How many different integer values can $m$ take?

Question

How many different integer values can $m$ take, given that the following expression is an integer?

$$\frac{\sqrt{8}+\sqrt{32}}{\sqrt{m}}$$

Well

$$\frac{\sqrt{8}+\sqrt{32}}{\sqrt{m}} = \frac{\sqrt{8}+2\sqrt{8}}{\sqrt{m}} = 3\frac{\sqrt{8}}{\sqrt{m}} = 3\sqrt{\frac{8}{m}}$$

So we can only choose $m\in \{2, 8\}$. However, are there other integer values that I might be missing?

EDIT: Writing

$$3\sqrt{\frac{8}{m}} = \sqrt{\frac{72}{m}}$$

It's now obvious that $m\in \{2, 8, 18, 72\}$.

Answer 1

That is $6\sqrt{\frac{2}{m}} \in \mathbb{Z}$, so $m$ have to be an even number; otherwise $6\sqrt{\frac{2}{m}}$ will not be an integer. So let $m = 2k$ then $6\sqrt{\frac{2}{m}} = 6 \frac{1}{\sqrt{k}} \in \mathbb{Z} \implies 6 = n\sqrt{k}$ for some $n \in \mathbb{Z}$. That is, $\sqrt{k}$ is divisible by $6$, so $k = 1,4,9,36$ and $m = 2k = 2,8,18,72$.

Answer 2

Your equation $$\frac{\sqrt{8}+\sqrt{32}}{\sqrt{m}}=n$$ for $m,n\in \mathbb{Z}$, $m\ge 0$, can be rewritten as

$$72=mn^2$$

Therefore, we look for square factors $n^2|72$. $72=2^3\cdot 3^2$, so by the fundamental theorem of arithmetic, $n^2\in \{1,2^2,3^2,2^2\cdot 3^2\}$, giving $m=72/n^2\in \{2,8,18,72\}$.