Number Theory Project The Interpretation of the definition Andre (JianYou) Wang Joint with JingYi Xue

Definition If with measure and is called positive definite Observation 1: Observation 2: That explained partly the etymology of positive.

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

Explanation of Definition Observation 1: this linear map can be represented as a matrix Observation 2: if the coefficient of is Then the new coefficient has this formula

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 find which further explained the term: positive polynomials

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:

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

By Bezout identity, we have a proper satisfy the equality,which fits our definition of being a map simultaneously. Under the map Under another map

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

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.

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

If we have, then All we’re supposed to check is that Without the loss of generality,we assume By definition, there is a matrix,such that,

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.

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.

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

Claim1: If then: Pf: Claim2: If p is an odd prime bigger than 7, then Pf:

cannot be true We know, so, Consider the set, where, Then the number in the set So Let If is also a solution

If contradict If is a solution

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

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

Discover that the two situation both are true, they coexist. i.e. if there exist and vice versa If let If Since 2p,3p one of then has to be true, then the other one is true

Claim5: can be represented Pf: Applying claim5, there exist (4 numbers are odd)

Claim 6 If,then can be represented, which is a simple corollary from Claim 5

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

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.

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,

When, is in the form which means We now apply the infinite descend to cause contradiction

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:

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)

Thank You!