Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $n$ be an odd prime. I know that the equation $x^n+y^n = z^2$ has no non-zero coprime solution in integers whenever $n \geq 5$, and that there are infinitely many solutions as soon as one drops the coprimality condition. As a curiosity, my question is: is there a way to parametrize all the non-coprime non-zero integer solutions to this equation (using different parametrization families, if necessary)? If it helps, take $n$ small, say $n = 5$ or $n = 7$.

share|improve this question
    
Do you know a single parametrization? –  joro Apr 4 at 9:14
    
Not off the top of my head, but I'm guessing (probably wrongly) that it shouldn't be too hard to come up with such a parametrization (not covering all solutions though)? –  user49135 Apr 4 at 9:33
    
@joro I was looking at your MO thread mathoverflow.net/questions/109125/… , and I was wondering whether your parametrised family for $n = 5$ to the equation $x^n + y^n = z^4$ covers all the non-zero solutions with $x \neq \pm y$. –  user49135 Apr 4 at 10:30
    
I doubt it is complete. Though it gives some solutions. –  joro Apr 4 at 10:43
2  
$(a(a^p+b^p))^p+(b(a^p+b^p))^p=((a^p+b^p)^{(p+1)/2})^2$. –  Gerry Myerson Apr 4 at 11:58

1 Answer 1

The answer to the question depends on the parity of $n$.

Suppose that $n=2k$ is even. Let $(x,y,z)$ be a nonzero solution to your equation, with $y\neq 0$. Then $(x/y,1,z/y^k)$ is also a solution of the equation. Hence to construct all the integral solutions you need to find all the rational points on the hyperelliptic curve $z^2=1+x^n$. For $n\geq 5$ there are only finitely many such points and for each point you find a one-dimensional family of solutions of the form $(tx,ty,t^k z)$ with $\gcd(x,y,z)=1$.

If $n=2k-1$ is odd then you can find all the integral solutions as follows. Let $x,y$ be coprime integers and write $x^n+y^n=ts^2$. Then $(tx,ty,t^k s)$ is a solution.

So $((x^n+y^n)x,(x^n+y^n)y,(x^n+y^n)^k)$, $x,y\in \mathbb{Z}$ parametrizes a subset of all the integral solutions. For fixed $x/y$ you will miss at most finitely many solutions. However, I doubt that you can give an algebraic description of all the integral solutions.

share|improve this answer

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.