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 $4$.

First part:
Consider three cases where $m$ is of the form $3k, 3k+1$ and $3k+2$ respectively. Squaring, you'll get only the first one (of the form $9k^2\equiv3k^2$) is the multiple of $3$ (the other two are $9k^2+6k+1=3(3k^2+2k)+1\equiv3k^2+1$ and $9k^2+12k+4=3(3k^2+4k)+1\equiv3k^2+4 \equiv3k^2+1$). Now convince yourself that for $m^2$ to be a multiple of $3$, $m$ must be of the form $3k$, i.e., $m$ is a multiple of $3$.
I put $\equiv$ signs as when dividing by $3$ these expressions are equivalent.
Other (elegant) way to state this problem is: Since $3$ is a prime and $3 | m^2=m\times m$, hence $3$ must be a factor at least one the factors of $m^2$; i.e. $3 | m$.
Second part:
As the wise people say: One counter-example is sufficient to disprove, take any $n$ which even and not a multiple of $4$ (like, $2$, $6$, $10$ etc). Clearly, $2^2=4$ is a multiple of $4$, but $2$ is not.
I have created one follow-up question here.
