number theory - Can $sqrt{n} + sqrt{m}$ be rational if neither $n,m$ are perfect squares?

Can the expression $\sqrt{n} + \sqrt{m}$ be rational if neither $n,m \in \mathbb{N}$ are perfect squares? It doesn't seem likely, the only way that could happen is if for example $\sqrt{m} = a-\sqrt{n}, \ \ a \in \mathbb{Q}$, which I don't think is possible, but how to show it?

Answer

Squaring we get, $m=a^2+n-2a\sqrt n\implies \sqrt n=\frac{a^2+n-m}{2a}$ which is rational