Prove that if $p$ and $q$ are distinct primes such that $pq\mid n^2$ then $pq\mid n$. Prove that if $p$ and $q$ are distinct primes such that $pq\mid n^2$ then $pq\mid n$.
My Attempt:
Assume that $pq\nmid n$. If $pq\nmid n$, then $p\nmid n$ or $q\nmid n$. Without loss of generality, assume that $p\nmid n$.
If $p\nmid n$, then $pq\nmid n^2$.
 A: First a quick lemma: If $(a,b)=1$ with $a\mid n$ and $b \mid n$, then $ab \mid n$.
Proof: Since $a \mid n$ and $b \mid n$, we have $n=ar$ and $n=bs$ for some $r,s \in \mathbb{Z}$. Since $(a,b)=1$, $ax+by=1$ for some $x,y \in \mathbb{Z}$, so 
\begin{align}
n=axn+byn=axbs+byar=ab(xs)+ab(yr)=ab(xs+yr),
\end{align}
so $ab \mid n$. 
Now it is easy to see via Euclid's Lemma that if $pq \mid n^2$, then $p \mid n$ and $q \mid n$, so since $p$ and $q$ are distinct primes, they are relatively prime, so $pq \mid n$ by our lemma. 
A: Remember euclids lemma... 
It does the job for me.
A: A proof by contraposition seems ideal here. 
Original claim: If $p$ and $q$ are distinct primes such that $pq\mid n^2$, then $pq\mid n$.
Contrapositive (equivalent claim): If $pq \nmid n$, then $pq\nmid n^2$.

Proof of contrapositive: First consider the Fundamental Theorem of Arithmetic (I assume you can use this result) which states the following:

Each integer greater than 1 can be written as a product of primes, and, except for the order in which these primes are written, this can be done in only one way.

By arranging the prime factors in increasing order, we see that each integer $n>1$ can be written in the form
$$
n = p_1^{e_1}p_2^{e_2}\ldots p_k^{e_k} = \prod_{i=1}^k p_i^{e_i}\quad (p_1 < p_2 < \cdots <p_k) \tag{1}
$$
where the primes $p_1,p_2,\ldots,p_k$ and the positive integers $e_1,e_2,\ldots,e_k$ are uniquely determined by $n$ (the form expressed in $(1)$ is thought of as the "prime decomposition" of $n$). 
Thus, assume the $n$ being considered in our problem is of the form presented in $(1)$. Given that $p$ and $q$ are distinct primes and that $pq\nmid n$, we can see that, at most, either $p$ or $q$ may be a factor of $n$ but not both (for example, if $p=2,q=5,n=2\cdot 3\cdot 7\cdot 11=462$, then we have that $pq=10$ and $n=462$ but $10\nmid 462$ even though $p$ is a factor of $n$ in this example). Consequently, we know that either $p$ or $q$ is a prime number that does not appear in the prime decomposition of $n$. 
Now consider the prime decomposition of $n^2$:
$$
n^2 = p_1^{e_1^2}p_2^{e_2^2}\ldots p_k^{e_k^2}.
$$
Notice that the prime numbers written out for $n$ are the same ones as those written out for $n^2$. Since one of $p$ or $q$ did not appear in the prime decomposition for $n$, whichever one did not appear in the prime decomposition of $n$ will be the same one that does not appear in the prime decomposition of $n^2$. That is to say, $pq \nmid n^2$, as desired. This concludes the proof.

Since the contrapositive has been proved, the original claim has been proved as well. The proof I gave above is not very clean and quite "hand-wavy", but I think it may be clearer than what else has been communicated perhaps. 
