Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
1  
$(x,y)=(1,-1)$ when $p$ odd gives solutions. –  Thomas Andrews Jun 1 '12 at 16:22
show 3 more comments

1 Answer 1

(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 $u$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.