Let $x, y, z$ be pairwise coprime positive integers. Does one have $x^5 + y^5 = z^p$ for any prime $p \geq 2$ ?
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1$\begingroup$ If Beal's conjecture is true, then $p$ must have to be $2$ so that $\text{gcd}(x,y,z)=1 \Rightarrow x,y,z$ are mutually co-prime. Hence, the problem becomes $x^5+y^5=z^2$ $\endgroup$– Alapan DasCommented Aug 9, 2020 at 8:25
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2$\begingroup$ In 1998 B Poonen had proven that the equation $x^5+y^5=z^2$ has no integer solution for $x,y,z$ co-prime. $\endgroup$– Alapan DasCommented Aug 9, 2020 at 8:42
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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" [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.
answered Aug 9, 2020 at 9:31
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6$\begingroup$ This is, in fact, quite usual in this business: many of these results (for small exponents) involve the computation of some Selmer group, which in turn depends on determining class groups (and unit groups) of certain algebraic number fields. Assuming GRH, class groups can be computed in reasonable time for much larger fields (in terms of degree and discriminant) than unconditionally. In the concrete case, what you need is that the class group of $\mathbb Q(\sqrt[p]{2})$ has odd order (and $p$ is not a Wieferich prime). $\endgroup$ Commented Aug 9, 2020 at 9:47