Beal's conjecture states that A, B, and C have a common prime factor.
Theorem. If A, B are odd and x, y are even, Beal's conjecture has no counterexample. Proof: Let D be odd, D > 1 and let w be even, w > 2. Then D^w == 9 (mod 24) while D == 0 (mod 3); otherwise, D^w == 1 (mod 24) (trivial). Any even C^z == {0; 8; 16} (mod 24): if C == 0 (mod 3), C^z == 0 (mod 24); if C == 1 (mod 3), C^z == 16 (mod 24); if C == 2 (mod 3), C^z == 8 (mod 24), while z is odd, and C^z == 16 (mod 24), while z is even (trivial). But C^z == (x'+y') (mod 24) where A^x = x' (mod 24), B^y = y' (mod 24); since (x'+y') = {2; 10; 18}, C^z == {2; 10; 18} (mod 24), which cannot be a counterexample to Beal's conjecture. - Sergey Pavlov, May 08 2021
