As a generalisation to the equation of Fermat, one can ask for rational solutions of $X^n+Y^n+Z^n=1$ (or almost equivalently integer solutions of $X^n+Y^n+Z^n=T^n$).


Contrary to the case of Fermat, the case where $n=3$ has infinitely many solutions, because the surface is rational. For $n=4$, we get a K3 surface and for $n\ge 5$ the surface should have finitely many points or the points should be contained in finitely many curves since it is of general type (Bombieri-Lang conjecture).


But are there some non-trivial solutions for $n\ge 5$ (and $n=4$)? Do we know if there are finitely many solutions for $n\ge 5$ ?


I guess that it should be classical, but I did not find it on this site or online, after googling "Fermat, surface, rational solutions", etc



Rational solutions of the Fermat equation $X^n+Y^n+Z^n=1$
It has been conjectured by Euler that this equation has no solutions in positive integers when $n\geq 4$.


When $n=4$, this was disproved by Elkies in the paper [Elkies, On A4+B4+C4=D4] in a very strong way: he proves that the rational points of this K3 surface are dense in the real points for the euclidean topology.


When $n\geq 5$ is odd, your surface contains lines, for instance the line $Z-T=X+Y=0$. Consequently, it has infinitely many rational points. Of course, this does not disprove Euler's conjecture, that required positive integers.


