if $x$ is rational and $x^2$ is natural, prove that $x$ is integer If $x$ is rational: There exists $\frac{a}{b}$ such that $a$, $b$ are integers.
If $x^2$ is natural:
$\left(\frac{a}{b}\right)^2$ is natural => $\frac{a^2}{b^2}$ is natural
Then $a^2$ divides $b^2$ => $a$ divides $b$
If $a$ divides $b$ and $a$, $b$ are integers, $\frac{a}{b}$ is integer
so $x$ is integer.
 A: What about this proof ?
Assume $x=\frac{p}{q}$ with coprime $p$ and $q$ (If $x$ is no integer, we can always find such a representation, otherwise nothing is to prove). Assume $x^2=\frac{p^2}{q^2}$ is a natural number. 
If $p^2$ and $q^2$ were not coprime, there would be a prime $s$ with $s|p^2$ and $s|q^2$. 
We could include $s|p$ and $s|q$, hence $p$ and $q$ would not be coprime , but this contradicts our assumption.
So, $p^2$ and $q^2$ are coprime, hence we can conclude $q^2=1$. 
This completes the proof.
A: As noted in the comments, you mean to write "$b^2$ divides $a^2$" in your third line. I wouldn't jump immediately from that to concluding though. So $b^2|a^2$ so there exists a $k$ such that $b^2k=a^2$.

Lemma: Let $x$ and $y$ be natural numbers. If $xy$ is a perfect square, then it is not the case that exactly one of $x$ and $y$ are perfect squares.

From this lemma, we see that $k$ must be a perfect square, so $k=\ell^2$. Now we can write $(b\ell)^2=a^2$ and conclude that $b\ell=a$, so $b|a$.
A: Let $x^2 = n \in \mathbb{N}\,$, then $x$ is a root of $x^2-n=0$. Since $x^2-n$ is a monic polynomial, any rational root must be an integer by the rational root theorem, so $x$ is an integer.
