A bound on potential positive solutions of x^n + y^n = z^n

Suppose (x,y,z) is a solution in \mathbb{N}^+ of the equation x^n + y^n = z^n, where n > 2. Show that x,y > n.

Note that z > y.

If x^n + y^n = z^n, then x^n = z^n - y^n = (z-y)(\sum_{k=0}^{n} {n \choose k} z^{n-k}y^k) \geq \sum_{k=0}^{n} {n \choose k} z^{n-k}y^k \geq {n \choose {n-1}} z y^{n-1} = nzy^{n-1} > ny^{n-1}. Thus x^n > ny^{n-1}. Likewise, y^n > nx^{n-1}.

Suppose, without loss of generality, that x > y. Then y^n > nx^{n-1} > ny^{n-1}, and so y > n. Thus x > n.

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