Solvability of two-variable quadratic equations with a parameter

Question

(a) Prove that there exist infinitely many values of integer parameter $a$ such that equation $$ 2 x^2+a x y+y^2+1 = 0 $$ is solvable in integers $(x,y)$.

(b) The same question for a similar equation $$ 2 x^2+a x y -y^2-1 = 0. $$

Here is a conditional proof. We will look for solutions with $x$ even. Considering the equations as quadratic in $y$, we need the determinant $(ax)^2\pm 8x^2-4$ to be a perfect square of an even integer, say $(2t)^2$. Equivalently, $$ t^2-(a^2\pm 8)(x/2)^2=-1. $$
Hence, we need to prove that there are infinitely many integers $d$ of the form $d=a^2\pm 8$ such that the negative Pell equation $t^2-du^2=-1$ is solvable. It is known to be solvable if, for example, $d$ is a prime of the form $4k+1$. By Bunyakovsky conjecture, $a^2+8$ (and similarly $a^2-8$) is a prime infinitely often. Obviously, if this is a prime then $a$ is odd, hence prime $a^2\pm 8$ is of the form $4k+1$, and the result follows.

I suspect that there should be a not-too-difficult unconditional proof, but currently do not see it, hence the question.

An update inspired by Stanley Yao Xiao's answer. In fact, there is a theorem proved in

Newman, M. (1977) A note on an equation related to the Pell equation, Amer. Math. Monthly, 84, 365–366.

stating that if $d = \prod_{i=1}^r p_i$, where $r=2$ or $r$ is odd, and all $p_i$ are primes congruent to $1$ modulo $4$ satisfying $\left(\frac{p_i}{p_j}\right)=-1$ for $i\neq j$, then the negative Pell equation $t^2-du^2=-1$ is solvable in integers. Hence, it would suffices to show that $a^2 \pm 8$ is of this form infinitely often. Stanley Yao Xiao's answer corresponds to the case $r=2$ of this observation. As correctly noted in the answer, the difficulty of this approach is in the condition $\left(\frac{p_i}{p_j}\right)=-1$.

Answer 1

Both equations are solvable in integers $(x,y)$ for infinitely many values of $a$. Specifically, the first equation is solvable for every $a$ of the form $a=-(2^n+3)$ for integer $n$, while the second one for every $a=(-2)^n-1$.

This can be seen using the following easy observation. For any integer $a$, and any even positive integer $k$, if $(x,y)$ is a solution to the equation $$ 2 x^2+a x y+y^2+k = 0, $$ then $(X,Y)=\left(\frac{x^2+k/2}{y},x\right)$ is a solution to equation $$ 2 X^2+a X Y+Y^2+k/2 = 0. $$ Indeed, $$ 2 X^2+a X Y+Y^2+k/2 = 2\left(\frac{x^2+k/2}{y}\right)^2+a\left(\frac{x^2+k/2}{y}\right)+x^2+k/2 = $$ $$ \frac{x^2+k/2}{y^2}(2(x^2+k/2)+axy+y^2) = \frac{x^2+k/2}{y^2}\cdot 0 = 0. $$ Now, starting with $k=2^n$ and applying this transformation $n$ times, we conclude that if the equation $$ 2x^2 + axy + y^2 + 2^n = 0 \quad \quad (1) $$ is solvable in $(x,y)$, then so is the equation $$ 2x^2 + axy + y^2 + 1 = 0 $$ with the same parameter $a$. It is left to note that (1) has a solution $(x,y)=(1,1)$ if $a=-(2^n+3)$.

Similarly, if for even integer $k$ pair $(x,y)$ is a solution to the equation $$ 2 x^2+a x y-y^2+k = 0, $$ then $(X,Y)=\left(\frac{x^2+k/2}{y},x\right)$ is a solution to equation $$ 2 X^2+a X Y-Y^2-k/2 = 0. $$ Starting with equation $$ 2x^2 + axy - y^2 - (-2)^n = 0 \quad \quad (2) $$ with solution $(x,y)$ and applying this n times, we conclude that equation $$ 2x^2 + axy - y^2 - 1 = 0 $$ is solvable in $(x,y)$ for the same $a$. But (2) has a solution $(x,y)=(1,1)$ for $a=(-2)^n-1$.

Answer 2

Following the argument in Fouvry and Kluners' seminal work on the negative Pell equation (On the negative Pell equation), it suffices to require that for $D_a = a^2 \pm 8$ that one has

$$\displaystyle \text{rk}_4(C_{D_a}) = 0,$$

or the $4$-rank of the narrow class group of the quadratic field $\mathbb{Q}(\sqrt{D_a})$ to vanish.

We restrict our attention to the case when $D_a$ is odd and has exactly two distinct prime divisors, say $p_1, p_2$. Necessarily both $p_1, p_2 \equiv 1 \pmod{4}$. We then have from the above cited paper that

$$\displaystyle 2^{\text{rk}_4(C_{D_a})} = \frac{1}{2 \cdot 2^{\omega(D_a)}} \sum_{d_0 d_1 d_2 d_3 = D_a} \left(\frac{d_0}{d_2}\right) \left(\frac{d_1}{d_3} \right). $$

Noting that $D_a = p_1 p_2$, the expression on the right simplifies to

$$\displaystyle \frac{1}{8} \left(12 +2 \left(\frac{p_1}{p_2} \right)+ 2 \left(\frac{p_2}{p_1} \right) \right) = \frac{1}{2} \left(3 + \left(\frac{p_1}{p_2} \right)\right).$$

Thus it suffices to answer the following question: are there infinitely many $a \in \mathbb{N}$ such that $a^2 \pm 8$ has exactly two (distinct) prime divisors $p_1, p_2$ such that $\left(\frac{p_1}{p_2} \right) = -1$?

This should be within reach of current technology, but I do not believe such a result is in the literature. Note that without the condition $\left(\frac{p_1}{p_2}\right) = -1$ then the question is answered by applying the weighted linear sieve for example, see Theorem 25.9 in Opera de Cribro.