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