<s>[INST] Can we find the integral solutions of $x(3y-5)=y^2+1$? 
$x(3y-5)=y^2+1$ 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.
 [/INST] 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.
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