Quadratic Integers: Some Properties and History

8/4/2019 Quadratic Integers: Some Properties and History

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Quadratic Integers: Some Properties and History

By Roger Bilisoly, PhD

Department of Mathematical Sciences

Central Connecticut State University (CCSU)

New Britain, Connecticut

Presented at CCSU

October 1, 2010

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2/51

What is a quadratic integer?

Quadratic integers are solutions of monic quadratics over . Let a, b .

Solvex ^ 2 2 a x a ^ 2 b ^ 2 0, xSolvex ^ 2 2 a x a ^ 2 2 b ^ 2 0, x

x a b, x a b

x a 2 b, x a 2 b

Let d be square-free (that is, it has no square divisors.)

Then quadratic integers are the following quadratic extensions of:

{a + b d : a, b} = [ d ] if d 2 or 3 (mod 4)

{a + b1 d 2 : a, b} = [ 1 d 2] if d 1 (mod 4)

AboveMathematica code shows[ 1 ] and [ 2 ].

See http://en.wikipedia.org/wiki/Quadratic_integer (accessed 9/2010.)

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2 CCSU Quadratic Integer SLIDES.nb


3/51

Quadratic integers are geometric lattices in 2

For example, [ 1 ] is a square lattice in .

max 7;

line Rangemax, max;lattice TableI n line, n, max, max Flatten;ListPlotMapRe, Im &, lattice, AspectRatio 1

6 4 2 2 4 6

6

4

2

2

4

6

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CCSU Quadratic Integer SLIDES.nb 3


4/51

Definition of a commutative ring with identity

For details, see a book on modern algebra such as

definition 1.1 of Chapter 3 of Hungerford (1980).

Example: (, +, *) is a commutative ring with identity because a, b, c

a + b (closure of addition)

(a + b) + c = a + (b + c) (addition is associative)

a + b = b + a (addition commutes)

! 0 s.t. a + 0 = 0 + a = a (unique two-sided additive identity)

a ! -a s.t. a + (-a) = (-a) + a = 0 (existence of unique additive inverse)

a*b (closure of multiplication)

(a*b)*c = a*(b*c) (multiplication is associative)

a*b = b*a (multiplication commutes)

! 1 s.t. a*1 = 1*a = a (unique two-sided multiplicative identity)

a*(b + c) = a*b + a*c and (a + b)*c = a*c + b*c (distributive law)

Thomas Hungerford,Algebra, Springer, 1980.

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5/51

Ideals of

Every ideal of consists of multiples of some integer, n.

Example. Multiples of 7 are an ideal in :

(7) = {..., -21, -14, -7, 0, 7, 14, 21, 28, ...} = {7 n: n } ,

In general, an ideal of a ring R satisfies the following:

I is closed under addition and multiplication.

(I, +, *) is a ring itself.

Any number in R times a number in I is still in I.

An ideal that consists of all multiples of one element in a ring is called aprincipal ideal.

An integral domain with only principal ideals is called a Principle Ideal Domain (PID).

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6/51

Ring homomorphisms and ideals

Theorem 2.8 of Hungerford (1980) states:

If f: R S is a homomorphism of rings, then the kernel of f is an ideal in R.

Conversely, if I is an ideal in R, then the map : R R/I given by r r + I

is an epimorphism of rings with kernel I. R/I is called a quotient ring.

Example: For n , /(n) has kernel n, which is equivalent to arithmetic (mod n).

Elements ofn can be written as a + (n), which shows why adding multiples ofn has

no effect in n.


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7/51

Famous example: Gaussian integers, [I]

[I] = { a + b I: a, b }, where I = 1 .

Mathematica shows us below how to add, multiply, anddivide these (typically division produces numbers in [I].)

a b I c d I ComplexExpanda b I c d I ComplexExpanda b I c d I ComplexExpand

a c b d

a c b d b c a d

a c

c2 d2

b d

c2 d2

b c

c2 d2

a d

c2 d2

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8/51

[I] has a norm

For = a + b I [I], the following is a norm: N( a + b I) = a2 + b2

Note that N() = 2, where | . | is the absolute value in .

Theorem N() = N()*N()

Proof Let = a + b I, = c + d I, and just compute:

a c b d ^ 2 b c a d ^ 2 Factor

a2 b2 c2 d2

Hence, if |, then N() | N().

Hence, if is a unit, then N() = 1 {1, I}.

So the multiplication in [I] is linked to that of.

See Theorem 1.2, Corollary 1.3, Theorem 2.4 of Keith Conrad's "The Gaussian Integers," which is available at http://www.math.u

conn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf.

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9/51

Characterizing all primes in [I]

The discussion in Section 9 of Conrad's "The Gaussian Integers"

is crowned by the following result:

Theorem 9.9. "Every prime in Z[i] is a unit multiple of the following primes:

i) 1 + I

ii) or, where N() = p is a prime in Z which is 1 mod 4.

iii) p, where p is a prime in Z with p 3 mod 4."

FactorInteger1 I, GaussianIntegers TrueFactorInteger1 000 000 009, GaussianIntegers TrueFactorInteger1 000 000 007, GaussianIntegers True

1 , 1

, 1, 3747 31400 , 1, 31 400 3747 , 1

1 000 000 007, 1

See Keith Conrad, "The Gaussian Integers," which is available at http://www.math.uconn.edu/~kconrad/blurbs/ugrad-

numthy/Zinotes.pdf.

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10/51

Brillhart Algorithm for solving p = a2 + b2

The last slides claims 1000000009 = (3747 + 31400 ) (3747 - 31400 ).

But how does one determine 3747 and 31400? Fermat gave this

problem as a challenge, and Hermite published an algorithm for it.

Brillhart's algorithm does this (Williams (1995), page 410):

(i) Find a solution ofz2 -1 (mod p) where 0 < z < p/2. This can be done

by using quadratic reciprocity: For p = 1000000009 and q = 11, (p/q) = (q/p) = -1

so that qp12 (q/p) = -1 (mod p).

(ii) Apply the Euclidean algorithm toz andp (in that order)

and determine the first remainder rk satisfying rk < p .

(iii)p = rk2 + rk1

2 .

Now apply this to 1000000009 in [I]. Below shows az that satisfies step (i).

p 1 000 000 009;

z PowerMod 11, p 1 4, pModz^2, p

569 522 298

1 000 000 008

Kenneth Williams, "Some Refinements of an Algorithm of Brillhart," Canadian Mathematical Society Conference Proceedings,

Volume 15, 1995.

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11/51

Brillhart Algorithm in Action

Now we implement step (ii).

dividea_, b_ : b, a b Floora b

brillhartp_, z_ : Moduletemp dividep, z,Whiletemp1 Sqrtp,

temp dividetemp1, temp2 ;Returntemp1 temp2 I

brillhart1 000 000 009, 569 522 298

31 400 3747

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12/51

The Euclidean Algorithm and GCDs in [I] using N()

The Euclidean algorithm in truncates the fractional part of the division.

For[I], we round to the nearest Gaussian integer. As Conrad points out,

the latter convention can also be applied to the Euclidean algorithm in .

euclid_ , _ : Module , , RoundRe RoundIm I; ; Return,

euclid28 17 , 34 19 , euclid34 19 , 9 17 , euclid9 17 , 1 6

, 9 17 , 2 , 1 6 , 3 , 0

Hence, 28 + 17 = (34 - 19 ) + 9 - 17 , and

34 - 19 = (2 + )(9 - 17 ) + (-1 + 6 ), and

9 - 17 = (-3 - )(-1 + 6 ) + 0.

So (-1 + 6 ) = GCD, as is any unit times this, such as - (-1 + 6 ) = 6 + .

ExtendedGCD28 17 I, 34 19 I

6 , 1 2 , 2

Note that = + where N() (1/2)N(). Numbers are from Example 3.2 of Keith Conrad's "The Gaussian Integers," which

is available at http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf.

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13/51

Unique factorization in [I]

A Euclidean domain Principal Ideal Domain Unique Factorization Domain.

This follows from Proposition 2.37, Corollary 2.38, and Theorem 2.52 of Weintraub (2008) .

Steven Weintraub, Factorization: Unique and Otherwise, A. K. Peters, 2008.

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14/51

Ideals in [I]

In[1]:= max 4;

circle Graphics Circle0, 0, 1;grid FlattenTablei, j, j, max, max, i, max, max, 1;Manipulatetrangrid Mapa, b, b, a. &, grid;

ListPlottrangrid, AspectRatio 1 Show, a, 5, 5, 1, b, 5, 5, 1

Out[4]=

a

b

20 10 10 20

20

10

10

20

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15/51

Pythagorean triples and [I]

It is well known that all Pythagorean triples are generated by

a2 b2, 2 a b, a2 b2

for a, b . However, the real

and imaginary part of(a + b I2 equals the first two entries ofthe above vector, so there's a bijection between squared

Gaussian integers all Pythagorean triples.

a b I ^ 2 Expand

a2 2 a b b2

1 2 I ^100

64431 646 909 858 924 948 087 806 774 847 687 61132 413 077 625 071 791 758 381 011 357 784



16/51

max 33;

grid FlattenTablei, j, j, max, max, i, max, max, 1;pythagorean Map1 ^ 2 2 ^2, 2 1 2 &, grid Union;ListPlot pythagorean, PlotRange max^2, max^2, max^2, max^2, AspectRatio 1

1000 500 500 1000

1000

500

500

1000

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17/51

Second example of quadratic integers: [ 2 ]

For a, b , N(a + b 2 ) = a2 - 2 b2 is a multiplicative norm.

f: [ 2 ] [ 2 ] defined by a + b 2 a - b 2 is an automorphism. f() is denoted .

[ 2 ] with norm, N(), is an Euclidean domain by Theorem 2.8 of Weintraub (2008)

So one can compute GCDs in [ 2 ].

Being an Euclidean domain [ 2 ] is a PID and UFD.

Since[ 2 ] is a PID, ideals are typically rotated, rectangular lattices.

However, there are differences between [ 2 ] and [ 1 ].


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18/51

What are the units of [ 2 ]?

That is, when is N(a + b 2 ) = a2 - 2 b2 = 1?

Note that if N(a + b 2 ) = 1, N(a b 2 n

) = 1n = 1.a2 - 2 b2 = 1 is called Pell's equation (mistakenly by Euler),

and all its solutions are generated by 1 2 n. We call

1 2 afundamental unit.

Table1 Sqrt2 ^ i Expand, i, 1, 20

1 2 , 3 2 2 , 7 5 2 , 17 12 2 , 41 29 2 , 99 70 2 , 239 169 2 , 577 408 2 ,

1393 985 2 , 3363 2378 2 , 8119 5741 2 , 19 601 13 860 2 , 47 321 33 4 61 2 ,

114 243 80 782 2 , 275 807 195 025 2 , 665 857 470 832 2 , 1 607 521 1 1 36 6 89 2 ,

3 880 899 2 7 44 210 2 , 9 369 319 6 6 25 109 2 , 22 619 537 15 9 94 428 2 Carefully speaking, each solution above corresponds to 4 solutions.

For example,N(1 + 2 ) = -1 implies that N(1 2 ) = -1.

Article 97 mentions Pell in Leonard Euler (Translated by Rev. John Hewlett),Elements of Algebra, Springer, 1984.

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19/51

Theorem about primes in [ 2 ]

From Proposition 13.1.3, and the discussion on pages 189-91 of

Ireland and Rosen (1990), the following is proved.

Theorem. Let d = 2, F = 4d = 8, , and p = prime in .

If (F/p) = 1, then p factors into (a + b d )(a - b d ) (that is, p splits.)

If (F/p) = -1, then p is prime in [ d ].

If (F/p) = 0, then p is a square (that is, p ramifies.)

Note thatp = 2 = 2 2, so 2 is a square in [ 2 ].

Kenneth Ireland and Michael Rosen,A Classical Introduction to Modern Number Theory, 2nd Edition, Springer, 1990.

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20/51

Examples of primes in [ 2 ]

Below shows which primes in are still primes in [ 2 ]

and which ones split. Note that the Reduce[] statement below solvesthe following generalized Pell's equation:x2- 2y2 =p.

DoIfJacobiSymbol2, p 1,

ans Reducex ^ 2 2 y ^ 2 p, x, y, Integers . C1 0 Last;f1 Partans, 1, 2 Partans, 2, 2 Sqrt2;f2 Partans, 1, 2 Partans, 2, 2 Sqrt2;Printf1, f2, f1 f2 Expand,Printp ,

p, TablePrimei, i, 1, 25



21/51

2

3

5

3 2 , 3 2 , 7

1113

5 2 2 , 5 2 2 , 17

19

5 2 , 5 2 , 23

29

7 3 2 , 7 3 2 , 31

37

7 2 2 , 7 2 2 , 41

43

7 2 , 7 2 , 47

53

59

61

67

11 5 2 , 11 5 2 , 71

9 2 2 , 9 2 2 , 73

9 2 , 9 2 , 79

83

11 4 2 , 11 4 2 , 89

13 6 2 , 13 6 2 , 97

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22/51

Prehistory of quadratic integers: Sum of two squares

According to page 14 of Edwards (1977), Diophantus'Arithmetic states

that 65 =12 82 42 + 72 as well as that this "is due to the fact that

65 is the product of 13 and 5, each of which numbers is the sum of two

squares." This suggests that Diophantus was aware of the identity below.

According to page 217 of Stillwell (1997), this identity was explicitly

stated by Abu Ja'far al-Khazin ca. 950, and Fibonacci proved it in his

1225 text, The Book of Squares (Liber Quadratorum).

Note that this identity is also the proof of N() = N()*N() in [I],

which was done in an earlier slide.

a c b d ^ 2 b c a d ^ 2 a ^ 2 b ^ 2 c ^ 2 d ^ 2 Expand

True

Harold Edwards, Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory , Springer, 1977.

John Stillwell,Numbers and Geometry, Springer UTM/RIM Series, 1997.

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23/51

Fermat and representing primes with quadratic forms

According to page 8 of Cox (1989), the first person to write about

the mathematics of what are now called quadratic integers is Fermat.

He mentions the link between primes and the sum of two squares

to Mersenne in 1640, and writes to Pascal aboutp =x2 + 2y2 and

=x2 + 3y2 in 1654. The below claims were noted in a letter to

Digby in 1658 (not with this modern notation, however.)

p 1 (mod 4) p =x2 +y2.

p 1 (mod 3) p =x2 + 3y2.

p 1 or 3 (mod 8) p =x2 + 2y2.

It took Euler 40 years to prove the above conjectures!

David Cox, Primes of the Form x2 + n y2: Fermat, Class Field Theory, and Complex Multiplication, Wiley, 1989.

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24/51

Identity for x2 + n y2

There are analogues of the identity that Diophantus knew, noted on

pages 17-18 of Edwards (1977). These were known by Fermat,who probably used them in his arguments of infinite descent.

The first factorization is given in Article 169 of Euler (1984).

The second factorization is given in Article 392 of Euler (1984).

See article 175 in Euler (1984) for the general case identity given below.

Factora ^ 2 b ^ 2 c ^ 2 d ^ 2, Extension IFactora ^ 2 2 b ^ 2 c ^ 2 2 d ^ 2, Extension Sqrt 2

a b a b c d c d

a 2 b a 2 b c 2 d c 2 d

a ^ 2 n b ^ 2 c ^ 2 n d ^ 2 a c n b d ^ 2 n b c a d ^ 2 Expand

True


Leonard Euler (Translated by Rev. John Hewlett), Elements of Algebra, Springer, 1984. This is based on an 1840 edition

published by Longman, Orme, and Co.

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25/51

Fermat, 1658, and Euler, 1744, on x2 + 5 y2

Fermat conjectured the following:

p, q 3, 7 (mod 20) p q = x

2

+ 5 y

2

From page 19 of Cox (1989):

For odd primes, p, not equal to 5, Euler proved:

p = x2 + 5 y2 p 1, 9 (mod 20)

2 p = x2 + 5 y2 p 3, 7 (mod 20)


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26/51

Lagrange's 1773-5 analysis of x2 + 5 y2

Page 33 of Cox (1989) states that Lagrange was able to prove the following:

For odd primesp, and for somex, y ,

1, 9 (mod 20) p = x2 5y2

3, 7 (mod 20) p = 2 x2 2x y 3y2.

Moreover, the formsx2 5y2 and 2 x2 2x y 3y2 are closely related.

, q 3, 7 (mod 20) p q = x2 + 5 y2 because of the first identity below, which is

(2.30), on page 37, of Cox (1989). And from an earlier slide (see 2nd identity below),

we know thatp, q 1, 9 (mod 20) p q = x2 + 5 y2.

Finally, it can be shown thatp 1, 9 (mod 20) and q 3, 7 (mod 20)

q = 2 x2

2x y 3y2

by the third identity below, which is on pages 18-19of the translator's introduction of Dedekind (1996).


Richard Dedekind, Theory of Algebraic Integers, Cambridge University Press, 1996. Translated by John Stillwell from 1877

edition.

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27/51

Lagrange's analysis summarized as C2

The above can be summarized by the following multiplication table of forms,

where form1 isx2 5y2 and form2 is 2 x2 2x y 3y2.

form1 form2

form1 form1 form2

form2 form2 form1

2 a ^ 2 2 a b 3 b ^ 2 2 x ^ 2 2 x y 3 y ^ 2 2 a x a y b x 3 b y ^ 2 5 a y b x ^ 2 Expand

a ^ 2 5 b ^ 2 x ^ 2 5 y ^ 2 a x 5 b y ^ 2 5 b x a y ^ 2 Expand

a ^ 2 5 b ^ 2 2 x ^ 2 2 x y 3 y ^ 2 2 a x b x 3 b y ^ 2 2 a x b x 3 b y a y 2 b x b y 3 a y 2 b x b y ^ 2 Expand

True

True

True



edition.

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28/51

Quadratic forms: Lagrange and Gauss

The material on the next few slides is from chapter 1, section 2A of Cox (1989),

which follows Lagrange's "Recherches d' Arithmetique" of 1773-75 and Gauss'

Disquisitiones Arithmeticaeof 1801.

(x,y) = a x2 + b x y + c y2, wherex, y, is called a quadratic form.

Note that a x2 + 2 b x y + c y2was the notation used in the 19th century.

is called primitive ifgcd(a, b, c) = 1.

m is represented byfifm =f(x, y) has a solution in integers.

m is properlyrepresented byf(x, y) ifgcd(x, y) = 1.

Formsfand g are equivalent iff(x, y) = g(p x + q y, r x + s y),

where p s - q r = 1 andp, q, r, s

Gauss makes the following distinction:

Formsfand g are properly equivalent when p s - q r = 1.

Formsfand g are improperly equivalent when p s - q r = -1.

Chapter 1, Section 2A of David Cox, Primes of the Form x2 + n y2: Fermat, Class Field Theory, and Complex Multiplication,

Wiley, 1989.

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29/51

Linear algebra and forms

Today we use the following language with quadratic forms and equivalence.

(x, y) = (x y)a b 2

b 2 cx

y=x

TAx .

Letx

y=

p q

r s

x

y= Px, thenfand g are equivalent iff(x) = g(Px).

Proper equivalence means det(P) = 1, and improper equivalence means det(P) = -1.

The discriminant offisD b2 4 a c 4 deta b 2

b 2 c

Also note that a change of coordinates on xTAx can be written as:

(PxT

AP x = xT

PT

AP x, and that det(PT

AP) = detP2

det(A) = det(A)since det(P) = 1. Thus iffand g are equivalent, they have the same discriminant.


Wiley, 1989.

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30/51

Reduced forms

Define a primitive, positive definite formf to be reduced if we have either

(1) |b| < a < c or

(2) 0 |b| = a c or

(3) 0 |b| a = c.

Theorem 2.8.Every primitive, positive definite form is properly equivalent

to a unique reduced form.

Recall thatfprimitive means gcd(a, b, c) = 1, and proper equivalence means

the transformation matrix P satisfies det(P) = 1.


Wiley, 1989.

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31/51

Gauss introduces class numbers

Gauss makes the following definitions in hisDisquisitiones Arithmeticae

(see page 29 of Cox (1989).)

Two forms are in the same class if they are properly equivalent.

Let h(D) = the number of reduced forms with discriminant D.

Gauss computed about 100 values ofh(D) in Article 303 of his

Disquisitiones Arithmeticae.


Wiley, 1989.

Carl Friedrich Gauss, Disquisitiones Arithmeticae,Yale University Press, 1965. Translated by Arthur A. Clarke.

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32/51

Table of h(D) and reduced forms

D h D Reduced Forms 4 1 x2 y2

8 1 x2 2 y2

12 1 x2 3 y2

20 2 x2 5 y2, 2 x2 2 x y 3 y2

28 1 x2 7 y2

56 4 x2 14 y2, 2 x2 7 y2, 3 x2 2 x y 5 y2

Above is equation 2.14, page 29 of Cox (1989). Gauss computed a more

extensive version of the first two rows here in Article 303 of Gauss (1965)

though in a different format.


Wiley, 1989.Carl Friedrich Gauss, Disquisitiones Arithmeticae,Yale University Press, 1965. Translated by Arthur A. Clarke.

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33/51

Gauss' conjecture for D < 0

Below is a plot of square-free D vs. class number.

Gauss conjectured that as D -, min h(D) .

This was proved by Heilbronn (1934) according to

http://mathworld.wolfram.com/GausssClassNumberConjecture.html.

class ; max 5000;DoIfSquareFreeQ d, AppendToclass, d, NumberFieldClassNumber Sqrtd, Null,

d, 1, max, 1ListPlotclass, AxesLabel "Squarefree D", "Class number"

1000 2000 3000 4000 5000Squarefree D

20

40

60

80

100

120

Class number

"Gauss's Class Number Conjecture" at WolframMathWorld. http://mathworld.wolfram.com/GausssClassNumberConjecture-

.html

accessed 9/2010.

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34/51

Gauss' conjecture for class number = 1, D > 0

Gauss conjectured that there are an infinite number of rings with

class number D for D > 0. This is still an open problem.

class ; max 5000;DoIfSquareFreeQ d,

AppendToclass, d, NumberFieldClassNumber Sqrtd, Null, d, 1, maxListPlotclass, AxesLabel "Squarefree D", "Class number"

1000 2000 3000 4000 5000Squarefree D

2

4

6

8

Class number

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35/51

Class number = 1 and unique factorization

From Alaca and Williams (2004):

Theorem 12.1.1.Let K be an algebraic number field. Then

h(K) = 1 OK is a PID OK is a UFD.

From Weintraub (2008):

Theorem 2.8. Let R = O( D ). Then R is a Euclidean domain for

D = -11, -7, -3, -2, -1, 2, 3, 5, 6, 7, 11, 13, 17, 21, and 29.

Proof: Weintraub proves this with elementary methods.

Note thatD = 33, 37, 41, 57, 73 also makeR a Euclidean domain, and

these are all the values ofD that makeR a Euclidean domain with

respect to its norm. However, note the following theorem:

Trivia: Theorem (Clark (1994)). The ring of integers of( 69 ) is

Euclidean but not norm-Euclidean.

From Wikipedia:

Stark-Heegner Theorem. -D = 1, 2, 3, 7, 11, 19, 43, 67, 163 is an

exhaustive list of D that make R = O( D ) a UFD.

Gauss conjectured that this was true in 1801. The proof was finally

finished in the 1950s and 1960s.

Saban Alaca and Kenneth Williams,Introductory Algebraic Number Theory, Cambridge, 2004.

David Clark, "A Quadratic Field which is Euclidean but not Norm-Euclidean," Manuscripta Mathematica, Vol 83, pages

327-330, 1994.

Steven Weintraub, Factorization : Unique and Otherwise, A. K. Peters, 2008.

"StarkHeegner theorem," http://en.wikipedia.org/wiki/Stark%E2%80%93Heegner_theorem.

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36/51

Non-unique factorization of irreducibles in [ 5 ]

A common example in books or on the web of

non-unique factorization is 2*3 = (1+ 5 ) (1- 5 ).For example, this appears on Wikipedia under

"Unique Factorization Domain." This is an old example,

one that appears in Dedekind's Theory of Algebraic Integers,

first example of equation (1), page 87. However, finding

examples usingMathematica does not take that much work.

The steps are:

(1) Produce irreducibles

(2) Multiple these irreducibles together

(3) Match up identical products

"Unique Factorization Domain" on Wikipedia at http://en.wikipedia.org/wiki/Unique_factorization_domain. Accessed 9/2010.Richard Dedekind, Theory of Algebraic Integers, Cambridge University Press, 1996. Translated by John Stillwell from 1877

edition.

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37/51


38/51

Examples of non-unique factorizations in [ 5 ]

6 3 26

1 5

1 5

4 2 5 1 5 1 5 4 2 5 2 5 2

3 3 5 3 1 5 3 3 5 2 5 1 5

9 3 39 2 5 2 5

8 2 5 3 5 1 5 8 2 5 4 5 2

2 4 5 3 5 1 5 2 4 5 1 2 5 2

11 5 3 5 2 5 11 5 1 2 5 1 5

9 3 5 4 5 1 5 9 3 5 3 5 39 3 5 1 2 5 1 5

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39/51

Dedekind's solution: Unique factorization of ideals

Dedekind showed that one could define divisibility for ideals of a ring

in his book Dedekind (1996). Here is an example of his

theory from page 173, Section 5.6 of Weintraub (2008).

Convert 6 = 2*3 = (1 + 5 )(1 - 5 ) to principal ideals in[ 5 ]:

(6) = (2)(3) = (1 + 5 )(1 - 5 ) (*)

LetI1 2, 1 5 andI2 3, 1 5 .

One can show that (2) =I12, (3) =I2I2, (1 + 5 ) =I1I2, and (1 - 5 ) =I1I2.

So (*) becomes (6) = (I12 I2I2) = I1I2 I1I2), which is just an reordering of the ideals.

Hence uniqueness (with respect to ideal factorization) is regained.

Richard Dedekind, Theory of Algebraic Integers, Cambridge University Press, 1996. Translated by John Stillwell from 1877edition.


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40/51

Plots of the ideals (1 + 5 ), I1, I2

max 10; plotmax 20;

m0 Graphics Circle0, 0, 1;grid FlattenTableTablei, j Sqrt d, j, max, max, i, max, max, 1;

a 1; b 1;

trangrid1 Mapa, b Sqrt5, b Sqrt5, a. &, grid;g1 ListPlot trangrid1, AspectRatio 1, PlotMarkers m0, 0.02,

PlotRange plotmax, plotmax, plotmax, plotmax

max 4; plotmax 20;

m0 Graphics Circle0, 0, 1;grid FlattenTableTablei, j Sqrt d, j, max, max, i, max, max, 1;

a1 2; b1 0; a2 1; b2 1;

trangrid1 Mapa1, b1 Sqrt5, b1 Sqrt5, a1. &, grid;trangrid2 Mapa2, b2 Sqrt5, b2 Sqrt5, a2. &, grid;trangrid Tablei j, i, trangrid1, j, trangrid2 Union;g ListPlot trangrid, AspectRatio 1, PlotMarkers m0, 0.012,

PlotRange plotmax, plotmax, plotmax, plotmax; Showg

a1 3; b1 0; a2 1; b2 1;

trangrid1 Mapa1, b1 Sqrt5, b1 Sqrt5, a1. &, grid;trangrid2 Mapa2, b2 Sqrt5, b2 Sqrt5, a2. &, grid;trangrid Tablei j, i, trangrid1, j, trangrid2 Union;g ListPlot trangrid, AspectRatio 1, PlotMarkers m0, 0.012,

PlotRange plotmax, plotmax, plotmax, plotmax; Showg

20 10 10 20

20

10

10

20



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20 10 10 20

20

10

10

20

20 10 10 20

20

10

10

20

As Stillwell (1998) points out on pages 244-5, the last two ideals are not rectangular lattices, hence

they are not principal ideals.


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References

Sukumar Adhikari,An Introduction to Commutative Algebra and Number Theory, CRC Press, Narosa Publishing House, 2001.

Saban Alaca and Kenneth Williams,Introductory Algebraic Number Theory, Cambridge, 2004.

Michael Artin, Chapter 11 ofAlgebra, Prentice Hall, 1991.

David Clark, "A Quadratic Field which is Euclidean but not Norm-Euclidean," Manuscripta Mathematica, Vol 83, pages

327-330, 1994.

Keith Conrad, "The Gaussian Integers" is available at http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf.



edition.


Leonard Euler (Translated by Rev. John Hewlett), Elements of Algebra, Springer, 1984. This is based on an 1840 edition

published by Longman, Orme, and Co.

Carl Friedrich Gauss, Disquisitiones Arithmeticae,Yale University Press, 1965. Translated by Arthur A. Clarke.

William Gilbert, "The Division Algorithm in Complex Bases," Canadian Mathematical Bulletin, Volume 39, 1996.William Gilbert, "Arithmetic in Complex Bases," Mathematics Magazine, Volume 57, 1984.



I. Katai and J. Szabo, "Canonical number systems for complex integers," Acta Sci. Math., Volume 37, 1975.

Donald Knuth, The Art of Computer Programming, Volume 2, Seminumerical Algorithms, 2nd Ed., Addison-Wesley, 1981.

William LeVeque, Fundamentals of Number Theory, Addison-Wesley, 1977.

William Stein,Algebraic Number Theory, A Computational Approach, 2008. Available at http://modular.math.washington.edu/-

books/ant/ant.pdf.



Wikipedia, http://www.wikipedia.org/.

Kenneth Williams, "Some Refinements of an Algorithm of Brillhart," Canadian Mathematical Society Conference Proceedings,

Volume 15, 1995.

WolframMathWorld, http://mathworld.wolfram.com/.

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Does Bill Gates do number theory on the side?

"Because both the system's privacy and the security of

digital money depend on encryption, a breakthrough in

mathematics or computer science that defeats the

cryptographic system could be a disaster. The obvious

mathematical breakthrough would be the

development of an easy way to factor large prime numbers."

(Gates 1995, The Road Ahead, p. 265).

We have seen above that some primes in do factor in [I] and [ 2 ].

For example, 97 = (4 + 9 ) (4 - 9 ) = 13 6 2 13 6 2 .

William Stein suggests "However, perhaps Gates is an algebraic

number theorist, and he really meant what he said: then we might imagine that

he meant factorization of primes of Z in rings of integers of number fields."

See page 47 of Stein (2008).

William Stein,Algebraic Number Theory, A Computational Approach, 2008. Available at http://modular.math.washington.edu/-

books/ant/ant.pdf.

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Characterizations of all ideals in [ 5 ]

Theorem 9.5 of Adhikari (2001).

"If I is a non-zero ideal of the ring [ 5 ] and a non-zero element of I ofminimal absolute value r, then either I is the principal ideal () with lattice

basis (, 5 ) or I is not a principal ideal and has the lattice basis

(, ( + 5 )/2). The second case occurs only when ( + 5 )/2

is an element of I."

Sukumar Adhikari,An Introduction to Commutative Algebra and Number Theory, CRC Press, Narosa Publishing House, 2001.

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How to find zfor the Brillhart-Hermite Algorithm?

This can be done by quadratic reciprocity: (p/q)(q/p) = (-1)^((p-1)/2 (q-1)/2)

(This is Theorem 1 of Ireland and Rosen (1990).)

For p = 1000000009 and q = 11, (p/q) (q/p) = 1

since (p-1) is divisible by 4, hence (p-1)/2 is even.

Note that (p/q) = (8/11) = (2/11)^3, but (2/11) = -1

by quadratic reciprocity: (2/q) = (-1)^((q^2-1)/8) = (-1)^15 = -1.

Hence (p/q) = (q/p) = -1.

By Proposition 5.1.2 of Ireland and Rosen (1990), q^((p-1)/2) (q/p) = -1 (mod p).

Hence z q^((p-1)/4) 11^250000002 (mod p).

p 1 000 000 009; q 11;

JacobiSymbolq, pz PowerMod q, p 1 4, p

1

569 522 298


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Modular arithmetic in [I]

Since the Euclidean algorithm works in [I], we are able to do

modular arithmetic in [I]. One example of this is that there's

an analog of Fermat's Little Theorem that holds in the Gaussian integers.

Below combines Keith Conrad's Theorems 7.12 and 7.14 of his

"The Gaussian Integers."

Theorem.Let be a Gaussian prime and note that the number of Gaussian integers

modulo is N() = . If 0 mod, then N 1 1 mod.


Keith Conrad, "The Gaussian Integers" is available at http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf.

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Gaussian integer bases

Gilbert (1996) points out that Knuth (1981) notes that one can write

complex numbers in binary notation for the base (-1 + I). Gilbert gives

the example (-2-I)/2 = 110.011I. Gilbert (1996) also notes thatusing the digits {0, 1, 2, ..., N(b)-1}, Katai and Szabo (1975) proved that

the bases (-n I), where n is a positive integer, are the only ones possible.

Finally, Gilbert (1984) shows how to add, subtract, multiply complex numbers

in this notation, while Gilbert (1996) shows how to divide in this notation.

William Gilbert, "The Division Algorithm in Complex Bases," Canadian Mathematical Bulletin, Volume 39, 1996.

William Gilbert, "Arithmetic in Complex Bases," Mathematics Magazine, Volume 57, 1984.

I. Katai and J. Szabo, "Canonical number systems for complex integers," Acta Sci. Math., Volume 37, 1975.

Donald Knuth, The Art of Computer Programming, Volume 2, Seminumerical Algorithms, 2nd Ed., Addison-Wesley, 1981.

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Rank plot for D > 0

class1 ; max 1000;DoIfSquareFreeQ d && NumberFieldClassNumber Sqrtd 1,

AppendToclass1, d, Null, d, 1, maxclass11 ;; 50ListPlotclass1, AxesLabel "Rank", D

1, 2, 3, 5, 6, 7, 11, 13, 14, 17, 19, 21, 22, 23, 29, 31, 33,37, 38, 41, 43, 46, 47, 53, 57, 59, 61, 62, 67, 69, 71, 73, 77, 83, 86,

89, 93, 94, 97, 101, 103, 107, 109, 113, 118, 127, 129, 131, 133, 134

50 100 150 200 250Rank

200

400

600

800

1000

D

"Gauss's Class Number Conjecture" at WolframMathWorld. http://mathworld.wolfram.com/GausssClassNumberConjecture-

.html

accessed 9/2010.

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Euler proves Fermat's conjectures for x2 + n y2

Page 12 of Cox (1989) says it took Euler 40 years from the time he first heard

about Fermat's conjectures concerningx

2

+ n y

2

, n = 1, 2, 3, until 1772, whenhe was able to prove all of them. His proof used his knowledge of quadratic

reciprocity, a result that he failed to prove in general. In Euler's notes

(see page 16 of Cox (1989)), he conjectures that (below uses modern notation):

(N/p) = 1 p = (mod 4N) for odd values of that need to be determined.


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Definite and indefinite forms

Equation (2.4) of Cox (1989) states: 4 a f(x, y) = (2a x + b y2 D y2.So ifD > 0, thenfcan be both positive and negative and is called indefinite.

IfD < 0 and a > 0, thenfis always positive and is called positive definite.

IfD < 0 and a < 0, thenfis always negative and is called negative definite.

From now on, we are only interested in positive definite forms such asx2 + n y2.


Wiley, 1989.

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[(1+ 3 )/2] vs.[ 3 ]

NEED CONCRETE EXAMPLE HERE.

Let = e2 i

3

= (-1 + 3 )/2. That is, is a cube root of unity. Let a, b below.

Exp2 Pi I 3Solvex ^ 2 2 a b x a ^ 2 a b b ^ 2 0, x Expand

2

3

x a b

2

1

2 3 b, x a

b

2

1

2 3 b

The real and imaginary parts of the solution above are either

both integer or both half-integers. Hence, the Eisenstein integers are

not[ 3 ] but are []. This is a technicality that Euler missed.

See http://en.wikipedia.org/wiki/Quadratic_integer and http://en.wikipedia.org/wiki/Eisenstein_integer (accessed 9/2010.)

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Quadratic Integers: Some Properties and History

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