How to determine solvability of binary quadratic Diophantine equations of the form $x^2-axy+bx-y+c=0$?

Question

I've been learning about Diophantine equations. I feel I have an understanding of standard linear equations and I am now working on binary, quadratic equations.

I'm trying to find integer solutions to equations of the form: $x^2-axy+bx-y+c=0$

For example: $x^2-29xy+27x-y+638373=0$

Before I take the time to try and find integers solutions, I'm hoping there is a trick to see if such solutions are even possible. Similar to how linear DE can determine solvability by $c|gcd(a,b)$.

Is there a simple way to determine the solvability of BQDE equations?

Answer 1

the quadratic form part has square discriminant, therefore factors, and this continues ..

Taking $z=x-ay,$
$$ (1+ax)(1-ab -az) = 1 + a^2 c $$

So, you need to factor$1 + a^2 c$ where both factors are $1 \pmod a$ or both $-1 \pmod a.$

If you get a pair of divisors that work, then check that $z \equiv x \pmod a$

Answer 2

In the more general case, the equation,

$$x^2-axy+bx-y+c=0$$

has infinitely many rational solutions,

$$\begin{align} x &= \frac{-n+(1-ab+a^2c)}{an}\\ y &= \frac{n^2+(ab-2)n+(1-ab+a^2c)}{a^2n} \end{align}$$

for arbitrary $n$.

But you prefer integer solutions. So if you are lucky enough such that your $a=\pm 1$, then choose $n = \pm 1$, and you have your integers regardless of $b,c$.