Do any integral solutions exist for $x^2-2=y^p$ for $p\geq3$?
Solutions to $x^2-2=y^p$ for $p\geq3$
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0Well, there will at least always be solutions at $(\pm\sqrt{2}, 0)$ – 2012-06-01
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0Obviously integers are meant, else you can solve for $x$. – 2012-06-01
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0Yes sorry, integral solutions. I have made the edit. – 2012-06-01
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0What'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.) – 2012-06-01
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0I still don't understand why random diophantine equations get lots of upvotes, especially since they're often just made up by the OP and there is no reason to suspect they are interesting for any reason. And, this one has a similar feel to another one just posted an hour ago that already has 9 upvotes (and now we see that person made up the question). After seeing that previous one, I thought about posting a similar one just to get a bunch of upvotes. – 2012-06-01
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1$(x,y)=(1,-1)$ when $p$ odd gives solutions. – 2012-06-01
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0@Graphth People like Diophantine equations because they are elementary. In particular, then, the "audience" for such questions is greater, hence the number of up votes is greater. A complex question about an intricacy in algebraic geometry schemes might be a "better" question, but its audience is smaller, so its votes will be lower. No way to avoid this, really, in an environment where all votes are equal. – 2012-06-01
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0@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
1 Answers
(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 $u$.