 Sharp-P and the Birch and Swinnerton-Dyer Conjecture

EasyChair Preprint no. 9368, version 3

Versions: 123history
5 pagesDate: November 27, 2022

Abstract

Assuming the Birch and Swinnerton-Dyer conjecture, an odd square-free integer $n$ is a congruent number if and only if the number of triplets of integers $(x, y, z)$ satisfying $2 \cdot x^{2} + y^{2} + 8 \cdot z^{2} = n$ is twice the number of triplets satisfying $2 \cdot x^{2} + y^{2} + 32 \cdot z^{2} = n$ due to Tunnell's theorem. However, we show these equations are instances of a variant of counting solutions of the homogeneous Diophantine equations of degree two which is a $\textit{\#P--complete}$ problem. Deciding whether $n$ is congruent or not is a problem in $NP$ since congruent numbers could be easily checked by a congruum, because of every congruent number is a product of a congruum and the square of a rational number. Certainly, every congruum is in the form of $4 \cdot m \cdot n \cdot (m^{2}-n^{2})$ (with $m>n$), where $m$ and $n$ are two distinct positive integers. We conjecture that if $P = NP$ and $FP \neq \#P$, then the Birch and Swinnerton-Dyer conjecture would be false.

Keyphrases: Boolean formula, completeness, complexity classes, polynomial time