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 $n = 7$.

  • $\begingroup$ Do you know a single parametrization? $\endgroup$ – joro Apr 4 '14 at 9:14
  • $\begingroup$ 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)? $\endgroup$ – user49135 Apr 4 '14 at 9:33
  • $\begingroup$ @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 $x \neq \pm y$. $\endgroup$ – user49135 Apr 4 '14 at 10:30
  • $\begingroup$ I doubt it is complete. Though it gives some solutions. $\endgroup$ – joro Apr 4 '14 at 10:43
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    $\begingroup$ $(a(a^p+b^p))^p+(b(a^p+b^p))^p=((a^p+b^p)^{(p+1)/2})^2$. $\endgroup$ – Gerry Myerson Apr 4 '14 at 11:58

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.


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