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

Question

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

Answer 1

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 $k>2$

Answer 2

$$\sqrt{9+4k^2}=m(\in \Bbb N)\\\implies 9=(m+2k)(m-2k),$$ now, each term in the right side is a divisor of $9$, with difference $4k$. Now, if $k\gt 2$, then the difference could have been $\gt 8$ or $\ge 9$, which is impossible.

Since, $9$, can't have two factors with difference more than $8$. So, a contradiction.

Answer 3

As Joffran poited out you have for $k > 2$ that $(2k+1)^2 = 4k^2 + 4k + 1 > 4k^2 +9 > 4k^2 = (2k)^2$. But then it would mean that $(2k+1) > \sqrt{4k^2 + 9} > 2k$ which is impossible since 2k + 1 and 2k are successive integers.

You can check by hand that the case $k = 1$ doesn't work.