TITLE: Rational solutions of the Fermat equation $X^n+Y^n+Z^n=1$
QUESTION [20 upvotes]: 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

REPLY [2 votes]: I found a parameterized solution of $X^5+Y^5+Z^5=T^5$ with $X$ and $T$ rational and $Y$ and $Z$ complex rational:
$(k^2-4k+1)^5+(2k-2+(k^2-2k+3)i)^5+(2k-2-(k^2-2k+3)i)^5=(k^2-3)^5$
There is an almost Desboves form of it, as follows:
$(b-a)^5+(a+ci)^5+(a-ci)^5=(b+a)^5$, where $2a^2+b^2=c^2$