Q:

How do I prove that $\sqrt{9+4k^2}$ holds integer value only for $k=0$ and $k=2$?

I've faced that sort of a problem while solving some other problem and it made me stuck for a while. It's vital to me to prove that for any other integer $k$ there can't be an integer output, i.e. a perfect square. 
Thanks in advance

A:

For $k>2, (2k+1)^2 = 4k^2+4k+1 \color{red}{ > 4k^2+9 > } 4k^2 = (2k)^2$
Since there are no integers between $2k$ and $2k+1$, $\sqrt{\color{red}{9+4k^2}}$ cannot be an integer when $k>2$

