Let $x, y, z$ be pairwise coprime positive integers. Does one have $x^5 + y^5 = z^p$ for any prime $p \geq 2$ ?



On the Diophantine equation $x^{5} + y^5 = z^p$
To the best of my knowledge, this is open for general $p$.
As mentioned by Alapan Das, Bjorn Poonen has solved the case
$p = 2$ and also $p = 3$ [B. Poonen, Some diophantine equations
of the form $x^n + y^n = z^m$, Acta Arith. 86 (1998), 193-205].
The case $p = 5$ is part of FLT. Sander Dahmen and Samir Siksek
[Perfect powers expressible as sums of two fifth or seventh
powers, Acta Arith. 164 (2014), 65-100] solve the cases
$p = 7$ and $p = 19$, and also, assuming GRH, $p = 11, 13$.
In my paper ["Chabauty without the Mordell-Weil group"](http://www.mathe2.uni-bayreuth.de/stoll/papers/ratpts-selmer-2016-12-07.pdf)
[In: G. Böckle, W. Decker, G: Malle (Eds.): Algorithmic and
Experimental Methods in Algebra, Geometry, and Number Theory,
Springer Verlag (2018)] I remove the GRH assumption on these
two cases and do also $p = 17$,
and I extend the range of primes for which the
equation can be solved under GRH to $p \le 53$. In all these
cases, no nontrivial solutions exist.


