z>x and z>y. Say y>x.

Then y^n > x^n = z^n - y^n

= (z-y)( n terms in z and y)

> (z-y)(ny^n-1)

Cancel y^n-1

y > n(z-y) >= z(z-y) >= z(1) = z, an impossibility.

blackjack

flooble's webmaster puzzle