Existence of rational points on generalized Fermat quintics

Question

Do there exist integers $(x,y,z)\neq (0,0,0)$ such that $$ (a) \quad 2x^5+3y^5=6z^5 $$ $$ (b) \quad x^5+3y^5=7z^5 $$

Here is a short motivation. Equation $ax^d + by^d=cz^d$ is trivial for $d=1$, solved by Lagrange for $d=2$, investigated in depth by Selmer https://doi.org/10.1007/BF02395746 for $d=3$. For $d=4$, at least the case $a=b=1$ is well-studied, see e.g. Section 6.6 of Cohen's famous book "Tools and Diophantine Equations, Vol. 1". However, I cannot find any references about the case $d=5$. The listed equations are the smallest ones with $d=5$ for which I cannot find neither small solution $(x,y,z)\neq (0,0,0)$ nor local obstructions modulo small $p$ (not sure for until which $p$ I should check), hence the question. If you can point me any references for the case $d\geq 5$ (or $d=4$ with general $a,b,c$), this would also be interesting.

Answer 1

Both curves have no rational points.

Curves of this shape map to genus 2 curves of the form $y^2 = a x^5 + b$ (one can make $a = 1$ if one likes), by quotienting out by the group of automorphisms generated by $(x:y:z) \mapsto (\zeta x:\zeta^{-1} y:z)$ where $\zeta$ is a primitive fifth root of unity (one can pick any two coordinates for the scaling to act on). For curves of this type, Magma can fairly easily compute the Mordell-Weil group (group of rational points on the Jacobian variety). In both cases, for the first curve I tried (using the quotient as above) the Mordell-Weil group has rank 1. One can then use Chabauty's method (also implemented in Magma) to determine the set of rational points on the genus 2 curve. In each case, there are three points, but they do not lift to rational points on the given curve.

(EDIT:) Using the quotients via the action on $x$ and $z$ leads to curves whose Jacobians have finite Mordell-Weil groups, leading to a simpler computation.

The three points one finds on the quotient curves correspond to $x = 0$, $y = 0$ and $z = 0$. So if these are the only points there, then one can easily check that there are no rational points on the original curves.

Answer 2

The first equation (a) can be shown to have no solutions by staring at the corresponding $abc$-Frey curve. The second seems harder (one can derive local information from the Frey curve, but not, as far as I can tell, a contradiction).