Let $a$ be a positive integer. Prove: if $\sqrt a ∈ \mathbb Q,$ then $\sqrt a ∈ \mathbb N.$ 
Let $a$ be a positive integer. Prove that if
  $\sqrt a ∈ \mathbb Q,$ then $\sqrt a ∈ \mathbb N.$

I know this has something to do with primes, but I'm having a hard time getting started.
 A: Suppose that $\sqrt a=p/q,$ where $p,q\in\mathbb Z.$ Then $$aq^2=p^2$$ and the Unique Factorization Theorem implies that the exponent of every prime factor of $a$ must be even, which means that $a$ is a perfect square and hence..
A: Hint:
You don't need to use primes. There is a very elementary proof.
Since $x$ is rational, there exist integers $n>0$ such that $nx$ is an integer. Let $q$ be the smallest of these, and  denote $y=x-\lfloor x\rfloor$ be the decimal part of $x$.
Prove $q'=qy$ is an integer and observe that  $q'x$ is an integer. Deduce $y=0$.
A: Let $\sqrt{a} = \frac{p}{q}$ where $p,q \in \mathbb{N}^*$ are relatively prime integers. Suppose for the sake of contradiction that $q > 1$. Notice that we have the following equality:
$$ a q^2 = p^2. $$ 
However if we pick a prime factor of $q$ say $s \mid q$, then $s \mid aq^2$ yet $s \nmid p^2$. Contradiction.
Remark:
We can prove by this way that for a positive integer $a$ and an integer $m > 1$, if $\sqrt[m]{a} \in \mathbb{Q}$ then $\sqrt[m]{a} \in \mathbb{N}$. 
