Integer solutions

We have $y^2-xy+13-2x=0$ , a quadratic in $y$ . For integer solutions, the discriminant of this quadratic must be the square of an integer, that is

$x^2+8x-52=m^2$

for some integer $m$ .

Completing the square and rearranging, we have

$(x+4)^2-m^2=68$

We can factorise the difference of two squares to get

$(x+4+m)(x+4-m)=68$

So we just need to look at the divisors (positive and negative) of $68$ . For a particular divisor $d$ , we would need to solve $x+4+m=d$ and $x+4-m=\frac{68}{d}$ . Adding these, we find $2x+8=d+\frac{68}{d}$ , so we are only interested in the cases when $d$ and $\frac{68}{d}$ have the same parity. They can't both be odd, as $68$ is even; so they must both be even. The cases are now easy to list:

It can easily be checked that these all lead to integer solutions for $y$ ; the solution pairs $(x,y)$ are $(-22,-19)$ , $(-22,-3)$ , $(14,-1)$ and $(14,15)$ , that is $\boxed4$ in total.

$x=\frac{y^2+13}{2+y}$

$\implies gcd(y^2+13,\pm (2+y))=2+y \implies gcd(2y-13,\pm(2+y))=2+y \implies gcd(17,\pm(2+y))=2+y$

where $2+y > 0$ .

therefore

$2+y= \pm 17 \ or \ \pm 1$

There would be $4$ possible pairs $(x,y)$ then. $(14,15),(-22,-19)(14,-1)(-22,-3)$