Prove that if the square of a number $m$ is a multiple of 3, then the number $m$ is also a multiple of 3. I'd like to prove that if ${m}^{2}$ is a multiple of $3$, then ${m}$ is also a multiple of $3$. Similarly, I'd like to disprove that if ${n}^{2}$ is a multiple of $4$, then ${n}$ is also a multiple of $4$.
Per the comment from @thisismuchhealthier, the context is that I'm studying the proof of the elementary theorem from analysis that there is no rational number whose square is $2$ and the related statements that $\sqrt{3}$ and $\sqrt{6}$ are both irrational, but there is a rational number whose square is $4$.

If $m = 3k+1$, then 
$$
m^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1
$$ 
and if $m = 3k+2$, then
$$
m^2 = 9k^2 + 12k + 4 = 3(3k^2 + 4k+1) + 1
$$
Thus, if $m^2$ is a multiple of $3$, then $m$ is a also a multiple of $3$.
(because if $m$ is not a multiple of $3$, then neither is $m^2$)
