how to prove if $m^{2}|n^{2}$ then m|n just hint please. How to prove the following?:  Let $m,n\in \mathbb{N}$ ; $ \;m^{2}\mid n^{2}\Longrightarrow \;\;m\mid n$
Just a hint please. I tried two ways but did not work.
 A: Prime factorization is the natural way to go. Another way is to let $d$ be the greatest common divisor of $m$ and $n$, and let $m=ad$, $n=bd$. Note that $a$ and $b$ are relatively prime.
Then from $m^2$ divides $n^2$ we conclude that $a^2$ divides $b^2$. We now show that $a^2=1$. To do this, note that if $a^2\gt 1$ then there is a prime $p$ that divides $a^2$. Use this to contradict the fact that $a$ and $b$ are relatively prime. 
A: Assume that $m^2 \mid n^2$ but $m\not\mid n$. Decompose $m$ and $n$ into their prime factors and exponents, i.e. $m = p_1^{a_1}\ldots p_N^{a_N}$, $n = p_1^{b_1}\ldots p_N^{b_N}$. If $m\not\mid n$, what can you say about these prime factors? And what can you say about the prime factors of $m^2$ and $n^2$? And then infer from these answers whether it is indeed possible that $m^2 \mid n^2$ but $m\not\mid n$.
A: Another way is to note that the integer $n^2/m^2$ is the square of a rational (namely, $n/m$), so we need to show that if $k$ is a positive integer and $\sqrt k$ is rational, then it is actually an integer. There are many proofs of this fact. Here is one that does not use prime factorizations:
If $\displaystyle \sqrt k=\frac nm$, we may start by assuming that $n,m$ have no common factors (else, divide both of them by their gcd. This does not change their quotient). Now, also 
 $$ \sqrt k=\frac k{\sqrt k}=\frac k{\frac nm}=\frac {mk}n. $$
Since  $n,m$ have no common factors, then there are integers $x,y$ with $nx+my=1$. Simple algebra verifies that since
 $$ \frac nm=\frac{km}n, $$
then also
 $$ \frac nm=\frac{km}n=\frac{ny+kmx}{my+nx}, $$
but the latter is an integer, since $nx+my=1$.
