Prove that if the square of a number $m$ is a multiple of 3, then the number $m$ is also a multiple of 3.

Hint $\ 3\mid (m\!-\!1)m(m\!+\!1)=\color{#c00}{m^3\!-m},\ $ so $\ 3\mid\color{#0a0}{m^3}\,\Rightarrow\, 3\mid \color{#0a0}{m^3}\!-(\color{#c00}{m^3\!-m}) = m$

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 $m^2$)

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.