Proof of the Ballantine-Merca Conjecture and theta function identities modulo $2$

By Proposition 2.1 we have $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$. Our condition $f_{a}\equiv f_{b}f_{c}\pmod{2}$ corresponds to the statement that the number of solutions $(y,z)\in\mathbb{N}^{2}$ to

is odd if and only $ax+1$ is a square.

First note that $(b^{\prime}+c^{\prime})\mid d$. This is because $a(b^{\prime}+c^{\prime})=db^{\prime}c^{\prime}$ and $\gcd(b^{\prime}c^{\prime},b^{\prime}+c^{\prime})=1$.

We consider any prime number $p\neq 2,3$ satisfying the following modulo constraints

Letting $\delta=\frac{p^{2}-1}{a}$, we substitute $x=\delta$ to Equation (3.1) and obtain $cy^{2}+bz^{2}=p^{2}(b+c),$ or $c^{\prime}y^{2}+b^{\prime}z^{2}=p^{2}(b^{\prime}+c^{\prime})$. Since $\big{(}\frac{-b^{\prime}c^{\prime}}{p}\big{)}=-1,$ it follows $p$ divides $y,z$. Furthermore, any solution $(y,z)$ of 3.1 must have $y>0$, as otherwise $b|y^{2}-1=1$, so $b=1$, which contradicts the identity $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$. Symmetrically, we must have $z>0$. Thus the only possible solution to 3.1 is $(y,z)=(p,p)$. Since $ax+1=a\delta+1=p^{2}$ is a square, 3.1 must have an odd number of solutions, thus $(y,z)=(p,p)$ must be a solution to 3.1. Therefore, we obtain $b\mid(p^{2}-1),\ c\mid(p^{2}-1),$ or equivalently

We have thus shown that for any prime $p$, 3.2 implies 3.3. We now attempt to construct some $p$ that satisfies 3.2 but not 3.3. We next introduce a technical lemma to establish the existence of a desirable $p$.

Before showing the lemma, we see how it finishes the proof of Proposition 3.1. We take $u=a=\frac{d}{b^{\prime}+c^{\prime}}b^{\prime}c^{\prime}$ and $v$ be the square-free reduction of $b^{\prime}c^{\prime}$. For every prime $q\mid(b^{\prime}+c^{\prime})$, we write $b^{\prime}+c^{\prime}=q^{\alpha}s,\ d=q^{\beta}d^{\prime}$ where $\gcd(s,q)=\gcd(d^{\prime},q)=1$. As $b^{\prime}+c^{\prime}|d$, we have $\alpha\leq\beta$. Note that

If $q\geq 3$, then Lemma 3.2 asserts that fixing $p\pmod{q^{\beta-\alpha}}$ suffices to ensure that $p$ satisfies (3.2). But for all such $p$, we must have $q^{\beta}\mid(p^{2}-1)$ by 3.3. This implication holds for all $p$ only if $q=3$ and $\beta=1$.

We assume the prime factorization of $v$ is $v=q_{1}q_{2}\cdots q_{m}$. Let the prime factorization of $u$ be $u=p_{1}^{a_{1}}\cdots p_{n}^{a_{n}}q_{1}^{l_{1}}q_{2}^{l_{2}}\cdots q_{m}^{l_{m}}$, where $p_{1},\cdots,p_{n}$ are prime factors distinct from $q_{1},\cdots,q_{m}$. Then we want to satisfy the congruence constraints

and $\big{(}\frac{-q_{1}q_{2}\cdots q_{m}}{p}\big{)}=-1$. Now we do case work. In each case, we describe a number of congruence conditions for $p$, and show that they guarantee that $p$ satisfy the above constraints. Then the Chinese remainder theorem shows that a modulo class mod $Q$ can be taken such that all the congruence conditions are satisfied.

Case 1: If there exists an $1\leq i\leq m$ such that $q_{i}$ is 3 mod 4. Then we could first satisfy 3.4 for all other prime divisors of $Q$ arbitrarily. In particular, we could choose $p\pmod{2^{v_{2}(Q)}}$ such that $v_{2}(p^{2}-1)=v_{2}(Q)$. Then by quadratic reciprocity,

where $c\in\{\pm 1\}$ is fixed. So we could adjust the sign of $p\pmod{q_{i}}$ to ensure that $p\equiv\pm 1\pmod{q_{i}^{l_{i}}}$ and $\big{(}\frac{-q_{1}q_{2}\cdots q_{m}}{p}\big{)}=-1$.

Case 2: If $v$ is odd and all $q_{i}$ are 1 mod 4. Then by quadratic reciprocity we have

Now note that to ensure $p^{2}\equiv 1\pmod{2^{v_{2}(u)}}$, we could take either of $p\equiv\pm 1\pmod{2^{v_{2}(Q)}}$. Thus we could first choose modulo classes that satisfy 3.4 for all odd prime divisors of $Q$, then choose the sign in $p\equiv\pm 1\pmod{2^{v_{2}(Q)}}$ to ensure the quadratic character $\big{(}\frac{-q_{1}q_{2}\cdots q_{m}}{p}\big{)}=-1$.

Let $q^{h}$ be any maximal power of a prime dividing $d$. It suffices to show $q^{h}\mid 24(b^{\prime}-c^{\prime})$. The case of $b^{\prime}+c^{\prime}$ not coprime to $q$ is handled in Proposition 3.1, so we assume $b^{\prime}+c^{\prime}$ is coprime to $q$.

By Proposition 3.1 and Proposition 3.3, we have $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$, $(b^{\prime}+c^{\prime})\mid 24$, and $d\mid 24(b^{\prime}-c^{\prime})$. Thus we have reduced all cases where $b^{\prime}\neq c^{\prime}$ to a finite computation, which we carried out to conclude the sporadic tuples above.