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.

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 .

• Tell us what you have tried, please. – Yves Daoust Mar 17 '16 at 11:42
• In the title, do you really mean "nautical" numbers, or is there a typo? – Giuseppe Negro Mar 17 '16 at 11:50
• Sorry that is a typo – Stranger Mar 17 '16 at 12:06

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.
• Because $a$ is natural so it can't have a denominator. (We assumed that the fraction was reduced). – C. Dubussy Mar 17 '16 at 13:55
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.