<s>[INST] Prove that there are no integer solutions to $x\left(y^{2}-1\right)=y\left(2+\frac{1}{x}\right)$ 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!
 [/INST] We have
$$xy^2-x = 2y+\frac{y}{x}$$
$$x^2y^2-x^2=2xy+y$$
$$x^2y^2-x^2-y-2xy = 0$$
Solve as a quadratic in $x$
$$(y^2-1)x^2-(2y)x-y = 0$$
Use quadratic formula
$$x = \frac{2y\pm \sqrt{4y^2+(4y^3-4y)}}{2(y^2-1)}$$
$$x = \frac{2y\pm \sqrt{4y^3+4y^2-4y}}{2y^2-2}$$
We can factor out a $2$ to get
$$x = \frac{y\pm \sqrt{y^3+y^2-y}}{y^2-1}$$
Take a look at the square root, the only rational root is $y = 0$ (by RRT), but testing this solution, $x = 0$, and the first expression has a $\frac{y}{x}$ in it, and obviously dividing by $0$ is illegal in this case.
Another way to see that $y = 0$ is the only rational root is to factor
$$y^3+y^2-y = y(y^2+y-1)$$
Then $y^2+y-1$ has no rational roots.
Therefore, there are no integer solutions.
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