Proving a mathematical statement about square roots and natural numbers 
Let $\alpha$ and $\beta$ be natural numbers. If $$\sqrt{\alpha} - \sqrt{\beta}$$ is a natural number, then $\sqrt{\alpha}$ and $\sqrt{\beta}$ are both natural numbers. 

Prove this statement please.
What I have tried; 
If the given proposition is true then so is the contraposition. Therefore, if $\sqrt \alpha$ or $\sqrt \beta$ aren't natural numbers then, $\sqrt \alpha - \sqrt \beta$ is not a natural number also. $$
\text{(Irrational number) - (rational number) = (irrational number)} $$
This is the end  I have tried but I just can't .
 A: Note that $$(\sqrt{\alpha} - \sqrt{\beta})(\sqrt{\alpha} + \sqrt{\beta})=a-b.$$ This implies that $(\sqrt{\alpha} + \sqrt{\beta})$ is rationnal. Hence $$\sqrt{a} = \frac{1}{2}((\sqrt{\alpha} - \sqrt{\beta})+(\sqrt{\alpha} + \sqrt{\beta}))$$ is rationnal (and the same for $b$). But if $\sqrt{a}$ is rationnel, there are $p,q \in \mathbb{N}$ (such that $p/q$ can't be further simplified) such that $$a = \frac{p^2}{q^2}.$$ Hence we must have $q=1$ which implies $a =p^2$. Finally $\sqrt{a} = p$ is natural. 
A: Let $\sqrt{a}-\sqrt{b} = m \in \mathbb{N}$. We then have
$$\sqrt{a} + \sqrt{b} = \dfrac{a-b}{\sqrt{a}-\sqrt{b}} = \dfrac{n}m \in \mathbb{Q}$$
Hence, we have
$\sqrt{a} = \dfrac{m+\dfrac{n}m}2 \in \mathbb{Q}$ and $\sqrt{b} = \dfrac{\dfrac{n}m-m}2 \in \mathbb{Q}$. Hence, both are rationals. Now a square of rational (non-natural) number will again be a rational non-natural number. This is fairly straightforward to prove by looking at $\sqrt{a}$ and $\sqrt{b}$ in reduced forms and obtaining their squares.
