119870lowast deg 119902

119894(119909) gt 0 119894 = 1 119898

(10)



119899= 120579 isin 119870

lowast= 119870 0 (11)








120574119894





119901be the


let

120593119901 Z 997888rarr Z

119901 (12)


119901(119911) equiv 119911(mod

119901) and 120593119901(119911) isin Z

119901


120593119901 Z [119909] 997888rarr Z

119901 [119909] (13)


119901and 120593

119901is




120593119901 119872119898119899 (Z) 997888rarr 119872

119898119899(Z119901) (14)


119898119899(Z)




For 119886 isin Z denote 120593119901(119886) = 119886

119901 For 119891(119909) isin Z[119909] denote

120593119901(119891(119909)) = 119891

119901(119909) isin Z

119901[119909] So if

119891 (119909) = 1198860119909119899+ sdot sdot sdot + 119886

119899 (15)

then119891119901(119909) = 120593

119901(119891 (119909)) = 120593

119901(1198860) 119909119899+ sdot sdot sdot + 120593

119901(119886119899)

= 1198860119901119909119899+ sdot sdot sdot + 119886

119899119901isin Z119901 [119909]

(16)


(Z) denote 120593119901(119860) = 119860

119901isin

119872119898119899

(Z119901) If 119860 = 119886

119894119895119898times119899

then 119860119901

= 120593119901(119886119894119895)119898times119899

=

119886119894119895119901

119898times119899





119901[119909] including






119901(119909) 119892119901(119909)) is not



119901(119909) 119892119901(119909) We


119901(119909) This 119890

119901(119909) is unique


119901(119909) by 119892

119901(119909) on field Z

119901)








120593119901(119891(119909)) and 119892

119901(119909) = 120593



Example 5 If 119891(119909) = 71199092+ 22 and 119892(119909) = 49119909

3+ 154119909


119901(119909) = 0 So



120601119899(119909) = prod

119896=1119899(119896119899)=1

(119909 minus 1198902119894120587119896119899

) (17)



radicsum119899

119894=01198862119894


119886119899and ℎ(119909) = 119888

0119909119896+ sdot sdot sdot + 119888



119899

sum119894=0

10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot10038161003816100381610038161003816100381610038161003816

1198880

1198860

10038161003816100381610038161003816100381610038161003816sdot1003817100381710038171003817119891 (119909)

1003817100381710038171003817 (18)



10038161003816100381610038161198881198941003816100381610038161003816 le 119873119891= 2119899minus1 1003817100381710038171003817119891 (119909)

1003817100381710038171003817 (19)




119894| | 119894 =

0 119899 le max|119886119894| | 119894 = 0 119899 le 119891(119909)






0119909119898+ sdot sdot sdot + 119887



10038161003816100381610038161198881198941003816100381610038161003816 le 119873119891119892

= 2min119899119898

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(20)




0and 1198870


119891119892and take








119901(119909)



119891119892


119901(119909) (if they






3+ 3119909 + 10 we



119901 gt 2 sdot 119873119891119892 (21)


119901(119888119894) =


119901(119888119894) = 119888119894+ 119901 gt 1199012




119901(119909) = 119887

0119901119909119896+ sdot sdot sdot + 119887

119896119901isin Z119901[119909]


119901(119909)

(03) Set 119894 = 0(04) While 119894 le 119896

(05) if 119887119894119901

lt 1199012

(06) set 119887119894= 119887119894119901

(07) else(08) set 119887

119894= 119887119894119901

minus 119901(10) set 119894 = 119894 + 1(11) Output 119889(119909) = 119887

0119909119896+ sdot sdot sdot + 119887

119896







119901(119909) we may not obtain 119889(119909)


2 sdot 119873119891119892

and for which

deg 119889 (119909) = deg 119890119901(119909) (22)




119901(119909) is gcd of119891

119901(119909) 119892119901(119909)








Example 11 For 119891(119909) = 1199092+ 4119909 + 3 and 119892(119909) = 119909

2+ 2119909 + 1


119890119901(119909) = gcd (119891

119901(119909) 119892

119901(119909)) = 2119909 + 2 isin Z

119901 [119909] (23)


119901(119909 + 1) = 2119909 + 2








119901 their fraction 119908119888

0


lc (119905 sdot 119890119901(119909)) = 119908 (24)












bound 119901 gt 2 sdot 119873119891119892





119891119892


119901is not



119889(119909) = gcd(119891(119909) 119892(119909)) = 1 Taking 119901 = 2 we get 1198912(119909) =

1199092+ 1 and 119892

2(119909) = 119909 + 1 In Z

2[119909] we have 119891

2(119909) = 119909

2+ 1 =

1199092+12= (119909+1)(119909+1) thus 119890

2(119909) = gcd(119891

2(119909) 1198922(119909)) = 119909+1


under 1205932






0




0= lc(119891(119909)) and 119887




119891119892by (20)



119901(119909) 119892

119901(119909) isin Z

119901[119909]

(10) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(11) If119863 le deg 119890119901(119909)


119901(119909)) = 119908



119901(119909)





0= lc(119891(119909)) and 119887

0= lc(119892(119909))





119891119892

(10) Set 119894 = 1(11) While 119894 = 119896 + 1



119901(119909) 119892

119901(119909) isin Z

119901[119909]

(14) calculate 119890119901= gcd(119891

119901(119909) 119892


119901[119909]

(15) if deg 119890119902119894= 0







0= lc(119891(119909)) and 119887






119902(119909) 119892

119902(119909) isin Z

119902[119909]

(09) Calculate 119890119902(119909) = gcd(119891

119902(119909) 119892


119902[119909]

(10) Set 119904(119902 119891 119892) = deg 119890119902(119909)

(11) Calculate119872119902119891119892


119902119891119892


119901(119909) 119892

119901(119909) isin Z

119901[119909]

(14) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(15) If119863 le deg 119890119901(119909)


119901(119909)) = 119908



119901(119909)








deg 119889 (119909) = deg 119889119901(119909) = deg gcd (119891

119901(119909) 119892

119901(119909))

= deg 119890119901(119909)

(26)

Proof Since 1198880



119901(119909) le

deg gcd(119891119901(119909) 119892119901(119909))


119891119901(119909)119889119901(119909) and 119892

119901(119909)119889119901(119909) in Z


izations of 119891119901(119909) and 119892

119901(119909) in UFD Z


deduce that

119890119901 (119909) asymp gcd (119891

119901 (119909) 119892119901 (119909))

asymp 119889119901(119909) sdot gcd(

119891119901(119909)

119889119901(119909)

119892119901(119909)

119889119901(119909)

)

(27)


119901(119909) And the



deg gcd(119891119901(119909)

119889119901 (119909)

119892119901(119909)

119889119901 (119909)

) gt 0 (28)

that is when 119891119901(119909)119889119901(119909) and 119892

119901(119909)119889119901(119909) are not


119901(119909)




120593119901 119872119898+119899

(Z) 997888rarr 119872119898+119899

(Z119901) (29)




) = 119878119891119901119889119901119892119901119889119901


119877119901= 120593119901 (119877) = 120593

119901(res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909)))

= res(119891119901(119909)

119889119901(119909)

119892119901(119909)

119889119901(119909)

)

(30)






119891119892


0= lc(119891(119909)) and 119887







119902119894(119909) 119892

119902119894(119909) isin Z

119902119894[119909]

(13) calculate 119890119902119894= gcd(119891

119902119894(119909) 119892


119901[119909]

(14) if deg 119890119902119894le 119904(119891 119892)

(15) set 119904(119891 119892) = deg 119890119902119894

(16) if deg 119890119902119894= 0





119901(119909) 119892

119901(119909) isin Z

119901[119909]

(23) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(24) If deg 119890119901= 119904(119891 119892)














119891119892

Then 119901 ∤ 119908 = gcd(1198860 1198870) since 119886

0 1198870le 119873119891119892 Calculate

119890119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) in Z





119901 we would

get deg ℎ119901(119909) gt deg 119889

119901(119909) = deg 119889(119909) = deg 119890

119901(119909) which










119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) check if deg 119890

119901(119909) is











and 119892(119909) = radic570

119873119891119892

= 2min86


1radic570

3

lt 512

(31)



(119909) 1198921031






119860119891119892

= radic(119899 + 1)119898(119898 + 1)

119899sdot 119873119898

119891119873119899

119892


119898(119898 + 1)

119899

sdot1003817100381710038171003817119891 (119909)

1003817100381710038171003817119898 1003817100381710038171003817119892 (119909)

1003817100381710038171003817119899

(32)


10038161003816100381610038161003816100381610038161003816res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909))10038161003816100381610038161003816100381610038161003816=10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le 119860119891119892 (33)


119891= 2119899minus1

119891(119909)

and 119873119892= 2119898minus1


10038171003817100381710038171003817100381710038171003817

119891 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119899 + 1)1198732

119891

10038171003817100381710038171003817100381710038171003817

119892 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119898 + 1)1198732

119892

(34)


119891119889119892119889 we get that

|119877| =10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le (radic(119899 + 1)119873

119891)119898

sdot (radic(119898 + 1)119873119892)119899

(35)



119891lt 1408 119873

119892lt 768

and119873119891119892

lt 512 Thus

|119877| le radic(8 + 1)6(6 + 1)

8sdot 1408

6sdot 5708= 120596

= 16505374299582118582810249858265119890 + 48

(36)



119901 gt 2 sdot 119860119891119892 (37)








119896 primes 119901119896 = 119901


1= 2 119901

2= 3 etc)


essential


119896 le 119899



119896 le 119860

119891119892


10 = 6 469 693 230





119896 also



11990130 = 31610054640417607788145206291544119890 + 46

lt 120596

11990131 = 4014476939333036189094441199026119890 + 48

gt 120596

(38)





119891


119891119892 Namely in




119891and 119873

119892appear 119898





119896 le


1 119901


119908 = gcd(1198860 1198870) where 119886

0= lc(119891(119909)) and 119887

0= lc(119892(119909))

If gcd(119891119901119894

(119909) 119892119901119894



(119909) 119892119901119894

(119909)) ge


119901119894

(119909) 119892119901119894

(119909)) = 1 for all 119894 = 1 119896 + 1 then119891119901119894

(119909) and 119892119901119894









2(119909) = 119909

8+1199096+1199094+1199093+1 1198922(119909) = 119909

6+1199094+119909+1 which


2+ 119909 + 1 = 1

And for 119901 = 3 we get 1198913(119909) = 119909

8+ 1199096+ 21199092+ 2119909 + 1

1198923(119909) = 2119909



119891=

2radic6119873119892= radic2 and119860




119901(119909) 119892119901(119909)) = 1 for 119901 = 2 3 5 7







119891 119873119892by Corollary 7

119873119891119892

by Corollary 9 and 119860119891119892


119896 le 119860


1199011 119901


119891119892 We do not know 119889(119909)



1 119901




119901119894




119901119894


119901119894


preimage 119896(119909) of 119905 sdot 119890119901119894



119901119894




numbers 1199011 119901


119894∤ 119908


2 sdot 119873119891119892


119901119894


1 119901

119896+1


(119909) ge


1 119901



119901119894


119891119892and


119901(119909) = 119872 then


119901(119909) = 119872 then we have





0119909119896+ sdot sdot sdot + 119888



119899and 119892(119909) = 119887

0119909119898+sdot sdot sdot+119887

119898

then |119888119894| le 2119896|11988801198860|119891(119909) and |119888

119894| le 2119896|11988801198870|119892(119909) Since

|11988801198860| is bounded by |gcd(119886

0 1198870)1198860| and |119888



10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot gcd (119886

0 1198870) sdotmin

1003817100381710038171003817119891 (119909)1003817100381710038171003817

100381610038161003816100381611988601003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(39)




119894| le 119872

119902119891119892 where

119872119902119891119892

= 2119904(119902119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(40)

119872119902119891119892







if 119901 gt 2 sdot 119872119902119891119892


min119891(119909)|1198860| 119892(119909)|119887





2(119909) = gcd(119891

2(119909) 1198922(119909)) =

1199092+ 119909 + 1 Then 119904(2 119891 119892) = deg 119890

2(119909) = 2 and

1198722119891119892

= 22sdot 1 sdotmin

radic113

1radic570

3 lt 3184 (41)

Take 119901 = 67 gt 2 sdot 1198722119891119892




deg 1198903(119909) = 0 and 119872







1 119902



(119909) 1198901199022

(119909)



119902119894

(119909) gt 0



119904 (119891 119892) = min deg 1198901199021(119909) deg 119890

119902119896+1

(119909) gt 0 (42)





119894of gcd(119891(119909) 119892(119909)) we get that |119888

119894| le 119872

119891119892

where

119872119891119892

= 2119904(119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(43)



119901(119909) in Z

119901[119909] If deg 119890

119901(119909) gt 119904(119891 119892) we










119894 Butwe





Competing Interests


Acknowledgments


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra












Z[119909]lowast









119891 (119909) = cont (119891 (119909)) sdot pp (119891 (119909))

119892 (119909) = cont (119892 (119909)) sdot pp (119892 (119909)) (5)


pp (119891 (119909) sdot 119892 (119909)) = pp (119891 (119909)) sdot pp (119892 (119909)) (6)






1205721

1sdot sdot sdot 119901120572119899

119899 So if deg119891(119909) = 0 then





119891 (119909) = cont (119891 (119909)) sdot pp (119891 (119909))

= ] 1199011205721

1sdot sdot sdot 119901120572119899

119899sdot 1199021205731

1(119909) sdot sdot sdot 119902

120573119898

119898(119909)

(7)


119894(119909) Eliminate


119894(119909) If




1sdot sdot sdot 119901120572119899





119892 (119909) = ]1015840 1199011205721015840

1

1sdot sdot sdot 1199011205721015840

119899

119899sdot 1199021205731015840

1

1(119909) sdot sdot sdot 119902

1205731015840

119898

119898(119909) (8)

then

gcd (119891 (119909) 119892 (119909)) = 120581 1199011205741

1sdot sdot sdot 119901120574119899

119899sdot 1199021205751

1(119909) sdot sdot sdot 119902

120575119898

119898(119909) (9)

where 120581 = plusmn1 120574119894= min120572

119894 1205721015840

119894 and 120575

119895= min120573

119895 1205731015840

119895



2+ 24119909 + 12









has a factorization

119891 (119909) = 120579 sdot 1199021205731

1(119909) sdot sdot sdot 119902

120573119898

119898(119909)

120579 isin 119870lowast deg 119902

119894(119909) gt 0 119894 = 1 119898

(10)



119899= 120579 isin 119870

lowast= 119870 0 (11)








120574119894





119901be the


let

120593119901 Z 997888rarr Z

119901 (12)


119901(119911) equiv 119911(mod

119901) and 120593119901(119911) isin Z

119901


120593119901 Z [119909] 997888rarr Z

119901 [119909] (13)


119901and 120593

119901is




120593119901 119872119898119899 (Z) 997888rarr 119872

119898119899(Z119901) (14)


119898119899(Z)




For 119886 isin Z denote 120593119901(119886) = 119886

119901 For 119891(119909) isin Z[119909] denote

120593119901(119891(119909)) = 119891

119901(119909) isin Z

119901[119909] So if

119891 (119909) = 1198860119909119899+ sdot sdot sdot + 119886

119899 (15)

then119891119901(119909) = 120593

119901(119891 (119909)) = 120593

119901(1198860) 119909119899+ sdot sdot sdot + 120593

119901(119886119899)

= 1198860119901119909119899+ sdot sdot sdot + 119886

119899119901isin Z119901 [119909]

(16)


(Z) denote 120593119901(119860) = 119860

119901isin

119872119898119899

(Z119901) If 119860 = 119886

119894119895119898times119899

then 119860119901

= 120593119901(119886119894119895)119898times119899

=

119886119894119895119901

119898times119899





119901[119909] including






119901(119909) 119892119901(119909)) is not



119901(119909) 119892119901(119909) We


119901(119909) This 119890

119901(119909) is unique


119901(119909) by 119892

119901(119909) on field Z

119901)








120593119901(119891(119909)) and 119892

119901(119909) = 120593



Example 5 If 119891(119909) = 71199092+ 22 and 119892(119909) = 49119909

3+ 154119909


119901(119909) = 0 So



120601119899(119909) = prod

119896=1119899(119896119899)=1

(119909 minus 1198902119894120587119896119899

) (17)



radicsum119899

119894=01198862119894


119886119899and ℎ(119909) = 119888

0119909119896+ sdot sdot sdot + 119888



119899

sum119894=0

10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot10038161003816100381610038161003816100381610038161003816

1198880

1198860

10038161003816100381610038161003816100381610038161003816sdot1003817100381710038171003817119891 (119909)

1003817100381710038171003817 (18)



10038161003816100381610038161198881198941003816100381610038161003816 le 119873119891= 2119899minus1 1003817100381710038171003817119891 (119909)

1003817100381710038171003817 (19)




119894| | 119894 =

0 119899 le max|119886119894| | 119894 = 0 119899 le 119891(119909)






0119909119898+ sdot sdot sdot + 119887



10038161003816100381610038161198881198941003816100381610038161003816 le 119873119891119892

= 2min119899119898

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(20)




0and 1198870


119891119892and take








119901(119909)



119891119892


119901(119909) (if they






3+ 3119909 + 10 we



119901 gt 2 sdot 119873119891119892 (21)


119901(119888119894) =


119901(119888119894) = 119888119894+ 119901 gt 1199012




119901(119909) = 119887

0119901119909119896+ sdot sdot sdot + 119887

119896119901isin Z119901[119909]


119901(119909)

(03) Set 119894 = 0(04) While 119894 le 119896

(05) if 119887119894119901

lt 1199012

(06) set 119887119894= 119887119894119901

(07) else(08) set 119887

119894= 119887119894119901

minus 119901(10) set 119894 = 119894 + 1(11) Output 119889(119909) = 119887

0119909119896+ sdot sdot sdot + 119887

119896







119901(119909) we may not obtain 119889(119909)


2 sdot 119873119891119892

and for which

deg 119889 (119909) = deg 119890119901(119909) (22)




119901(119909) is gcd of119891

119901(119909) 119892119901(119909)








Example 11 For 119891(119909) = 1199092+ 4119909 + 3 and 119892(119909) = 119909

2+ 2119909 + 1


119890119901(119909) = gcd (119891

119901(119909) 119892

119901(119909)) = 2119909 + 2 isin Z

119901 [119909] (23)


119901(119909 + 1) = 2119909 + 2








119901 their fraction 119908119888

0


lc (119905 sdot 119890119901(119909)) = 119908 (24)












bound 119901 gt 2 sdot 119873119891119892





119891119892


119901is not



119889(119909) = gcd(119891(119909) 119892(119909)) = 1 Taking 119901 = 2 we get 1198912(119909) =

1199092+ 1 and 119892

2(119909) = 119909 + 1 In Z

2[119909] we have 119891

2(119909) = 119909

2+ 1 =

1199092+12= (119909+1)(119909+1) thus 119890

2(119909) = gcd(119891

2(119909) 1198922(119909)) = 119909+1


under 1205932






0




0= lc(119891(119909)) and 119887




119891119892by (20)



119901(119909) 119892

119901(119909) isin Z

119901[119909]

(10) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(11) If119863 le deg 119890119901(119909)


119901(119909)) = 119908



119901(119909)





0= lc(119891(119909)) and 119887

0= lc(119892(119909))





119891119892

(10) Set 119894 = 1(11) While 119894 = 119896 + 1



119901(119909) 119892

119901(119909) isin Z

119901[119909]

(14) calculate 119890119901= gcd(119891

119901(119909) 119892


119901[119909]

(15) if deg 119890119902119894= 0







0= lc(119891(119909)) and 119887






119902(119909) 119892

119902(119909) isin Z

119902[119909]

(09) Calculate 119890119902(119909) = gcd(119891

119902(119909) 119892


119902[119909]

(10) Set 119904(119902 119891 119892) = deg 119890119902(119909)

(11) Calculate119872119902119891119892


119902119891119892


119901(119909) 119892

119901(119909) isin Z

119901[119909]

(14) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(15) If119863 le deg 119890119901(119909)


119901(119909)) = 119908



119901(119909)








deg 119889 (119909) = deg 119889119901(119909) = deg gcd (119891

119901(119909) 119892

119901(119909))

= deg 119890119901(119909)

(26)

Proof Since 1198880



119901(119909) le

deg gcd(119891119901(119909) 119892119901(119909))


119891119901(119909)119889119901(119909) and 119892

119901(119909)119889119901(119909) in Z


izations of 119891119901(119909) and 119892

119901(119909) in UFD Z


deduce that

119890119901 (119909) asymp gcd (119891

119901 (119909) 119892119901 (119909))

asymp 119889119901(119909) sdot gcd(

119891119901(119909)

119889119901(119909)

119892119901(119909)

119889119901(119909)

)

(27)


119901(119909) And the



deg gcd(119891119901(119909)

119889119901 (119909)

119892119901(119909)

119889119901 (119909)

) gt 0 (28)

that is when 119891119901(119909)119889119901(119909) and 119892

119901(119909)119889119901(119909) are not


119901(119909)




120593119901 119872119898+119899

(Z) 997888rarr 119872119898+119899

(Z119901) (29)




) = 119878119891119901119889119901119892119901119889119901


119877119901= 120593119901 (119877) = 120593

119901(res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909)))

= res(119891119901(119909)

119889119901(119909)

119892119901(119909)

119889119901(119909)

)

(30)






119891119892


0= lc(119891(119909)) and 119887







119902119894(119909) 119892

119902119894(119909) isin Z

119902119894[119909]

(13) calculate 119890119902119894= gcd(119891

119902119894(119909) 119892


119901[119909]

(14) if deg 119890119902119894le 119904(119891 119892)

(15) set 119904(119891 119892) = deg 119890119902119894

(16) if deg 119890119902119894= 0





119901(119909) 119892

119901(119909) isin Z

119901[119909]

(23) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(24) If deg 119890119901= 119904(119891 119892)














119891119892

Then 119901 ∤ 119908 = gcd(1198860 1198870) since 119886

0 1198870le 119873119891119892 Calculate

119890119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) in Z





119901 we would

get deg ℎ119901(119909) gt deg 119889

119901(119909) = deg 119889(119909) = deg 119890

119901(119909) which










119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) check if deg 119890

119901(119909) is











and 119892(119909) = radic570

119873119891119892

= 2min86


1radic570

3

lt 512

(31)



(119909) 1198921031






119860119891119892

= radic(119899 + 1)119898(119898 + 1)

119899sdot 119873119898

119891119873119899

119892


119898(119898 + 1)

119899

sdot1003817100381710038171003817119891 (119909)

1003817100381710038171003817119898 1003817100381710038171003817119892 (119909)

1003817100381710038171003817119899

(32)


10038161003816100381610038161003816100381610038161003816res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909))10038161003816100381610038161003816100381610038161003816=10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le 119860119891119892 (33)


119891= 2119899minus1

119891(119909)

and 119873119892= 2119898minus1


10038171003817100381710038171003817100381710038171003817

119891 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119899 + 1)1198732

119891

10038171003817100381710038171003817100381710038171003817

119892 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119898 + 1)1198732

119892

(34)


119891119889119892119889 we get that

|119877| =10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le (radic(119899 + 1)119873

119891)119898

sdot (radic(119898 + 1)119873119892)119899

(35)



119891lt 1408 119873

119892lt 768

and119873119891119892

lt 512 Thus

|119877| le radic(8 + 1)6(6 + 1)

8sdot 1408

6sdot 5708= 120596

= 16505374299582118582810249858265119890 + 48

(36)



119901 gt 2 sdot 119860119891119892 (37)








119896 primes 119901119896 = 119901


1= 2 119901

2= 3 etc)


essential


119896 le 119899



119896 le 119860

119891119892


10 = 6 469 693 230





119896 also



11990130 = 31610054640417607788145206291544119890 + 46

lt 120596

11990131 = 4014476939333036189094441199026119890 + 48

gt 120596

(38)





119891


119891119892 Namely in




119891and 119873

119892appear 119898





119896 le


1 119901


119908 = gcd(1198860 1198870) where 119886

0= lc(119891(119909)) and 119887

0= lc(119892(119909))

If gcd(119891119901119894

(119909) 119892119901119894



(119909) 119892119901119894

(119909)) ge


119901119894

(119909) 119892119901119894

(119909)) = 1 for all 119894 = 1 119896 + 1 then119891119901119894

(119909) and 119892119901119894









2(119909) = 119909

8+1199096+1199094+1199093+1 1198922(119909) = 119909

6+1199094+119909+1 which


2+ 119909 + 1 = 1

And for 119901 = 3 we get 1198913(119909) = 119909

8+ 1199096+ 21199092+ 2119909 + 1

1198923(119909) = 2119909



119891=

2radic6119873119892= radic2 and119860




119901(119909) 119892119901(119909)) = 1 for 119901 = 2 3 5 7







119891 119873119892by Corollary 7

119873119891119892

by Corollary 9 and 119860119891119892


119896 le 119860


1199011 119901


119891119892 We do not know 119889(119909)



1 119901




119901119894




119901119894


119901119894


preimage 119896(119909) of 119905 sdot 119890119901119894



119901119894




numbers 1199011 119901


119894∤ 119908


2 sdot 119873119891119892


119901119894


1 119901

119896+1


(119909) ge


1 119901



119901119894


119891119892and


119901(119909) = 119872 then


119901(119909) = 119872 then we have





0119909119896+ sdot sdot sdot + 119888



119899and 119892(119909) = 119887

0119909119898+sdot sdot sdot+119887

119898

then |119888119894| le 2119896|11988801198860|119891(119909) and |119888

119894| le 2119896|11988801198870|119892(119909) Since

|11988801198860| is bounded by |gcd(119886

0 1198870)1198860| and |119888



10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot gcd (119886

0 1198870) sdotmin

1003817100381710038171003817119891 (119909)1003817100381710038171003817

100381610038161003816100381611988601003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(39)




119894| le 119872

119902119891119892 where

119872119902119891119892

= 2119904(119902119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(40)

119872119902119891119892







if 119901 gt 2 sdot 119872119902119891119892


min119891(119909)|1198860| 119892(119909)|119887





2(119909) = gcd(119891

2(119909) 1198922(119909)) =

1199092+ 119909 + 1 Then 119904(2 119891 119892) = deg 119890

2(119909) = 2 and

1198722119891119892

= 22sdot 1 sdotmin

radic113

1radic570

3 lt 3184 (41)

Take 119901 = 67 gt 2 sdot 1198722119891119892




deg 1198903(119909) = 0 and 119872







1 119902



(119909) 1198901199022

(119909)



119902119894

(119909) gt 0



119904 (119891 119892) = min deg 1198901199021(119909) deg 119890

119902119896+1

(119909) gt 0 (42)





119894of gcd(119891(119909) 119892(119909)) we get that |119888

119894| le 119872

119891119892

where

119872119891119892

= 2119904(119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(43)



119901(119909) in Z

119901[119909] If deg 119890

119901(119909) gt 119904(119891 119892) we










119894 Butwe





Competing Interests


Acknowledgments


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















has a factorization

119891 (119909) = 120579 sdot 1199021205731

1(119909) sdot sdot sdot 119902

120573119898

119898(119909)

120579 isin 119870lowast deg 119902

119894(119909) gt 0 119894 = 1 119898

(10)



119899= 120579 isin 119870

lowast= 119870 0 (11)








120574119894





119901be the


let

120593119901 Z 997888rarr Z

119901 (12)


119901(119911) equiv 119911(mod

119901) and 120593119901(119911) isin Z

119901


120593119901 Z [119909] 997888rarr Z

119901 [119909] (13)


119901and 120593

119901is




120593119901 119872119898119899 (Z) 997888rarr 119872

119898119899(Z119901) (14)


119898119899(Z)




For 119886 isin Z denote 120593119901(119886) = 119886

119901 For 119891(119909) isin Z[119909] denote

120593119901(119891(119909)) = 119891

119901(119909) isin Z

119901[119909] So if

119891 (119909) = 1198860119909119899+ sdot sdot sdot + 119886

119899 (15)

then119891119901(119909) = 120593

119901(119891 (119909)) = 120593

119901(1198860) 119909119899+ sdot sdot sdot + 120593

119901(119886119899)

= 1198860119901119909119899+ sdot sdot sdot + 119886

119899119901isin Z119901 [119909]

(16)


(Z) denote 120593119901(119860) = 119860

119901isin

119872119898119899

(Z119901) If 119860 = 119886

119894119895119898times119899

then 119860119901

= 120593119901(119886119894119895)119898times119899

=

119886119894119895119901

119898times119899





119901[119909] including






119901(119909) 119892119901(119909)) is not



119901(119909) 119892119901(119909) We


119901(119909) This 119890

119901(119909) is unique


119901(119909) by 119892

119901(119909) on field Z

119901)








120593119901(119891(119909)) and 119892

119901(119909) = 120593



Example 5 If 119891(119909) = 71199092+ 22 and 119892(119909) = 49119909

3+ 154119909


119901(119909) = 0 So



120601119899(119909) = prod

119896=1119899(119896119899)=1

(119909 minus 1198902119894120587119896119899

) (17)



radicsum119899

119894=01198862119894


119886119899and ℎ(119909) = 119888

0119909119896+ sdot sdot sdot + 119888



119899

sum119894=0

10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot10038161003816100381610038161003816100381610038161003816

1198880

1198860

10038161003816100381610038161003816100381610038161003816sdot1003817100381710038171003817119891 (119909)

1003817100381710038171003817 (18)



10038161003816100381610038161198881198941003816100381610038161003816 le 119873119891= 2119899minus1 1003817100381710038171003817119891 (119909)

1003817100381710038171003817 (19)




119894| | 119894 =

0 119899 le max|119886119894| | 119894 = 0 119899 le 119891(119909)






0119909119898+ sdot sdot sdot + 119887



10038161003816100381610038161198881198941003816100381610038161003816 le 119873119891119892

= 2min119899119898

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(20)




0and 1198870


119891119892and take








119901(119909)



119891119892


119901(119909) (if they






3+ 3119909 + 10 we



119901 gt 2 sdot 119873119891119892 (21)


119901(119888119894) =


119901(119888119894) = 119888119894+ 119901 gt 1199012




119901(119909) = 119887

0119901119909119896+ sdot sdot sdot + 119887

119896119901isin Z119901[119909]


119901(119909)

(03) Set 119894 = 0(04) While 119894 le 119896

(05) if 119887119894119901

lt 1199012

(06) set 119887119894= 119887119894119901

(07) else(08) set 119887

119894= 119887119894119901

minus 119901(10) set 119894 = 119894 + 1(11) Output 119889(119909) = 119887

0119909119896+ sdot sdot sdot + 119887

119896







119901(119909) we may not obtain 119889(119909)


2 sdot 119873119891119892

and for which

deg 119889 (119909) = deg 119890119901(119909) (22)




119901(119909) is gcd of119891

119901(119909) 119892119901(119909)








Example 11 For 119891(119909) = 1199092+ 4119909 + 3 and 119892(119909) = 119909

2+ 2119909 + 1


119890119901(119909) = gcd (119891

119901(119909) 119892

119901(119909)) = 2119909 + 2 isin Z

119901 [119909] (23)


119901(119909 + 1) = 2119909 + 2








119901 their fraction 119908119888

0


lc (119905 sdot 119890119901(119909)) = 119908 (24)












bound 119901 gt 2 sdot 119873119891119892





119891119892


119901is not



119889(119909) = gcd(119891(119909) 119892(119909)) = 1 Taking 119901 = 2 we get 1198912(119909) =

1199092+ 1 and 119892

2(119909) = 119909 + 1 In Z

2[119909] we have 119891

2(119909) = 119909

2+ 1 =

1199092+12= (119909+1)(119909+1) thus 119890

2(119909) = gcd(119891

2(119909) 1198922(119909)) = 119909+1


under 1205932






0




0= lc(119891(119909)) and 119887




119891119892by (20)



119901(119909) 119892

119901(119909) isin Z

119901[119909]

(10) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(11) If119863 le deg 119890119901(119909)


119901(119909)) = 119908



119901(119909)





0= lc(119891(119909)) and 119887

0= lc(119892(119909))





119891119892

(10) Set 119894 = 1(11) While 119894 = 119896 + 1



119901(119909) 119892

119901(119909) isin Z

119901[119909]

(14) calculate 119890119901= gcd(119891

119901(119909) 119892


119901[119909]

(15) if deg 119890119902119894= 0







0= lc(119891(119909)) and 119887






119902(119909) 119892

119902(119909) isin Z

119902[119909]

(09) Calculate 119890119902(119909) = gcd(119891

119902(119909) 119892


119902[119909]

(10) Set 119904(119902 119891 119892) = deg 119890119902(119909)

(11) Calculate119872119902119891119892


119902119891119892


119901(119909) 119892

119901(119909) isin Z

119901[119909]

(14) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(15) If119863 le deg 119890119901(119909)


119901(119909)) = 119908



119901(119909)








deg 119889 (119909) = deg 119889119901(119909) = deg gcd (119891

119901(119909) 119892

119901(119909))

= deg 119890119901(119909)

(26)

Proof Since 1198880



119901(119909) le

deg gcd(119891119901(119909) 119892119901(119909))


119891119901(119909)119889119901(119909) and 119892

119901(119909)119889119901(119909) in Z


izations of 119891119901(119909) and 119892

119901(119909) in UFD Z


deduce that

119890119901 (119909) asymp gcd (119891

119901 (119909) 119892119901 (119909))

asymp 119889119901(119909) sdot gcd(

119891119901(119909)

119889119901(119909)

119892119901(119909)

119889119901(119909)

)

(27)


119901(119909) And the



deg gcd(119891119901(119909)

119889119901 (119909)

119892119901(119909)

119889119901 (119909)

) gt 0 (28)

that is when 119891119901(119909)119889119901(119909) and 119892

119901(119909)119889119901(119909) are not


119901(119909)




120593119901 119872119898+119899

(Z) 997888rarr 119872119898+119899

(Z119901) (29)




) = 119878119891119901119889119901119892119901119889119901


119877119901= 120593119901 (119877) = 120593

119901(res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909)))

= res(119891119901(119909)

119889119901(119909)

119892119901(119909)

119889119901(119909)

)

(30)






119891119892


0= lc(119891(119909)) and 119887







119902119894(119909) 119892

119902119894(119909) isin Z

119902119894[119909]

(13) calculate 119890119902119894= gcd(119891

119902119894(119909) 119892


119901[119909]

(14) if deg 119890119902119894le 119904(119891 119892)

(15) set 119904(119891 119892) = deg 119890119902119894

(16) if deg 119890119902119894= 0





119901(119909) 119892

119901(119909) isin Z

119901[119909]

(23) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(24) If deg 119890119901= 119904(119891 119892)














119891119892

Then 119901 ∤ 119908 = gcd(1198860 1198870) since 119886

0 1198870le 119873119891119892 Calculate

119890119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) in Z





119901 we would

get deg ℎ119901(119909) gt deg 119889

119901(119909) = deg 119889(119909) = deg 119890

119901(119909) which










119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) check if deg 119890

119901(119909) is











and 119892(119909) = radic570

119873119891119892

= 2min86


1radic570

3

lt 512

(31)



(119909) 1198921031






119860119891119892

= radic(119899 + 1)119898(119898 + 1)

119899sdot 119873119898

119891119873119899

119892


119898(119898 + 1)

119899

sdot1003817100381710038171003817119891 (119909)

1003817100381710038171003817119898 1003817100381710038171003817119892 (119909)

1003817100381710038171003817119899

(32)


10038161003816100381610038161003816100381610038161003816res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909))10038161003816100381610038161003816100381610038161003816=10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le 119860119891119892 (33)


119891= 2119899minus1

119891(119909)

and 119873119892= 2119898minus1


10038171003817100381710038171003817100381710038171003817

119891 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119899 + 1)1198732

119891

10038171003817100381710038171003817100381710038171003817

119892 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119898 + 1)1198732

119892

(34)


119891119889119892119889 we get that

|119877| =10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le (radic(119899 + 1)119873

119891)119898

sdot (radic(119898 + 1)119873119892)119899

(35)



119891lt 1408 119873

119892lt 768

and119873119891119892

lt 512 Thus

|119877| le radic(8 + 1)6(6 + 1)

8sdot 1408

6sdot 5708= 120596

= 16505374299582118582810249858265119890 + 48

(36)



119901 gt 2 sdot 119860119891119892 (37)








119896 primes 119901119896 = 119901


1= 2 119901

2= 3 etc)


essential


119896 le 119899



119896 le 119860

119891119892


10 = 6 469 693 230





119896 also



11990130 = 31610054640417607788145206291544119890 + 46

lt 120596

11990131 = 4014476939333036189094441199026119890 + 48

gt 120596

(38)





119891


119891119892 Namely in




119891and 119873

119892appear 119898





119896 le


1 119901


119908 = gcd(1198860 1198870) where 119886

0= lc(119891(119909)) and 119887

0= lc(119892(119909))

If gcd(119891119901119894

(119909) 119892119901119894



(119909) 119892119901119894

(119909)) ge


119901119894

(119909) 119892119901119894

(119909)) = 1 for all 119894 = 1 119896 + 1 then119891119901119894

(119909) and 119892119901119894









2(119909) = 119909

8+1199096+1199094+1199093+1 1198922(119909) = 119909

6+1199094+119909+1 which


2+ 119909 + 1 = 1

And for 119901 = 3 we get 1198913(119909) = 119909

8+ 1199096+ 21199092+ 2119909 + 1

1198923(119909) = 2119909



119891=

2radic6119873119892= radic2 and119860




119901(119909) 119892119901(119909)) = 1 for 119901 = 2 3 5 7







119891 119873119892by Corollary 7

119873119891119892

by Corollary 9 and 119860119891119892


119896 le 119860


1199011 119901


119891119892 We do not know 119889(119909)



1 119901




119901119894




119901119894


119901119894


preimage 119896(119909) of 119905 sdot 119890119901119894



119901119894




numbers 1199011 119901


119894∤ 119908


2 sdot 119873119891119892


119901119894


1 119901

119896+1


(119909) ge


1 119901



119901119894


119891119892and


119901(119909) = 119872 then


119901(119909) = 119872 then we have





0119909119896+ sdot sdot sdot + 119888



119899and 119892(119909) = 119887

0119909119898+sdot sdot sdot+119887

119898

then |119888119894| le 2119896|11988801198860|119891(119909) and |119888

119894| le 2119896|11988801198870|119892(119909) Since

|11988801198860| is bounded by |gcd(119886

0 1198870)1198860| and |119888



10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot gcd (119886

0 1198870) sdotmin

1003817100381710038171003817119891 (119909)1003817100381710038171003817

100381610038161003816100381611988601003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(39)




119894| le 119872

119902119891119892 where

119872119902119891119892

= 2119904(119902119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(40)

119872119902119891119892







if 119901 gt 2 sdot 119872119902119891119892


min119891(119909)|1198860| 119892(119909)|119887





2(119909) = gcd(119891

2(119909) 1198922(119909)) =

1199092+ 119909 + 1 Then 119904(2 119891 119892) = deg 119890

2(119909) = 2 and

1198722119891119892

= 22sdot 1 sdotmin

radic113

1radic570

3 lt 3184 (41)

Take 119901 = 67 gt 2 sdot 1198722119891119892




deg 1198903(119909) = 0 and 119872







1 119902



(119909) 1198901199022

(119909)



119902119894

(119909) gt 0



119904 (119891 119892) = min deg 1198901199021(119909) deg 119890

119902119896+1

(119909) gt 0 (42)





119894of gcd(119891(119909) 119892(119909)) we get that |119888

119894| le 119872

119891119892

where

119872119891119892

= 2119904(119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(43)



119901(119909) in Z

119901[119909] If deg 119890

119901(119909) gt 119904(119891 119892) we










119894 Butwe





Competing Interests


Acknowledgments


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra













120593119901(119891(119909)) and 119892

119901(119909) = 120593



Example 5 If 119891(119909) = 71199092+ 22 and 119892(119909) = 49119909

3+ 154119909


119901(119909) = 0 So



120601119899(119909) = prod

119896=1119899(119896119899)=1

(119909 minus 1198902119894120587119896119899

) (17)



radicsum119899

119894=01198862119894


119886119899and ℎ(119909) = 119888

0119909119896+ sdot sdot sdot + 119888



119899

sum119894=0

10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot10038161003816100381610038161003816100381610038161003816

1198880

1198860

10038161003816100381610038161003816100381610038161003816sdot1003817100381710038171003817119891 (119909)

1003817100381710038171003817 (18)



10038161003816100381610038161198881198941003816100381610038161003816 le 119873119891= 2119899minus1 1003817100381710038171003817119891 (119909)

1003817100381710038171003817 (19)




119894| | 119894 =

0 119899 le max|119886119894| | 119894 = 0 119899 le 119891(119909)






0119909119898+ sdot sdot sdot + 119887



10038161003816100381610038161198881198941003816100381610038161003816 le 119873119891119892

= 2min119899119898

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(20)




0and 1198870


119891119892and take








119901(119909)



119891119892


119901(119909) (if they






3+ 3119909 + 10 we



119901 gt 2 sdot 119873119891119892 (21)


119901(119888119894) =


119901(119888119894) = 119888119894+ 119901 gt 1199012




119901(119909) = 119887

0119901119909119896+ sdot sdot sdot + 119887

119896119901isin Z119901[119909]


119901(119909)

(03) Set 119894 = 0(04) While 119894 le 119896

(05) if 119887119894119901

lt 1199012

(06) set 119887119894= 119887119894119901

(07) else(08) set 119887

119894= 119887119894119901

minus 119901(10) set 119894 = 119894 + 1(11) Output 119889(119909) = 119887

0119909119896+ sdot sdot sdot + 119887

119896







119901(119909) we may not obtain 119889(119909)


2 sdot 119873119891119892

and for which

deg 119889 (119909) = deg 119890119901(119909) (22)




119901(119909) is gcd of119891

119901(119909) 119892119901(119909)








Example 11 For 119891(119909) = 1199092+ 4119909 + 3 and 119892(119909) = 119909

2+ 2119909 + 1


119890119901(119909) = gcd (119891

119901(119909) 119892

119901(119909)) = 2119909 + 2 isin Z

119901 [119909] (23)


119901(119909 + 1) = 2119909 + 2








119901 their fraction 119908119888

0


lc (119905 sdot 119890119901(119909)) = 119908 (24)












bound 119901 gt 2 sdot 119873119891119892





119891119892


119901is not



119889(119909) = gcd(119891(119909) 119892(119909)) = 1 Taking 119901 = 2 we get 1198912(119909) =

1199092+ 1 and 119892

2(119909) = 119909 + 1 In Z

2[119909] we have 119891

2(119909) = 119909

2+ 1 =

1199092+12= (119909+1)(119909+1) thus 119890

2(119909) = gcd(119891

2(119909) 1198922(119909)) = 119909+1


under 1205932






0




0= lc(119891(119909)) and 119887




119891119892by (20)



119901(119909) 119892

119901(119909) isin Z

119901[119909]

(10) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(11) If119863 le deg 119890119901(119909)


119901(119909)) = 119908



119901(119909)





0= lc(119891(119909)) and 119887

0= lc(119892(119909))





119891119892

(10) Set 119894 = 1(11) While 119894 = 119896 + 1



119901(119909) 119892

119901(119909) isin Z

119901[119909]

(14) calculate 119890119901= gcd(119891

119901(119909) 119892


119901[119909]

(15) if deg 119890119902119894= 0







0= lc(119891(119909)) and 119887






119902(119909) 119892

119902(119909) isin Z

119902[119909]

(09) Calculate 119890119902(119909) = gcd(119891

119902(119909) 119892


119902[119909]

(10) Set 119904(119902 119891 119892) = deg 119890119902(119909)

(11) Calculate119872119902119891119892


119902119891119892


119901(119909) 119892

119901(119909) isin Z

119901[119909]

(14) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(15) If119863 le deg 119890119901(119909)


119901(119909)) = 119908



119901(119909)








deg 119889 (119909) = deg 119889119901(119909) = deg gcd (119891

119901(119909) 119892

119901(119909))

= deg 119890119901(119909)

(26)

Proof Since 1198880



119901(119909) le

deg gcd(119891119901(119909) 119892119901(119909))


119891119901(119909)119889119901(119909) and 119892

119901(119909)119889119901(119909) in Z


izations of 119891119901(119909) and 119892

119901(119909) in UFD Z


deduce that

119890119901 (119909) asymp gcd (119891

119901 (119909) 119892119901 (119909))

asymp 119889119901(119909) sdot gcd(

119891119901(119909)

119889119901(119909)

119892119901(119909)

119889119901(119909)

)

(27)


119901(119909) And the



deg gcd(119891119901(119909)

119889119901 (119909)

119892119901(119909)

119889119901 (119909)

) gt 0 (28)

that is when 119891119901(119909)119889119901(119909) and 119892

119901(119909)119889119901(119909) are not


119901(119909)




120593119901 119872119898+119899

(Z) 997888rarr 119872119898+119899

(Z119901) (29)




) = 119878119891119901119889119901119892119901119889119901


119877119901= 120593119901 (119877) = 120593

119901(res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909)))

= res(119891119901(119909)

119889119901(119909)

119892119901(119909)

119889119901(119909)

)

(30)






119891119892


0= lc(119891(119909)) and 119887







119902119894(119909) 119892

119902119894(119909) isin Z

119902119894[119909]

(13) calculate 119890119902119894= gcd(119891

119902119894(119909) 119892


119901[119909]

(14) if deg 119890119902119894le 119904(119891 119892)

(15) set 119904(119891 119892) = deg 119890119902119894

(16) if deg 119890119902119894= 0





119901(119909) 119892

119901(119909) isin Z

119901[119909]

(23) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(24) If deg 119890119901= 119904(119891 119892)














119891119892

Then 119901 ∤ 119908 = gcd(1198860 1198870) since 119886

0 1198870le 119873119891119892 Calculate

119890119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) in Z





119901 we would

get deg ℎ119901(119909) gt deg 119889

119901(119909) = deg 119889(119909) = deg 119890

119901(119909) which










119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) check if deg 119890

119901(119909) is











and 119892(119909) = radic570

119873119891119892

= 2min86


1radic570

3

lt 512

(31)



(119909) 1198921031






119860119891119892

= radic(119899 + 1)119898(119898 + 1)

119899sdot 119873119898

119891119873119899

119892


119898(119898 + 1)

119899

sdot1003817100381710038171003817119891 (119909)

1003817100381710038171003817119898 1003817100381710038171003817119892 (119909)

1003817100381710038171003817119899

(32)


10038161003816100381610038161003816100381610038161003816res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909))10038161003816100381610038161003816100381610038161003816=10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le 119860119891119892 (33)


119891= 2119899minus1

119891(119909)

and 119873119892= 2119898minus1


10038171003817100381710038171003817100381710038171003817

119891 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119899 + 1)1198732

119891

10038171003817100381710038171003817100381710038171003817

119892 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119898 + 1)1198732

119892

(34)


119891119889119892119889 we get that

|119877| =10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le (radic(119899 + 1)119873

119891)119898

sdot (radic(119898 + 1)119873119892)119899

(35)



119891lt 1408 119873

119892lt 768

and119873119891119892

lt 512 Thus

|119877| le radic(8 + 1)6(6 + 1)

8sdot 1408

6sdot 5708= 120596

= 16505374299582118582810249858265119890 + 48

(36)



119901 gt 2 sdot 119860119891119892 (37)








119896 primes 119901119896 = 119901


1= 2 119901

2= 3 etc)


essential


119896 le 119899



119896 le 119860

119891119892


10 = 6 469 693 230





119896 also



11990130 = 31610054640417607788145206291544119890 + 46

lt 120596

11990131 = 4014476939333036189094441199026119890 + 48

gt 120596

(38)





119891


119891119892 Namely in




119891and 119873

119892appear 119898





119896 le


1 119901


119908 = gcd(1198860 1198870) where 119886

0= lc(119891(119909)) and 119887

0= lc(119892(119909))

If gcd(119891119901119894

(119909) 119892119901119894



(119909) 119892119901119894

(119909)) ge


119901119894

(119909) 119892119901119894

(119909)) = 1 for all 119894 = 1 119896 + 1 then119891119901119894

(119909) and 119892119901119894









2(119909) = 119909

8+1199096+1199094+1199093+1 1198922(119909) = 119909

6+1199094+119909+1 which


2+ 119909 + 1 = 1

And for 119901 = 3 we get 1198913(119909) = 119909

8+ 1199096+ 21199092+ 2119909 + 1

1198923(119909) = 2119909



119891=

2radic6119873119892= radic2 and119860




119901(119909) 119892119901(119909)) = 1 for 119901 = 2 3 5 7







119891 119873119892by Corollary 7

119873119891119892

by Corollary 9 and 119860119891119892


119896 le 119860


1199011 119901


119891119892 We do not know 119889(119909)



1 119901




119901119894




119901119894


119901119894


preimage 119896(119909) of 119905 sdot 119890119901119894



119901119894




numbers 1199011 119901


119894∤ 119908


2 sdot 119873119891119892


119901119894


1 119901

119896+1


(119909) ge


1 119901



119901119894


119891119892and


119901(119909) = 119872 then


119901(119909) = 119872 then we have





0119909119896+ sdot sdot sdot + 119888



119899and 119892(119909) = 119887

0119909119898+sdot sdot sdot+119887

119898

then |119888119894| le 2119896|11988801198860|119891(119909) and |119888

119894| le 2119896|11988801198870|119892(119909) Since

|11988801198860| is bounded by |gcd(119886

0 1198870)1198860| and |119888



10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot gcd (119886

0 1198870) sdotmin

1003817100381710038171003817119891 (119909)1003817100381710038171003817

100381610038161003816100381611988601003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(39)




119894| le 119872

119902119891119892 where

119872119902119891119892

= 2119904(119902119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(40)

119872119902119891119892







if 119901 gt 2 sdot 119872119902119891119892


min119891(119909)|1198860| 119892(119909)|119887





2(119909) = gcd(119891

2(119909) 1198922(119909)) =

1199092+ 119909 + 1 Then 119904(2 119891 119892) = deg 119890

2(119909) = 2 and

1198722119891119892

= 22sdot 1 sdotmin

radic113

1radic570

3 lt 3184 (41)

Take 119901 = 67 gt 2 sdot 1198722119891119892




deg 1198903(119909) = 0 and 119872







1 119902



(119909) 1198901199022

(119909)



119902119894

(119909) gt 0



119904 (119891 119892) = min deg 1198901199021(119909) deg 119890

119902119896+1

(119909) gt 0 (42)





119894of gcd(119891(119909) 119892(119909)) we get that |119888

119894| le 119872

119891119892

where

119872119891119892

= 2119904(119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(43)



119901(119909) in Z

119901[119909] If deg 119890

119901(119909) gt 119904(119891 119892) we










119894 Butwe





Competing Interests


Acknowledgments


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119901(119909) = 119887

0119901119909119896+ sdot sdot sdot + 119887

119896119901isin Z119901[119909]


119901(119909)

(03) Set 119894 = 0(04) While 119894 le 119896

(05) if 119887119894119901

lt 1199012

(06) set 119887119894= 119887119894119901

(07) else(08) set 119887

119894= 119887119894119901

minus 119901(10) set 119894 = 119894 + 1(11) Output 119889(119909) = 119887

0119909119896+ sdot sdot sdot + 119887

119896







119901(119909) we may not obtain 119889(119909)


2 sdot 119873119891119892

and for which

deg 119889 (119909) = deg 119890119901(119909) (22)




119901(119909) is gcd of119891

119901(119909) 119892119901(119909)








Example 11 For 119891(119909) = 1199092+ 4119909 + 3 and 119892(119909) = 119909

2+ 2119909 + 1


119890119901(119909) = gcd (119891

119901(119909) 119892

119901(119909)) = 2119909 + 2 isin Z

119901 [119909] (23)


119901(119909 + 1) = 2119909 + 2








119901 their fraction 119908119888

0


lc (119905 sdot 119890119901(119909)) = 119908 (24)












bound 119901 gt 2 sdot 119873119891119892





119891119892


119901is not



119889(119909) = gcd(119891(119909) 119892(119909)) = 1 Taking 119901 = 2 we get 1198912(119909) =

1199092+ 1 and 119892

2(119909) = 119909 + 1 In Z

2[119909] we have 119891

2(119909) = 119909

2+ 1 =

1199092+12= (119909+1)(119909+1) thus 119890

2(119909) = gcd(119891

2(119909) 1198922(119909)) = 119909+1


under 1205932






0




0= lc(119891(119909)) and 119887




119891119892by (20)



119901(119909) 119892

119901(119909) isin Z

119901[119909]

(10) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(11) If119863 le deg 119890119901(119909)


119901(119909)) = 119908



119901(119909)





0= lc(119891(119909)) and 119887

0= lc(119892(119909))





119891119892

(10) Set 119894 = 1(11) While 119894 = 119896 + 1



119901(119909) 119892

119901(119909) isin Z

119901[119909]

(14) calculate 119890119901= gcd(119891

119901(119909) 119892


119901[119909]

(15) if deg 119890119902119894= 0







0= lc(119891(119909)) and 119887






119902(119909) 119892

119902(119909) isin Z

119902[119909]

(09) Calculate 119890119902(119909) = gcd(119891

119902(119909) 119892


119902[119909]

(10) Set 119904(119902 119891 119892) = deg 119890119902(119909)

(11) Calculate119872119902119891119892


119902119891119892


119901(119909) 119892

119901(119909) isin Z

119901[119909]

(14) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(15) If119863 le deg 119890119901(119909)


119901(119909)) = 119908



119901(119909)








deg 119889 (119909) = deg 119889119901(119909) = deg gcd (119891

119901(119909) 119892

119901(119909))

= deg 119890119901(119909)

(26)

Proof Since 1198880



119901(119909) le

deg gcd(119891119901(119909) 119892119901(119909))


119891119901(119909)119889119901(119909) and 119892

119901(119909)119889119901(119909) in Z


izations of 119891119901(119909) and 119892

119901(119909) in UFD Z


deduce that

119890119901 (119909) asymp gcd (119891

119901 (119909) 119892119901 (119909))

asymp 119889119901(119909) sdot gcd(

119891119901(119909)

119889119901(119909)

119892119901(119909)

119889119901(119909)

)

(27)


119901(119909) And the



deg gcd(119891119901(119909)

119889119901 (119909)

119892119901(119909)

119889119901 (119909)

) gt 0 (28)

that is when 119891119901(119909)119889119901(119909) and 119892

119901(119909)119889119901(119909) are not


119901(119909)




120593119901 119872119898+119899

(Z) 997888rarr 119872119898+119899

(Z119901) (29)




) = 119878119891119901119889119901119892119901119889119901


119877119901= 120593119901 (119877) = 120593

119901(res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909)))

= res(119891119901(119909)

119889119901(119909)

119892119901(119909)

119889119901(119909)

)

(30)






119891119892


0= lc(119891(119909)) and 119887







119902119894(119909) 119892

119902119894(119909) isin Z

119902119894[119909]

(13) calculate 119890119902119894= gcd(119891

119902119894(119909) 119892


119901[119909]

(14) if deg 119890119902119894le 119904(119891 119892)

(15) set 119904(119891 119892) = deg 119890119902119894

(16) if deg 119890119902119894= 0





119901(119909) 119892

119901(119909) isin Z

119901[119909]

(23) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(24) If deg 119890119901= 119904(119891 119892)














119891119892

Then 119901 ∤ 119908 = gcd(1198860 1198870) since 119886

0 1198870le 119873119891119892 Calculate

119890119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) in Z





119901 we would

get deg ℎ119901(119909) gt deg 119889

119901(119909) = deg 119889(119909) = deg 119890

119901(119909) which










119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) check if deg 119890

119901(119909) is











and 119892(119909) = radic570

119873119891119892

= 2min86


1radic570

3

lt 512

(31)



(119909) 1198921031






119860119891119892

= radic(119899 + 1)119898(119898 + 1)

119899sdot 119873119898

119891119873119899

119892


119898(119898 + 1)

119899

sdot1003817100381710038171003817119891 (119909)

1003817100381710038171003817119898 1003817100381710038171003817119892 (119909)

1003817100381710038171003817119899

(32)


10038161003816100381610038161003816100381610038161003816res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909))10038161003816100381610038161003816100381610038161003816=10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le 119860119891119892 (33)


119891= 2119899minus1

119891(119909)

and 119873119892= 2119898minus1


10038171003817100381710038171003817100381710038171003817

119891 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119899 + 1)1198732

119891

10038171003817100381710038171003817100381710038171003817

119892 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119898 + 1)1198732

119892

(34)


119891119889119892119889 we get that

|119877| =10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le (radic(119899 + 1)119873

119891)119898

sdot (radic(119898 + 1)119873119892)119899

(35)



119891lt 1408 119873

119892lt 768

and119873119891119892

lt 512 Thus

|119877| le radic(8 + 1)6(6 + 1)

8sdot 1408

6sdot 5708= 120596

= 16505374299582118582810249858265119890 + 48

(36)



119901 gt 2 sdot 119860119891119892 (37)








119896 primes 119901119896 = 119901


1= 2 119901

2= 3 etc)


essential


119896 le 119899



119896 le 119860

119891119892


10 = 6 469 693 230





119896 also



11990130 = 31610054640417607788145206291544119890 + 46

lt 120596

11990131 = 4014476939333036189094441199026119890 + 48

gt 120596

(38)





119891


119891119892 Namely in




119891and 119873

119892appear 119898





119896 le


1 119901


119908 = gcd(1198860 1198870) where 119886

0= lc(119891(119909)) and 119887

0= lc(119892(119909))

If gcd(119891119901119894

(119909) 119892119901119894



(119909) 119892119901119894

(119909)) ge


119901119894

(119909) 119892119901119894

(119909)) = 1 for all 119894 = 1 119896 + 1 then119891119901119894

(119909) and 119892119901119894









2(119909) = 119909

8+1199096+1199094+1199093+1 1198922(119909) = 119909

6+1199094+119909+1 which


2+ 119909 + 1 = 1

And for 119901 = 3 we get 1198913(119909) = 119909

8+ 1199096+ 21199092+ 2119909 + 1

1198923(119909) = 2119909



119891=

2radic6119873119892= radic2 and119860




119901(119909) 119892119901(119909)) = 1 for 119901 = 2 3 5 7







119891 119873119892by Corollary 7

119873119891119892

by Corollary 9 and 119860119891119892


119896 le 119860


1199011 119901


119891119892 We do not know 119889(119909)



1 119901




119901119894




119901119894


119901119894


preimage 119896(119909) of 119905 sdot 119890119901119894



119901119894




numbers 1199011 119901


119894∤ 119908


2 sdot 119873119891119892


119901119894


1 119901

119896+1


(119909) ge


1 119901



119901119894


119891119892and


119901(119909) = 119872 then


119901(119909) = 119872 then we have





0119909119896+ sdot sdot sdot + 119888



119899and 119892(119909) = 119887

0119909119898+sdot sdot sdot+119887

119898

then |119888119894| le 2119896|11988801198860|119891(119909) and |119888

119894| le 2119896|11988801198870|119892(119909) Since

|11988801198860| is bounded by |gcd(119886

0 1198870)1198860| and |119888



10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot gcd (119886

0 1198870) sdotmin

1003817100381710038171003817119891 (119909)1003817100381710038171003817

100381610038161003816100381611988601003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(39)




119894| le 119872

119902119891119892 where

119872119902119891119892

= 2119904(119902119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(40)

119872119902119891119892







if 119901 gt 2 sdot 119872119902119891119892


min119891(119909)|1198860| 119892(119909)|119887





2(119909) = gcd(119891

2(119909) 1198922(119909)) =

1199092+ 119909 + 1 Then 119904(2 119891 119892) = deg 119890

2(119909) = 2 and

1198722119891119892

= 22sdot 1 sdotmin

radic113

1radic570

3 lt 3184 (41)

Take 119901 = 67 gt 2 sdot 1198722119891119892




deg 1198903(119909) = 0 and 119872







1 119902



(119909) 1198901199022

(119909)



119902119894

(119909) gt 0



119904 (119891 119892) = min deg 1198901199021(119909) deg 119890

119902119896+1

(119909) gt 0 (42)





119894of gcd(119891(119909) 119892(119909)) we get that |119888

119894| le 119872

119891119892

where

119872119891119892

= 2119904(119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(43)



119901(119909) in Z

119901[119909] If deg 119890

119901(119909) gt 119904(119891 119892) we










119894 Butwe





Competing Interests


Acknowledgments


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra












0= lc(119891(119909)) and 119887




119891119892by (20)



119901(119909) 119892

119901(119909) isin Z

119901[119909]

(10) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(11) If119863 le deg 119890119901(119909)


119901(119909)) = 119908



119901(119909)





0= lc(119891(119909)) and 119887

0= lc(119892(119909))





119891119892

(10) Set 119894 = 1(11) While 119894 = 119896 + 1



119901(119909) 119892

119901(119909) isin Z

119901[119909]

(14) calculate 119890119901= gcd(119891

119901(119909) 119892


119901[119909]

(15) if deg 119890119902119894= 0







0= lc(119891(119909)) and 119887






119902(119909) 119892

119902(119909) isin Z

119902[119909]

(09) Calculate 119890119902(119909) = gcd(119891

119902(119909) 119892


119902[119909]

(10) Set 119904(119902 119891 119892) = deg 119890119902(119909)

(11) Calculate119872119902119891119892


119902119891119892


119901(119909) 119892

119901(119909) isin Z

119901[119909]

(14) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(15) If119863 le deg 119890119901(119909)


119901(119909)) = 119908



119901(119909)








deg 119889 (119909) = deg 119889119901(119909) = deg gcd (119891

119901(119909) 119892

119901(119909))

= deg 119890119901(119909)

(26)

Proof Since 1198880



119901(119909) le

deg gcd(119891119901(119909) 119892119901(119909))


119891119901(119909)119889119901(119909) and 119892

119901(119909)119889119901(119909) in Z


izations of 119891119901(119909) and 119892

119901(119909) in UFD Z


deduce that

119890119901 (119909) asymp gcd (119891

119901 (119909) 119892119901 (119909))

asymp 119889119901(119909) sdot gcd(

119891119901(119909)

119889119901(119909)

119892119901(119909)

119889119901(119909)

)

(27)


119901(119909) And the



deg gcd(119891119901(119909)

119889119901 (119909)

119892119901(119909)

119889119901 (119909)

) gt 0 (28)

that is when 119891119901(119909)119889119901(119909) and 119892

119901(119909)119889119901(119909) are not


119901(119909)




120593119901 119872119898+119899

(Z) 997888rarr 119872119898+119899

(Z119901) (29)




) = 119878119891119901119889119901119892119901119889119901


119877119901= 120593119901 (119877) = 120593

119901(res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909)))

= res(119891119901(119909)

119889119901(119909)

119892119901(119909)

119889119901(119909)

)

(30)






119891119892


0= lc(119891(119909)) and 119887







119902119894(119909) 119892

119902119894(119909) isin Z

119902119894[119909]

(13) calculate 119890119902119894= gcd(119891

119902119894(119909) 119892


119901[119909]

(14) if deg 119890119902119894le 119904(119891 119892)

(15) set 119904(119891 119892) = deg 119890119902119894

(16) if deg 119890119902119894= 0





119901(119909) 119892

119901(119909) isin Z

119901[119909]

(23) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(24) If deg 119890119901= 119904(119891 119892)














119891119892

Then 119901 ∤ 119908 = gcd(1198860 1198870) since 119886

0 1198870le 119873119891119892 Calculate

119890119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) in Z





119901 we would

get deg ℎ119901(119909) gt deg 119889

119901(119909) = deg 119889(119909) = deg 119890

119901(119909) which










119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) check if deg 119890

119901(119909) is











and 119892(119909) = radic570

119873119891119892

= 2min86


1radic570

3

lt 512

(31)



(119909) 1198921031






119860119891119892

= radic(119899 + 1)119898(119898 + 1)

119899sdot 119873119898

119891119873119899

119892


119898(119898 + 1)

119899

sdot1003817100381710038171003817119891 (119909)

1003817100381710038171003817119898 1003817100381710038171003817119892 (119909)

1003817100381710038171003817119899

(32)


10038161003816100381610038161003816100381610038161003816res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909))10038161003816100381610038161003816100381610038161003816=10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le 119860119891119892 (33)


119891= 2119899minus1

119891(119909)

and 119873119892= 2119898minus1


10038171003817100381710038171003817100381710038171003817

119891 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119899 + 1)1198732

119891

10038171003817100381710038171003817100381710038171003817

119892 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119898 + 1)1198732

119892

(34)


119891119889119892119889 we get that

|119877| =10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le (radic(119899 + 1)119873

119891)119898

sdot (radic(119898 + 1)119873119892)119899

(35)



119891lt 1408 119873

119892lt 768

and119873119891119892

lt 512 Thus

|119877| le radic(8 + 1)6(6 + 1)

8sdot 1408

6sdot 5708= 120596

= 16505374299582118582810249858265119890 + 48

(36)



119901 gt 2 sdot 119860119891119892 (37)








119896 primes 119901119896 = 119901


1= 2 119901

2= 3 etc)


essential


119896 le 119899



119896 le 119860

119891119892


10 = 6 469 693 230





119896 also



11990130 = 31610054640417607788145206291544119890 + 46

lt 120596

11990131 = 4014476939333036189094441199026119890 + 48

gt 120596

(38)





119891


119891119892 Namely in




119891and 119873

119892appear 119898





119896 le


1 119901


119908 = gcd(1198860 1198870) where 119886

0= lc(119891(119909)) and 119887

0= lc(119892(119909))

If gcd(119891119901119894

(119909) 119892119901119894



(119909) 119892119901119894

(119909)) ge


119901119894

(119909) 119892119901119894

(119909)) = 1 for all 119894 = 1 119896 + 1 then119891119901119894

(119909) and 119892119901119894









2(119909) = 119909

8+1199096+1199094+1199093+1 1198922(119909) = 119909

6+1199094+119909+1 which


2+ 119909 + 1 = 1

And for 119901 = 3 we get 1198913(119909) = 119909

8+ 1199096+ 21199092+ 2119909 + 1

1198923(119909) = 2119909



119891=

2radic6119873119892= radic2 and119860




119901(119909) 119892119901(119909)) = 1 for 119901 = 2 3 5 7







119891 119873119892by Corollary 7

119873119891119892

by Corollary 9 and 119860119891119892


119896 le 119860


1199011 119901


119891119892 We do not know 119889(119909)



1 119901




119901119894




119901119894


119901119894


preimage 119896(119909) of 119905 sdot 119890119901119894



119901119894




numbers 1199011 119901


119894∤ 119908


2 sdot 119873119891119892


119901119894


1 119901

119896+1


(119909) ge


1 119901



119901119894


119891119892and


119901(119909) = 119872 then


119901(119909) = 119872 then we have





0119909119896+ sdot sdot sdot + 119888



119899and 119892(119909) = 119887

0119909119898+sdot sdot sdot+119887

119898

then |119888119894| le 2119896|11988801198860|119891(119909) and |119888

119894| le 2119896|11988801198870|119892(119909) Since

|11988801198860| is bounded by |gcd(119886

0 1198870)1198860| and |119888



10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot gcd (119886

0 1198870) sdotmin

1003817100381710038171003817119891 (119909)1003817100381710038171003817

100381610038161003816100381611988601003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(39)




119894| le 119872

119902119891119892 where

119872119902119891119892

= 2119904(119902119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(40)

119872119902119891119892







if 119901 gt 2 sdot 119872119902119891119892


min119891(119909)|1198860| 119892(119909)|119887





2(119909) = gcd(119891

2(119909) 1198922(119909)) =

1199092+ 119909 + 1 Then 119904(2 119891 119892) = deg 119890

2(119909) = 2 and

1198722119891119892

= 22sdot 1 sdotmin

radic113

1radic570

3 lt 3184 (41)

Take 119901 = 67 gt 2 sdot 1198722119891119892




deg 1198903(119909) = 0 and 119872







1 119902



(119909) 1198901199022

(119909)



119902119894

(119909) gt 0



119904 (119891 119892) = min deg 1198901199021(119909) deg 119890

119902119896+1

(119909) gt 0 (42)





119894of gcd(119891(119909) 119892(119909)) we get that |119888

119894| le 119872

119891119892

where

119872119891119892

= 2119904(119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(43)



119901(119909) in Z

119901[119909] If deg 119890

119901(119909) gt 119904(119891 119892) we










119894 Butwe





Competing Interests


Acknowledgments


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra












0= lc(119891(119909)) and 119887






119902(119909) 119892

119902(119909) isin Z

119902[119909]

(09) Calculate 119890119902(119909) = gcd(119891

119902(119909) 119892


119902[119909]

(10) Set 119904(119902 119891 119892) = deg 119890119902(119909)

(11) Calculate119872119902119891119892


119902119891119892


119901(119909) 119892

119901(119909) isin Z

119901[119909]

(14) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(15) If119863 le deg 119890119901(119909)


119901(119909)) = 119908



119901(119909)








deg 119889 (119909) = deg 119889119901(119909) = deg gcd (119891

119901(119909) 119892

119901(119909))

= deg 119890119901(119909)

(26)

Proof Since 1198880



119901(119909) le

deg gcd(119891119901(119909) 119892119901(119909))


119891119901(119909)119889119901(119909) and 119892

119901(119909)119889119901(119909) in Z


izations of 119891119901(119909) and 119892

119901(119909) in UFD Z


deduce that

119890119901 (119909) asymp gcd (119891

119901 (119909) 119892119901 (119909))

asymp 119889119901(119909) sdot gcd(

119891119901(119909)

119889119901(119909)

119892119901(119909)

119889119901(119909)

)

(27)


119901(119909) And the



deg gcd(119891119901(119909)

119889119901 (119909)

119892119901(119909)

119889119901 (119909)

) gt 0 (28)

that is when 119891119901(119909)119889119901(119909) and 119892

119901(119909)119889119901(119909) are not


119901(119909)




120593119901 119872119898+119899

(Z) 997888rarr 119872119898+119899

(Z119901) (29)




) = 119878119891119901119889119901119892119901119889119901


119877119901= 120593119901 (119877) = 120593

119901(res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909)))

= res(119891119901(119909)

119889119901(119909)

119892119901(119909)

119889119901(119909)

)

(30)






119891119892


0= lc(119891(119909)) and 119887







119902119894(119909) 119892

119902119894(119909) isin Z

119902119894[119909]

(13) calculate 119890119902119894= gcd(119891

119902119894(119909) 119892


119901[119909]

(14) if deg 119890119902119894le 119904(119891 119892)

(15) set 119904(119891 119892) = deg 119890119902119894

(16) if deg 119890119902119894= 0





119901(119909) 119892

119901(119909) isin Z

119901[119909]

(23) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(24) If deg 119890119901= 119904(119891 119892)














119891119892

Then 119901 ∤ 119908 = gcd(1198860 1198870) since 119886

0 1198870le 119873119891119892 Calculate

119890119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) in Z





119901 we would

get deg ℎ119901(119909) gt deg 119889

119901(119909) = deg 119889(119909) = deg 119890

119901(119909) which










119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) check if deg 119890

119901(119909) is











and 119892(119909) = radic570

119873119891119892

= 2min86


1radic570

3

lt 512

(31)



(119909) 1198921031






119860119891119892

= radic(119899 + 1)119898(119898 + 1)

119899sdot 119873119898

119891119873119899

119892


119898(119898 + 1)

119899

sdot1003817100381710038171003817119891 (119909)

1003817100381710038171003817119898 1003817100381710038171003817119892 (119909)

1003817100381710038171003817119899

(32)


10038161003816100381610038161003816100381610038161003816res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909))10038161003816100381610038161003816100381610038161003816=10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le 119860119891119892 (33)


119891= 2119899minus1

119891(119909)

and 119873119892= 2119898minus1


10038171003817100381710038171003817100381710038171003817

119891 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119899 + 1)1198732

119891

10038171003817100381710038171003817100381710038171003817

119892 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119898 + 1)1198732

119892

(34)


119891119889119892119889 we get that

|119877| =10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le (radic(119899 + 1)119873

119891)119898

sdot (radic(119898 + 1)119873119892)119899

(35)



119891lt 1408 119873

119892lt 768

and119873119891119892

lt 512 Thus

|119877| le radic(8 + 1)6(6 + 1)

8sdot 1408

6sdot 5708= 120596

= 16505374299582118582810249858265119890 + 48

(36)



119901 gt 2 sdot 119860119891119892 (37)








119896 primes 119901119896 = 119901


1= 2 119901

2= 3 etc)


essential


119896 le 119899



119896 le 119860

119891119892


10 = 6 469 693 230





119896 also



11990130 = 31610054640417607788145206291544119890 + 46

lt 120596

11990131 = 4014476939333036189094441199026119890 + 48

gt 120596

(38)





119891


119891119892 Namely in




119891and 119873

119892appear 119898





119896 le


1 119901


119908 = gcd(1198860 1198870) where 119886

0= lc(119891(119909)) and 119887

0= lc(119892(119909))

If gcd(119891119901119894

(119909) 119892119901119894



(119909) 119892119901119894

(119909)) ge


119901119894

(119909) 119892119901119894

(119909)) = 1 for all 119894 = 1 119896 + 1 then119891119901119894

(119909) and 119892119901119894









2(119909) = 119909

8+1199096+1199094+1199093+1 1198922(119909) = 119909

6+1199094+119909+1 which


2+ 119909 + 1 = 1

And for 119901 = 3 we get 1198913(119909) = 119909

8+ 1199096+ 21199092+ 2119909 + 1

1198923(119909) = 2119909



119891=

2radic6119873119892= radic2 and119860




119901(119909) 119892119901(119909)) = 1 for 119901 = 2 3 5 7







119891 119873119892by Corollary 7

119873119891119892

by Corollary 9 and 119860119891119892


119896 le 119860


1199011 119901


119891119892 We do not know 119889(119909)



1 119901




119901119894




119901119894


119901119894


preimage 119896(119909) of 119905 sdot 119890119901119894



119901119894




numbers 1199011 119901


119894∤ 119908


2 sdot 119873119891119892


119901119894


1 119901

119896+1


(119909) ge


1 119901



119901119894


119891119892and


119901(119909) = 119872 then


119901(119909) = 119872 then we have





0119909119896+ sdot sdot sdot + 119888



119899and 119892(119909) = 119887

0119909119898+sdot sdot sdot+119887

119898

then |119888119894| le 2119896|11988801198860|119891(119909) and |119888

119894| le 2119896|11988801198870|119892(119909) Since

|11988801198860| is bounded by |gcd(119886

0 1198870)1198860| and |119888



10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot gcd (119886

0 1198870) sdotmin

1003817100381710038171003817119891 (119909)1003817100381710038171003817

100381610038161003816100381611988601003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(39)




119894| le 119872

119902119891119892 where

119872119902119891119892

= 2119904(119902119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(40)

119872119902119891119892







if 119901 gt 2 sdot 119872119902119891119892


min119891(119909)|1198860| 119892(119909)|119887





2(119909) = gcd(119891

2(119909) 1198922(119909)) =

1199092+ 119909 + 1 Then 119904(2 119891 119892) = deg 119890

2(119909) = 2 and

1198722119891119892

= 22sdot 1 sdotmin

radic113

1radic570

3 lt 3184 (41)

Take 119901 = 67 gt 2 sdot 1198722119891119892




deg 1198903(119909) = 0 and 119872







1 119902



(119909) 1198901199022

(119909)



119902119894

(119909) gt 0



119904 (119891 119892) = min deg 1198901199021(119909) deg 119890

119902119896+1

(119909) gt 0 (42)





119894of gcd(119891(119909) 119892(119909)) we get that |119888

119894| le 119872

119891119892

where

119872119891119892

= 2119904(119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(43)



119901(119909) in Z

119901[119909] If deg 119890

119901(119909) gt 119904(119891 119892) we










119894 Butwe





Competing Interests


Acknowledgments


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra














119891119892


0= lc(119891(119909)) and 119887







119902119894(119909) 119892

119902119894(119909) isin Z

119902119894[119909]

(13) calculate 119890119902119894= gcd(119891

119902119894(119909) 119892


119901[119909]

(14) if deg 119890119902119894le 119904(119891 119892)

(15) set 119904(119891 119892) = deg 119890119902119894

(16) if deg 119890119902119894= 0





119901(119909) 119892

119901(119909) isin Z

119901[119909]

(23) Calculate 119890119901(119909) = gcd(119891

119901(119909) 119892


119901[119909]

(24) If deg 119890119901= 119904(119891 119892)














119891119892

Then 119901 ∤ 119908 = gcd(1198860 1198870) since 119886

0 1198870le 119873119891119892 Calculate

119890119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) in Z





119901 we would

get deg ℎ119901(119909) gt deg 119889

119901(119909) = deg 119889(119909) = deg 119890

119901(119909) which










119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) check if deg 119890

119901(119909) is











and 119892(119909) = radic570

119873119891119892

= 2min86


1radic570

3

lt 512

(31)



(119909) 1198921031






119860119891119892

= radic(119899 + 1)119898(119898 + 1)

119899sdot 119873119898

119891119873119899

119892


119898(119898 + 1)

119899

sdot1003817100381710038171003817119891 (119909)

1003817100381710038171003817119898 1003817100381710038171003817119892 (119909)

1003817100381710038171003817119899

(32)


10038161003816100381610038161003816100381610038161003816res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909))10038161003816100381610038161003816100381610038161003816=10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le 119860119891119892 (33)


119891= 2119899minus1

119891(119909)

and 119873119892= 2119898minus1


10038171003817100381710038171003817100381710038171003817

119891 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119899 + 1)1198732

119891

10038171003817100381710038171003817100381710038171003817

119892 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119898 + 1)1198732

119892

(34)


119891119889119892119889 we get that

|119877| =10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le (radic(119899 + 1)119873

119891)119898

sdot (radic(119898 + 1)119873119892)119899

(35)



119891lt 1408 119873

119892lt 768

and119873119891119892

lt 512 Thus

|119877| le radic(8 + 1)6(6 + 1)

8sdot 1408

6sdot 5708= 120596

= 16505374299582118582810249858265119890 + 48

(36)



119901 gt 2 sdot 119860119891119892 (37)








119896 primes 119901119896 = 119901


1= 2 119901

2= 3 etc)


essential


119896 le 119899



119896 le 119860

119891119892


10 = 6 469 693 230





119896 also



11990130 = 31610054640417607788145206291544119890 + 46

lt 120596

11990131 = 4014476939333036189094441199026119890 + 48

gt 120596

(38)





119891


119891119892 Namely in




119891and 119873

119892appear 119898





119896 le


1 119901


119908 = gcd(1198860 1198870) where 119886

0= lc(119891(119909)) and 119887

0= lc(119892(119909))

If gcd(119891119901119894

(119909) 119892119901119894



(119909) 119892119901119894

(119909)) ge


119901119894

(119909) 119892119901119894

(119909)) = 1 for all 119894 = 1 119896 + 1 then119891119901119894

(119909) and 119892119901119894









2(119909) = 119909

8+1199096+1199094+1199093+1 1198922(119909) = 119909

6+1199094+119909+1 which


2+ 119909 + 1 = 1

And for 119901 = 3 we get 1198913(119909) = 119909

8+ 1199096+ 21199092+ 2119909 + 1

1198923(119909) = 2119909



119891=

2radic6119873119892= radic2 and119860




119901(119909) 119892119901(119909)) = 1 for 119901 = 2 3 5 7







119891 119873119892by Corollary 7

119873119891119892

by Corollary 9 and 119860119891119892


119896 le 119860


1199011 119901


119891119892 We do not know 119889(119909)



1 119901




119901119894




119901119894


119901119894


preimage 119896(119909) of 119905 sdot 119890119901119894



119901119894




numbers 1199011 119901


119894∤ 119908


2 sdot 119873119891119892


119901119894


1 119901

119896+1


(119909) ge


1 119901



119901119894


119891119892and


119901(119909) = 119872 then


119901(119909) = 119872 then we have





0119909119896+ sdot sdot sdot + 119888



119899and 119892(119909) = 119887

0119909119898+sdot sdot sdot+119887

119898

then |119888119894| le 2119896|11988801198860|119891(119909) and |119888

119894| le 2119896|11988801198870|119892(119909) Since

|11988801198860| is bounded by |gcd(119886

0 1198870)1198860| and |119888



10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot gcd (119886

0 1198870) sdotmin

1003817100381710038171003817119891 (119909)1003817100381710038171003817

100381610038161003816100381611988601003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(39)




119894| le 119872

119902119891119892 where

119872119902119891119892

= 2119904(119902119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(40)

119872119902119891119892







if 119901 gt 2 sdot 119872119902119891119892


min119891(119909)|1198860| 119892(119909)|119887





2(119909) = gcd(119891

2(119909) 1198922(119909)) =

1199092+ 119909 + 1 Then 119904(2 119891 119892) = deg 119890

2(119909) = 2 and

1198722119891119892

= 22sdot 1 sdotmin

radic113

1radic570

3 lt 3184 (41)

Take 119901 = 67 gt 2 sdot 1198722119891119892




deg 1198903(119909) = 0 and 119872







1 119902



(119909) 1198901199022

(119909)



119902119894

(119909) gt 0



119904 (119891 119892) = min deg 1198901199021(119909) deg 119890

119902119896+1

(119909) gt 0 (42)





119894of gcd(119891(119909) 119892(119909)) we get that |119888

119894| le 119872

119891119892

where

119872119891119892

= 2119904(119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(43)



119901(119909) in Z

119901[119909] If deg 119890

119901(119909) gt 119904(119891 119892) we










119894 Butwe





Competing Interests


Acknowledgments


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119901(119909) = gcd(119891

119901(119909) 119892119901(119909)) check if deg 119890

119901(119909) is











and 119892(119909) = radic570

119873119891119892

= 2min86


1radic570

3

lt 512

(31)



(119909) 1198921031






119860119891119892

= radic(119899 + 1)119898(119898 + 1)

119899sdot 119873119898

119891119873119899

119892


119898(119898 + 1)

119899

sdot1003817100381710038171003817119891 (119909)

1003817100381710038171003817119898 1003817100381710038171003817119892 (119909)

1003817100381710038171003817119899

(32)


10038161003816100381610038161003816100381610038161003816res(

119891 (119909)

119889 (119909)119892 (119909)

119889 (119909))10038161003816100381610038161003816100381610038161003816=10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le 119860119891119892 (33)


119891= 2119899minus1

119891(119909)

and 119873119892= 2119898minus1


10038171003817100381710038171003817100381710038171003817

119891 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119899 + 1)1198732

119891

10038171003817100381710038171003817100381710038171003817

119892 (119909)

119889 (119909)

10038171003817100381710038171003817100381710038171003817le radic(119898 + 1)1198732

119892

(34)


119891119889119892119889 we get that

|119877| =10038161003816100381610038161003816119878119891119889119892119889

10038161003816100381610038161003816le (radic(119899 + 1)119873

119891)119898

sdot (radic(119898 + 1)119873119892)119899

(35)



119891lt 1408 119873

119892lt 768

and119873119891119892

lt 512 Thus

|119877| le radic(8 + 1)6(6 + 1)

8sdot 1408

6sdot 5708= 120596

= 16505374299582118582810249858265119890 + 48

(36)



119901 gt 2 sdot 119860119891119892 (37)








119896 primes 119901119896 = 119901


1= 2 119901

2= 3 etc)


essential


119896 le 119899



119896 le 119860

119891119892


10 = 6 469 693 230





119896 also



11990130 = 31610054640417607788145206291544119890 + 46

lt 120596

11990131 = 4014476939333036189094441199026119890 + 48

gt 120596

(38)





119891


119891119892 Namely in




119891and 119873

119892appear 119898





119896 le


1 119901


119908 = gcd(1198860 1198870) where 119886

0= lc(119891(119909)) and 119887

0= lc(119892(119909))

If gcd(119891119901119894

(119909) 119892119901119894



(119909) 119892119901119894

(119909)) ge


119901119894

(119909) 119892119901119894

(119909)) = 1 for all 119894 = 1 119896 + 1 then119891119901119894

(119909) and 119892119901119894









2(119909) = 119909

8+1199096+1199094+1199093+1 1198922(119909) = 119909

6+1199094+119909+1 which


2+ 119909 + 1 = 1

And for 119901 = 3 we get 1198913(119909) = 119909

8+ 1199096+ 21199092+ 2119909 + 1

1198923(119909) = 2119909



119891=

2radic6119873119892= radic2 and119860




119901(119909) 119892119901(119909)) = 1 for 119901 = 2 3 5 7







119891 119873119892by Corollary 7

119873119891119892

by Corollary 9 and 119860119891119892


119896 le 119860


1199011 119901


119891119892 We do not know 119889(119909)



1 119901




119901119894




119901119894


119901119894


preimage 119896(119909) of 119905 sdot 119890119901119894



119901119894




numbers 1199011 119901


119894∤ 119908


2 sdot 119873119891119892


119901119894


1 119901

119896+1


(119909) ge


1 119901



119901119894


119891119892and


119901(119909) = 119872 then


119901(119909) = 119872 then we have





0119909119896+ sdot sdot sdot + 119888



119899and 119892(119909) = 119887

0119909119898+sdot sdot sdot+119887

119898

then |119888119894| le 2119896|11988801198860|119891(119909) and |119888

119894| le 2119896|11988801198870|119892(119909) Since

|11988801198860| is bounded by |gcd(119886

0 1198870)1198860| and |119888



10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot gcd (119886

0 1198870) sdotmin

1003817100381710038171003817119891 (119909)1003817100381710038171003817

100381610038161003816100381611988601003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(39)




119894| le 119872

119902119891119892 where

119872119902119891119892

= 2119904(119902119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(40)

119872119902119891119892







if 119901 gt 2 sdot 119872119902119891119892


min119891(119909)|1198860| 119892(119909)|119887





2(119909) = gcd(119891

2(119909) 1198922(119909)) =

1199092+ 119909 + 1 Then 119904(2 119891 119892) = deg 119890

2(119909) = 2 and

1198722119891119892

= 22sdot 1 sdotmin

radic113

1radic570

3 lt 3184 (41)

Take 119901 = 67 gt 2 sdot 1198722119891119892




deg 1198903(119909) = 0 and 119872







1 119902



(119909) 1198901199022

(119909)



119902119894

(119909) gt 0



119904 (119891 119892) = min deg 1198901199021(119909) deg 119890

119902119896+1

(119909) gt 0 (42)





119894of gcd(119891(119909) 119892(119909)) we get that |119888

119894| le 119872

119891119892

where

119872119891119892

= 2119904(119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(43)



119901(119909) in Z

119901[119909] If deg 119890

119901(119909) gt 119904(119891 119892) we










119894 Butwe





Competing Interests


Acknowledgments


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra













119896 primes 119901119896 = 119901


1= 2 119901

2= 3 etc)


essential


119896 le 119899



119896 le 119860

119891119892


10 = 6 469 693 230





119896 also



11990130 = 31610054640417607788145206291544119890 + 46

lt 120596

11990131 = 4014476939333036189094441199026119890 + 48

gt 120596

(38)





119891


119891119892 Namely in




119891and 119873

119892appear 119898





119896 le


1 119901


119908 = gcd(1198860 1198870) where 119886

0= lc(119891(119909)) and 119887

0= lc(119892(119909))

If gcd(119891119901119894

(119909) 119892119901119894



(119909) 119892119901119894

(119909)) ge


119901119894

(119909) 119892119901119894

(119909)) = 1 for all 119894 = 1 119896 + 1 then119891119901119894

(119909) and 119892119901119894









2(119909) = 119909

8+1199096+1199094+1199093+1 1198922(119909) = 119909

6+1199094+119909+1 which


2+ 119909 + 1 = 1

And for 119901 = 3 we get 1198913(119909) = 119909

8+ 1199096+ 21199092+ 2119909 + 1

1198923(119909) = 2119909



119891=

2radic6119873119892= radic2 and119860




119901(119909) 119892119901(119909)) = 1 for 119901 = 2 3 5 7







119891 119873119892by Corollary 7

119873119891119892

by Corollary 9 and 119860119891119892


119896 le 119860


1199011 119901


119891119892 We do not know 119889(119909)



1 119901




119901119894




119901119894


119901119894


preimage 119896(119909) of 119905 sdot 119890119901119894



119901119894




numbers 1199011 119901


119894∤ 119908


2 sdot 119873119891119892


119901119894


1 119901

119896+1


(119909) ge


1 119901



119901119894


119891119892and


119901(119909) = 119872 then


119901(119909) = 119872 then we have





0119909119896+ sdot sdot sdot + 119888



119899and 119892(119909) = 119887

0119909119898+sdot sdot sdot+119887

119898

then |119888119894| le 2119896|11988801198860|119891(119909) and |119888

119894| le 2119896|11988801198870|119892(119909) Since

|11988801198860| is bounded by |gcd(119886

0 1198870)1198860| and |119888



10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot gcd (119886

0 1198870) sdotmin

1003817100381710038171003817119891 (119909)1003817100381710038171003817

100381610038161003816100381611988601003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(39)




119894| le 119872

119902119891119892 where

119872119902119891119892

= 2119904(119902119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(40)

119872119902119891119892







if 119901 gt 2 sdot 119872119902119891119892


min119891(119909)|1198860| 119892(119909)|119887





2(119909) = gcd(119891

2(119909) 1198922(119909)) =

1199092+ 119909 + 1 Then 119904(2 119891 119892) = deg 119890

2(119909) = 2 and

1198722119891119892

= 22sdot 1 sdotmin

radic113

1radic570

3 lt 3184 (41)

Take 119901 = 67 gt 2 sdot 1198722119891119892




deg 1198903(119909) = 0 and 119872







1 119902



(119909) 1198901199022

(119909)



119902119894

(119909) gt 0



119904 (119891 119892) = min deg 1198901199021(119909) deg 119890

119902119896+1

(119909) gt 0 (42)





119894of gcd(119891(119909) 119892(119909)) we get that |119888

119894| le 119872

119891119892

where

119872119891119892

= 2119904(119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(43)



119901(119909) in Z

119901[119909] If deg 119890

119901(119909) gt 119904(119891 119892) we










119894 Butwe





Competing Interests


Acknowledgments


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119891 119873119892by Corollary 7

119873119891119892

by Corollary 9 and 119860119891119892


119896 le 119860


1199011 119901


119891119892 We do not know 119889(119909)



1 119901




119901119894




119901119894


119901119894


preimage 119896(119909) of 119905 sdot 119890119901119894



119901119894




numbers 1199011 119901


119894∤ 119908


2 sdot 119873119891119892


119901119894


1 119901

119896+1


(119909) ge


1 119901



119901119894


119891119892and


119901(119909) = 119872 then


119901(119909) = 119872 then we have





0119909119896+ sdot sdot sdot + 119888



119899and 119892(119909) = 119887

0119909119898+sdot sdot sdot+119887

119898

then |119888119894| le 2119896|11988801198860|119891(119909) and |119888

119894| le 2119896|11988801198870|119892(119909) Since

|11988801198860| is bounded by |gcd(119886

0 1198870)1198860| and |119888



10038161003816100381610038161198881198941003816100381610038161003816 le 2119896sdot gcd (119886

0 1198870) sdotmin

1003817100381710038171003817119891 (119909)1003817100381710038171003817

100381610038161003816100381611988601003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(39)




119894| le 119872

119902119891119892 where

119872119902119891119892

= 2119904(119902119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(40)

119872119902119891119892







if 119901 gt 2 sdot 119872119902119891119892


min119891(119909)|1198860| 119892(119909)|119887





2(119909) = gcd(119891

2(119909) 1198922(119909)) =

1199092+ 119909 + 1 Then 119904(2 119891 119892) = deg 119890

2(119909) = 2 and

1198722119891119892

= 22sdot 1 sdotmin

radic113

1radic570

3 lt 3184 (41)

Take 119901 = 67 gt 2 sdot 1198722119891119892




deg 1198903(119909) = 0 and 119872







1 119902



(119909) 1198901199022

(119909)



119902119894

(119909) gt 0



119904 (119891 119892) = min deg 1198901199021(119909) deg 119890

119902119896+1

(119909) gt 0 (42)





119894of gcd(119891(119909) 119892(119909)) we get that |119888

119894| le 119872

119891119892

where

119872119891119892

= 2119904(119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(43)



119901(119909) in Z

119901[119909] If deg 119890

119901(119909) gt 119904(119891 119892) we










119894 Butwe





Competing Interests


Acknowledgments


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119894of gcd(119891(119909) 119892(119909)) we get that |119888

119894| le 119872

119891119892

where

119872119891119892

= 2119904(119891119892)

sdot gcd (1198860 1198870)

sdotmin1003817100381710038171003817119891 (119909)

100381710038171003817100381710038161003816100381610038161198860

1003816100381610038161003816

1003817100381710038171003817119892 (119909)1003817100381710038171003817

100381610038161003816100381611988701003816100381610038161003816

(43)



119901(119909) in Z

119901[119909] If deg 119890

119901(119909) gt 119904(119891 119892) we










119894 Butwe





Competing Interests


Acknowledgments


References



















Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Research Article On Degrees of Modular Common Divisors and...

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