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?

If $\sqrt{n} + \sqrt{m}$ is rational, then since
($\sqrt{n} + \sqrt{m})(\sqrt{n} - \sqrt{m}) = n - m,$
$\sqrt{n} - \sqrt{m}$ is rational.  Thus
$\sqrt{n}, \sqrt{m}$ are rational, n,m are squares.  
