Can we find the integral solutions of $x(3y-5)=y^2+1$?

Question

$x(3y-5)=y^2+1$ has no integral solution. True or False?

Answer False, as $(-1,1)$ satisfies it.

I checked on WolframAlpha that the given hyperbola has four integral solutions. $(-1,1),(-1,-4),(5,13),(5,2)$.

I wonder if we can find that without hit and try.

My Attempt:

$y^2-3xy+1+5x=0$ is a quadratic in $y$. So,

$$y=\frac{3x\pm\sqrt{9x^2-4-20x}}{2}$$

For $y$ to be integer, the discriminant should be a perfect square, that too of an even number. Also, $x$ should be even so that the denominator $2$ can be cancelled out.

But wolfram states only odd $x$.

What's my error? Also, how to proceed next?

I tried equating the discriminant with $(2k)^2$ but couldn't conclude.

Answer 1

Hint. This is a linear equation in $x$, so $$ x = \frac{y^2+1}{3y-5} = \frac{1}{3}\frac{3y^2+3}{3y-5} = \frac13\frac{y(3y-5)+(5/3)(3y-5)+3+25/3}{3y-5} $$ Therefore $$ 9x = 3y+5+\frac{34}{3y-5} \implies \frac{34}{3y-5}=9x-3y-5.$$ The right hand side of the equation is an integer, if $x$ and $y$ are integers.