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?

Nice way to see thinks
Assume that, $$(\sqrt{n}+\sqrt{m})=\frac{p}{q}$$
Then we have
$$(\sqrt{n}+\sqrt{m})=\frac{p}{q}\in\Bbb Q \implies 
n+m+2\sqrt{nm} =(\sqrt{n}+\sqrt{m})^2 =\frac{p^2}{q^2}\in\Bbb Q\\\implies \sqrt{nm} =\frac{n+m}{2}+\frac{p^2}{2q^2}\in\Bbb Q $$

But if $ nm $ is not a perfect square then   $\sqrt{nm}\not \in\Bbb Q ,$ (This can be easily prove using the fundamental theorem of number theory: Decomposition into prime numbers)
  Hence  in this case we have $$\sqrt{n}+\sqrt{m}\not \in\Bbb Q$$

Remark $~~~~~$1. $mn$ can be a perfect square even though neither $n$ nor $m$ is a perfect square. (see the example below)


*We can still have $\sqrt{n}+\sqrt{m}\not\in \Bbb Q$ even if $mn$ is perfect square.(see the example below)


Example: $n= 3$ and $ m = 12$ are not perfect square and $ nm = 36 =6^2.$
Moreover, 
$$\sqrt{n}+\sqrt{m} = \sqrt{3}+\sqrt{12} =3\sqrt 3 \not \in\Bbb Q$$
