Number Theory ProjectThe Interpretation of the definition

Andre (JianYou) WangJoint with JingYi Xue

Definition

• If with measureand is called positive definiteObservation 1:Observation 2: That explained partly the etymology of

positive .

2 2f ax bxy cy 2 4 0d b ac 0a

0c

2 24 2 0 0af ax by dy f

Definition

• If there exists a linear mapping with integer coefficient between two different positive definite polynomials, these two are called equivalent to each other. i.e. there is a map

such that it maps

, , 1x mX nY y pX qY mq np

f g

Explanation of Definition

• Observation 1: this linear map can be represented as a matrix

Observation 2: if the coefficient of isThen the new coefficient has this formula

m n

p q

f , ,a b c

2 2

2 2

'

' 2 2

'

a am bmp cp

b amn bmq bnp pqc

c an bnq cq

• Observation 3: this definition is actually fitting into the definition of equivalence relation, namely, it meets reflex, transitive, and symmetric. That’s why we can define equivalent class, each equivalent class is disjoint, in particular, the element in the same equivalent class has same measure.

• Observation 4:we may findwhich further explained the term: positive

polynomials

0, 0, 0b c a

• Observation 5: the composition of two maps

is actually the multiplication of the matrix .

Observation 6: the restriction is the perfect condition to ensure the map has an inverse which agrees with the law of inverse in matrix, namely:

m n

p q

r s

u v

1mp nq

det 1X

• Claim 1: In every equivalent class, there is a polynomial with coefficients satisfy the condition

• Proof: By well-ordering principle, let be the minimal value represented by the equivalent class, and let be an element in that class,

then

0 b a c 'a

', ', 'a b c

2 2' ' 'a a m b mp c p gcd , 1m p

• By Bezout identity, we have a propersatisfy the equality ,which fits our

definition of being a map simultaneously.Under the map

Under another map

,q n

1mq np

', ', ' , '', ''a b c a b cm n

p q

1

0 1

x

, '', '' , ,a b c a b c

• For this new there is a properwhich will enable the tuple to satisfy the claim. The claim is proven.

, ,a b c x

• Claim 2: Using Claim we can easily assume that for each measure , there are only finitely many equivalent class, and we can actually compute the upper bound for coefficient.

• This suggests that this sort of polynomial has finite classes.

d

22 2 24 4 33

dd ac b a b a a

• Claim 3: The number of the equivalent class with a fixed measure, is the number of the tuples that satisfy

• Proof: Using Claim 1, we know that for every class there is at least one element satisfy this condition .All I need to do is to show that two polynomials in the same class can not be in the set of tuples together

, ,a b c2 4 ,0d b ac b a c

• If we have , thenAll we’re supposed to check is that

Without the loss of generality ,we assumeBy definition, there is a matrix ,such that,

f gf g

f ga am n

p q

( , , ) ( , , )f f f g g ga b c a b c

• Clearly, can’t work.• In the other case, , the similar argument

goes to the rest entries.• Now we prove that there is one to one and onto

relationship between these two sets, which boils down to the truth, the sizes are equal.

f ga a1mp

2 2 2 2

2 2 2( )

1,0

f g g g g g g g

g g g g g g

a a m b mp c p a a m b mp c p

a m a mp a p a m p a mp a mp

mp

Recap

• These 3 Claims provide a rather efficient way to determine the equivalent classes of positive definite quadratic polynomials, in that this fact can help us classify different class with no ambiguity

• In light of this, problem 3.2, and the entire problem 4 is just a simple corollary.

Moreover

• Conjecture 1: if the number sets represented by the former is not the number set represented by the latter.

• Phenomenon: Different equivalent classes represents number in a distinct way, some number sets are disjoint, some are intertwined, and some are contained in the other.

f g

• Conjecture 2: If and represents a single different number , then they represents infinite many different number.

• We are convinced that these phenomenon are closely related to the problem 5 and problem 6

f g

• Claim1: If• then: • Pf:

• Claim2: If p is an odd prime bigger than 7, then• • Pf:

2 2 , ( , ) 1ax bxy cy kp k p 21, 4

dd b ac

p

2 2

2 2

(2 ) 4

(2 ) (mod )

ax by dy apk

dy ax by p

2 25p x y 1,9(mod 20)p 20 4 1 5 1 5

" " : 1p p p p p p

5 1(mod 4), 1, 1,4(mod5)

5

pIf p p

p

1,9(mod 20)p

• cannot be true• • We know , so , • Consider the set , where , • Then the number in the set • So

• Let• If is also a solution

3(mod 4)If p 3,7(mod 20)p

" "5

1p

0s2

0 5(mod )s p

0u vs 0 ,u v p 2

1p p

1 1 0 2 2 0(mod )u v s u v s p 2 2

1 2 1 2

2 21 2 1 2

( ) 5( ) 0(mod )

0 ( ) 5( ) 6

u u v v p

u u v v p

1 2 1 2 'u u t v v t 2 2' 5t t p ( , ')

5

tt

• If is also a solution • If contradict• If contradict• If is a solution

2 2' 4t t p '( , )2 2

t t

2 2' 3t t p 2 2,3(mod5)t 2 2' 2t t p 2 2,3(mod5)t

2 2't t p ( , ')t t

• Claim3: Any can be represented by• Any two number can be both represented by • , the product also can be represented.• Pf: The first prop is obvious• the second: let

2n 2 25x y

2 25x y

2 21 15A x y 2 2

2 25B x y

2 21 2 1 2 1 2 1 2( 5 ) 5( )AB x x y y x y y x

• Claim4: If , then 2p , 3p can be represented.

• Pf: the first part of the argument of claim3 is also valid

• then (claim 2)• If is also a solution(contradict

to claim 2)• If is also a solution

contradiction

3,7(mod 20)p

2 2't t p 2 2' 5t t p ( , ')

5

tt

2 2' 4t t p '( , )2 2

t t

• Discover that the two situation both are true, they coexist. i.e. if there exist

• and vice versa• If let • If • If

Since 2p,3p one of then has to be true, then the other one is true

2 21 15 2u v p

2 22 25 3u v p

2 21 15 2u v p 1 1 1 12 1, 2 1u x v y

2 21 1 1 15( 1) (5 2) 3y x y x p

2 21 15 3u v p 1 1, 1(mod 3)u v

1 1 13 1, 3 1u x v y 2 21 1 1 15( ) (5 2) 2y x y x p

1 1 13 1, 3 1u x v y 2 21 1 1 15( ) (5 2) 2y x y x p

1 1 13 1, 3 1u x v y

1 1 13 1, 3 1u x v y

2 21 1 1 15( ) (5 2) 2y x y x p

2 21 1 1 15( ) (5 2) 2y x y x p

• Claim5: can be represented• Pf: Applying claim5, there exist

• (4 numbers are odd)

3,7(mod 20)p pq

2 21 15 2u v p

2 22 25 2u v p

2 21 2 1 2 1 2 2 15( ) 5( )

2 2

u u v v u v u vpq

• Claim 6• If ,then can be

represented, which is a simple corollary from Claim 5

3,7 mod 20p 2p

• Theorem: Integer can be represented in the form of if and only if

Where :

n2 25x y

21 2 1 22 5 m nn a p p p q q q

1 2, , , 1,9 mod 20mp p p 1 2, , , 3,7 mod 20nq q q

0 mod 2n , 1,0

• First step: can be represented, can

be represented, can be evenly paired, and each pair is in the form, 5 is in the form.

The product thus can be represneted.

1 2, , mp p p 2a1 2, , , , 2mq q q

• Second step: prove the converse is also true. If the converse is not true, it suffices to say

that can be represented when where can be represented,

n n Bp

B 3,7,2 mod 20p

• When , is in the formwhich means

We now apply the infinite descend to cause contradiction

3,7 mod 20p 3p

2 2 2 2 2 23 5 5x y u v s t

2 23 mod5

0 mod5

x e

x e

2 2 2 23 5 5x y e f

• Likewise, the other case can be dealt with in almost the same way.

Thus, there is a contradiction to our previous assumption which means the converse is alos true and the theorem is proven

Problem 6.2

Proposition:

2 2 2 2 2 22 7 , 2 7 (2 7 ) 14( )A x y B s t AB xs yt xt ys

• Moreover, by using the method in claim2, we can give a proof to the conjecture 1:

• If p is prime, then

• Pf: By using , it’s easy to know the satisfied primes above.

• only to focus on • Here’s the result: • (call A, B)

2,7p 2 2 2 2

2 2

14 2 7 1,9,15,23,25,39(mod56)

p 1(mod8)

3 14 3,5,13,19,27,45(mod56)

p x y or x y p

these are primes

p x y p

2 214 ,2 , 14x y p p p

141

p

1,9,15,23,25,39(mod56)p 3,5,13,19,27,45(mod56)p

2 2

2 2 2 21 1

2 2

2 2 2 21 1 1 1

2 2 2 2

2 2 2 2

2 2 2

1 14 ,

2 14 2 , 2 , 2 7

3 14 3 ,

4 14 4 , 2 , 2 , 14

5 14 5 , (mod5)

(5 1) 14(5 1) 5 ( 14 3) 14( ) 3

(5 1) 14(5 1) 5 ( 14 3) 14(

x y p A

x y p x x p x y A

x y p B

x y p x x y y p x y A

x y p x y B

u v p u v u v p

u v p u v u v

2

2 2 2 2

2 2 2 2

) 3

(5 2) 14(5 2) 5 ( 14 6) 14( ) 3

(5 2) 14(5 2) 5 ( 14 6) 14( ) 3

p

u v p u v u v p

u v p u v u v p

2 2 2 2 2 21 1 1

2 2 2 2

2 2 2 2

2 2 2 21 1

2 2 2 21 1 1 1

1

6 14 6 , 2 ,3 2 7 , (mod3)

2(3 1) 7(3 1) 3 (2 2 3) 14( ) 3

2(3 1) 7(3 1) 3 (2 2 3) 14( ) 3

7 14 7 , 7 ,2 7

8 14 8 , 2 , 2 ,2 14

x y p x x p x y x y B

u v p x y x y p

u v p x y x y p

x y p x x y x p A

x y p x x y y p x y

x

2 22 2 1

2 2

2 2 2 2 2 21 1 1

2 21 1 2 1 2 1

1

2 2 2 2

2 , 2 7

9 14 9

10 14 10 , 2 ,5 2 7 ,5 | ( )

5 | ,5 | , 5, 5 , 5 ,2 7 1,

1(mod5) 2(mod5)

5 2(5 1) 7(5 2) ( 4 7 2) 14( 2 1) 3

5 2

x p x y A

x y p A

x y p x x p x y x y

a x y p x x y y x y contradiction

b x y

p u v u v u v p

p

2 2 2 2(5 1) 7(5 2) (4 7 2) 14( 2 1) 3u v u v u v p

1

2 2 2 2

2 2 2 2

2 2

2

2 2 21 1 1 1

2(mod5) 1(mod5)

5 2(5 2) 7(5 1) (4 7 3) 14( 2 ) 3

5 2(5 2) 7(5 1) ( 4 7 3) 14( 2 ) 3

11 14 11

1,3,4,5,9(mod11)

12 14 12 , 2 , 2 ,3 14

c x y

p u v u v u v p

p u v u v u v p B

x y p

x contradiction

x y p x x y y p x y

2

2 2

2 21 1 1 1

2 2

2 2

2 2

2 2 2 21 1

,

13 14 13

13 | ,13 | , 13, 14 1, 1, 0

1, 1(mod13)

3, 3(mod13)

4, 4(mod13)

14 14 14 , 14 , 14 ,

B

x y p B

a x y p x y x y

b x y

c x y

d x y

x y p x x p x y A

• Thank You!