Is it possible that although fermat's last theorem implies $x^p + y^p = z^p$ has no solution in integers $x,y,z$ and $p\geq3$, for some $q>p, x^p + y^p \equiv z^p \pmod q$. Like maybe $x^p \equiv 2 \pmod q$, $y^p$ is $1 \pmod q$ and $z^p \equiv 3 \pmod q$?
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3Yes, for all $p$, and for all sufficiently large $q$, there are nontrivial solutions to $x^p+y^p\equiv z^p\bmod q$. First proved by Pellet. – Gerry Myerson Jan 06 '20 at 00:20
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Related: https://math.stackexchange.com/questions/914195 – Bart Michels Jan 06 '20 at 10:20
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A. E. Pellet, Mémoire sur la théorie algébrique des équations, Bull. Soc. Math. France 15 (1887) 61-102, showed that for every prime $p$ there is a number $q_0(p)$ such that if $q$ is prime and $q\ge q_0(p)$ then $x^p+y^p+z^p\equiv0\bmod q$ has nontrivial solutions. (Later authors found explicit values for $q_0(p)$, and slicker proofs than that of Pellet, but it does appear that Pellet was the first one to get a result).
Gerry Myerson
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As Gerry Myerson says there are slick proofs of this: One particularly fun one you may enjoy is due to Schur, who invoked graph theoretic colouring results (a la Ramsey theory). You can find many accounts of this online by searching "Schur Fermat's last theorem" but here is a link to one anyway: http://people.math.harvard.edu/~sebv/155r-fall-2019/schur.pdf
Gerry Myerson
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Alex J Best
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