I Since z^n is bigger that y^n, it must follow that z exceeds y.
II Let x^n<=y^n; then it must follow that z^n<=2y^n
We are going to consider the ratio of y and z, with y=(z1), the most favourable case (see Note).
III z^(z+n)/(z1)^(z+n)=((z1)/z)^(nz)
IV {((z1)/z)^(nz)} lim(z>±infinity) ((1+z)/z)^(nz) = e
V But e>2; hence there are no solutions if the power (z+n) exceeds z.
Note: Let, say, y=z/2; then z^(z+n)/(z/2)^(z+n)=2^(z+n)

Posted by broll
on 20110622 01:37:42 