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

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    @Graphth: True, but sort of interesting questions *not* made up by the OP also get a lot of upvotes, and one could argue that they require even less effort to pose. The upvote distinction should be between questions that show effort towards an analysis and questions that don't.2012-06-01

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(Beginning of answer.)

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 $u$.