Suppose that $x,y,z$ are natural numbers and $$x^x + y^y = 2 z^z$$ Prove that $x=y=z$.
2026-04-12 05:52:41.1775973161
Exponential diophantine equation of the form $x^x + y^y = 2 z^z$
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Suppose that $z<y$: $$z<y \implies z+1 \leq y $$ then $$2z^z < z^{z+1}+(z+1)z^z\leq y^y$$ therefore the first assumption is false, so that we can conclude $x=y=z$.