TITLE: Prove that there are no integer solutions to $x\left(y^{2}-1\right)=y\left(2+\frac{1}{x}\right)$
QUESTION [6 upvotes]: I have struggled on this problem for quite a bit of time now, asked some of my peers and teachers, and I am yet to find the solution. Here is the problem:

Prove that there are no integer solutions to the equation $$x\left(y^{2}-1\right)=y\left(2+\frac{1}{x}\right)$$

Here is what I have tried:

Expanding, moving things around, factoring (I wasn't able to factor it into something useful)
Expanding, converting to a cubic equation (Too difficult to solve)
Expanding, converting to a quadratic, using the quadratic formula (I was unable to simplify it enough)

It would be great if you guys could help!

REPLY [2 votes]: We are given $x(y^{2}-1)=y\left(2+\dfrac{1}{x}\right)$ with $x,y\in\Bbb Z$. The presence of the $1/x$ term implies $x\neq0$ and hence $y\neq0$. Multiplying through by $x$ gives $$x^2(y^2-1)=y(2x+1).$$Note that $2x+1$ is odd. Hence $y$ cannot be odd, because then $y^2-1$ would be even, and our equation would equate an even to an odd number. So $y$ is even. Hence $x^2$ is even, and therefore so is $x$. It follows that $y$ is divisible by $4$. Then $|(y^2-1)/y|=|y-1/y|>3$, while $|(2x+1)/x^2|=|2/x+1/x^2|<2$. Consequently our equation cannot be satisfied.