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!
 A: Rewriting the equation as $y/x=x(y^2-1)-2y$, we see that we must have $x\mid y$ (since the right hand side is an integer). So letting $y=xu$ (with $x\not=0$), we get
$$u=x(x^2u^2-1)-2xu$$
which implies $x\mid u$ and $u\mid x$, so $u=\sigma x$ with $\sigma=\pm1$. But this gives
$$\sigma x=x(x^4-1)-2\sigma x^2$$
which simplifies (on cancelling out an $x$) to
$$x^4-2\sigma x-1-\sigma=0$$
and neither $x^4-2x-2=0$ nor $x^4+2x=0$ has any (nonzero) integer roots.
A: Well you can't have $x=0$ so multiply through by $x$ to obtain $$x^2(y^2-1)=y(2x+1)$$
Then either you have $y=\pm 1$ [or $y=0$] (which you can exclude) or the left-hand side is positive.
Now compare the terms in $x$ on either side (careful that $2x+1$ may be negative) and the terms in $y$ on either side (with similar care).
A: Welcome to MSE. You can solve for $y$ using the quadratic formula:
$$
y = \frac{2x+1\pm\sqrt{4 x^4+4 x^2+4 x+1}}{2 x^2}
$$Credit to J.W. Tanner for salvaging this answer.  For $x\ge 1$, $4x^4+4x^2+4x+1$ is between $(2x^2+1)^2$ and $(2x^2+2)^2$, so its square root is not an integer. Similarly, for $x\le-1$, it is between $4x^4$ and $(2x^2+1)^2$, and we can rule out the case $x=0$ in the original equation. Then there are no integer solutions.
A: 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.
A: 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.
A: While you mention that graphing it doesn't actually provide a proof, it might help with recognizing where things are interesting. If we graph the equation at Desmos, we get:
https://www.desmos.com/calculator/tplmejuuj0
This graph makes it obvious there are no integer solutions other than $(0,0)$, which we must eliminate because we can't have $x=0$. But how to prove this? I think a proof by contradiction is our best bet.
Assume $x, y \in \mathbb Z $. Then the left side $x(y^2-1)$ is always an integer.
We already know $x \neq 0$
First, consider $x = \pm 1$. We have $y^2 - 1 = 3y$ or $1-y^2=y$. Neither $y^2-3y-1$ nor $y^2+y-1$ has a rational root (By the rational root theorem, $y$ can only be $\pm 1$, and neither choice gives us a zero).
Second, consider $x$ is any other integer. Therefore $2+1/x$ is not an integer. Since we know the left side must be an integer, for the right side to also be an integer, $y$ must be an integer multiple of $x$, or $y=kx, k \in \mathbb Z$. In which case we have:
$$ x(k^2x^2-1) = 2kx +k $$
$$ k^2x^3-x = 2kx+k $$
$$k^2x^3-2kx -x-k = 0 $$
By the rational root theorem, any integer root must be one of $\{\pm1,\pm k,k^2\}$. Since none of those roots make the left side equal to zero for integer $k$, there are no integer roots for $|x| > 1$.
We have eliminated all possible integer solutions for $x$. Therefore there is no solution with $x,y \in \mathbb Z$.
A bit complicated, but I hope it helps.
