Consider a non-square integer $n$. If its square root was rational, then we would have $$\sqrt n=\frac{a}{b}$$ for some $a,b\in\mathbb{Z}$ and so $a^2=nb^2$. But this is impossible, because $n$ is non-square while $b^2$ is square, so that $nb^2$ cannot be equal to a square $a^2$. Therefore $\sqrt n$ is irrational whenever $n$ is a non-square integer.
I am just wondering because it seems like the standard "Pythagorean" proof that $\sqrt 2$ is irrational could be much simplified by just noting that $a^2=2b^2$ admits no integer solutions.