# Square root of a natural number to square root of another natural number

Is there such integers $x,y$ which they're not perfect squares and they're not equal, such that:

$\sqrt{x}^\sqrt{y}$ is actually an integer? Or rational number?

• I added the tag of number theory to the question. I do not know whether elementary number theory tag is more proper. Sep 25, 2013 at 19:09

By Gelfond-Schneider theorem, you can show that $\sqrt{x}^\sqrt{y}$ is irrational.