# Solutions to $x^2-2=y^p$ for $p\geq3$

Do any integral solutions exist for $x^2-2=y^p$ for $p\geq3$?

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Well, there will at least always be solutions at $(\pm\sqrt{2}, 0)$ –  Arthur Jun 1 '12 at 16:07
Obviously integers are meant, else you can solve for $x$. –  sdcvvc Jun 1 '12 at 16:08
Yes sorry, integral solutions. I have made the edit. –  rckrd Jun 1 '12 at 16:09
What's the source of the problem? (If it comes from a competition, for example, that strongly suggests there's a relatively elementary solution and that would be helpful.) –  Qiaochu Yuan Jun 1 '12 at 16:16
$(x,y)=(1,-1)$ when $p$ odd gives solutions. –  Thomas Andrews Jun 1 '12 at 16:22

As I mentioned in comments, there is the trivial solution $(x,y)=(1,-1)$ when $p$ is odd.

Assuming you are looking for positive $y$, we might start as follows.

If $p$ is even, then it is clearly not possible, because $x^2-2 = z^2$ has no solutions, and a solution which exists for $y^p$ would yield a solution for $z=y^{p/2}$.

More generally, if you have proven there is no positive solution for $p$ a prime, you have no solution for any $p\geq 3$.

So you can restrict yourself to $p$ an odd prime.

By unique factorization in $\mathbb Z[\sqrt 2]$, this would mean that $x+\sqrt{2} = u(a+b\sqrt{2})^p$ where $a,b$ are some integers and $u$ is a unit of $\mathbb Z[\sqrt 2]$. You can actually restrict to $u = (1+\sqrt{2})^k$ with $0\leq k < p$.

If $u=1$, this can't happen, because $(a+b\sqrt{2})^p = m + n\sqrt{2}$ where $$n=\sum_{k=0}^{\lfloor \frac{p-1}2\rfloor}\binom p {2k+1} 2^k a^{p-2k-1}b^{2k+1}$$

We want $n=1$. But $n$ is divisible by $b$ so $b=\pm 1$.

Since $p$ is odd, $p-2k-1$ is always even, so every term of $\frac{n}{b}$ is positive unless $a=0$. Since there is more than one term when $p\geq 3$, this means $\frac{n}{b}>1$ and hence $n\neq 1$.

So we know $u\neq 1$. I'm not entirely sure how to proceed here for other $u$.

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