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!

  • $\begingroup$ Have you tried to graph it? $\endgroup$
    – Novice
    Aug 19, 2020 at 20:20
  • $\begingroup$ I have, but graphing does not provide a formal proof. $\endgroup$ Aug 19, 2020 at 20:24

6 Answers 6


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


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.


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).

  • $\begingroup$ I’m probably missing something... I don’t understand the arguments of last paragraph. Can you elaborate? $\endgroup$ Aug 19, 2020 at 20:36
  • $\begingroup$ @mathcounterexamples.net I didn't make an argument, I suggested a direction. If you can prove, for example, that $x^2\gt |2x+1|$ and $y^2-1\gt |y|$ for most integers, you will have that the product on the left is greater than the product on the right. In this kind of argument you then simply need to deal with the limited number of exceptions. $\endgroup$ Aug 20, 2020 at 5:55
  • $\begingroup$ Another observation is that $y^2-1$ has different parity from $y$. Since $2x+1$ is odd, $y$ cannot be odd (otherwise you have a factor $2$ on the left (in fact a factor $8$), but not on the right). You therefore have that $x=2a$ and $y=2b$ are both even and you get $4a^2(4b^2-1)=2b(4a+1)$ or $2a^2(4b^2-1)=b(4a+1)$, from which $b$ must be even (and $y$ therefore divisible by $4$). $\endgroup$ Aug 20, 2020 at 6:03

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.

  • $\begingroup$ Thanks so much! I don't know what the rational root theorem is. I looked it up, and took some info in, but it would be great if you could explain how we would use RRT for this! $\endgroup$ Aug 19, 2020 at 20:23
  • $\begingroup$ Thanks for the credit, Integrand. Should you also rule out $x=\pm1$? Maybe we meant $x\ge1$ and $x\le-1$? $\endgroup$ Aug 19, 2020 at 21:59

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.


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.

  • $\begingroup$ I think your solution (and mine) are incorrect. Just because a polynomial has no rational roots does not mean its square root is always irrational. For instance, $x^3-4x+1$ at $x=3,4,10$ $\endgroup$
    – Integrand
    Aug 19, 2020 at 20:31
  • 1
    $\begingroup$ Yes, you are probably correct @Integrand... I'm going to look at this again $\endgroup$ Aug 19, 2020 at 20:34
  • 1
    $\begingroup$ @Integrand: could you salvage your answer by showing that $\sqrt{4x^4+4x^2+4x+1}$ is between $2x^2+1$ and $2x^2+2$ for $x>1$ and $2x^2$ and $2x^2+1$ for $x<-1$ and therefore not an integer? $\endgroup$ Aug 19, 2020 at 20:49

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:


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.

  • $\begingroup$ Since you mentioned it in earlier comments that you weren't sure about the rational root theorem, it states that for any rational root p/q, p must have a factor in common with the first coefficient, and q must have a factor in common with the last coefficient. Here we only really care about p because we only want integers. Also I realized Barry Cipra's answer is a less complicated version of what I wrote! $\endgroup$ Aug 19, 2020 at 21:55

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