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 $4$.
 A: Hint $\ 3\mid (m\!-\!1)m(m\!+\!1)=\color{#c00}{m^3\!-m},\ $  so $\ 3\mid\color{#0a0}{m^3}\,\Rightarrow\, 3\mid \color{#0a0}{m^3}\!-(\color{#c00}{m^3\!-m}) = m$
A: 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 $m^2$)
A: 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.
A: First proof (I asume $m \in N$, is natural):
$m$ can be written as a product of different powers of prim numberts $p_i$
$m = p_1^{n_1}*p_2^{n_2}*...$
So $m^2 = p_1^{2n_1}*p_2^{2n_2}*... = 2^{2*n_1}*3^{2n_2}*...$.
Thus of $m^2$ is a multiple of $3$ it must be $n_2 >0$. Therefore $p_1^{n_1}*p_2^{n_2}*...$ is also a multiple of $3$.
Second disproof:
By counter example $n^2 = 4$ so it is a mutliple of 4. But $n = \pm \sqrt 4  = \pm 2$ is not a multiple of $4$ 
A: We prove the contrapositive: If $3\nmid m$ then $3\nmid m^2$.
If $3\nmid m$, then $\gcd(3,m)=1$ because $3$ is prime. By Bézout's theorem, $mx+3y=1$ for integers $x,y$.
Squaring yields $m^2x^2+3(3y^2+2mxy)=1$, that is, $m^2X+3Y=1$ hence $\gcd(m^2,3)=1$.
Note: this solution does not rely on FTA or the uniqueness of quotient and remainder, but does extend to all primes.
Correction (Thank you, Bill): The common proof of Bézout's theorem does rely on uniqueness of quotient and remainder.
A: Hint: To prove that if 3 divides $n^2$ then it divides $n$ you can prove that if 3 doesn't divide $n$, it doesn't divide $n^2$.
A: if $n$ is not a mutiple of three (it is $1$ or $2$ $\mod 3$), then $n^2$ is $1^2=1$ or $2^2=1$ $\mod3$ none congruent to three.
for the second question, look at $2$ and $2^2=4$
