If $U, V, W$ and m are natural numbers such that $U^m + V^m = W^m$ , then $m$< max$( U , V ,W)$

Question

If $U, V, W$ and m are natural numbers such that $U^m + V^m = W^m$ , then $m$< max$( U , V ,W)$...How to prove this?

It can easily proved that $m$ $\neq$ max$( U , V ,W)$ and $m$ > max$( U , V ,W)$ for some $m , U , V , W$. But how can I prove that it will happen for any choice of $m , U , V , W$.

Can anyone please help me?

Answer 1

I'll assume that $0$ is not considered a natural number. If $U$, $V$ and $W$ are natural numbers such that $$U^m+V^m=W^m,$$ then by Fermat's last theorem $m\leq2$. Hence it suffices to show that there are no solutions to $$U+V=W\qquad\text{ and }\qquad U^2+V^2=W^2,$$ with $U,V,W\leq1$ and $U,V,W\leq2$, respectively.

Answer 2

Assume without loss of generality that $U \ge V$. If $U$ and $V$ are both greater than zero, then $W>U$ and in particular $W \ge U+1$. Now assume $m > W$. Can you prove that $W^m \ge (U+1)^m > 2U^m \ge U^m + V^m$?