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

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

Assume $m$ is a non-square integer. Then $\sqrt{m}$ is irrational, and if $x=\sqrt{m}+\sqrt{n}$, then

$$(x-\sqrt{m})^2=x^2-2x\sqrt{m}+m=n$$

Or

$$\frac{x^2+m-n}{2x}=\sqrt{m}$$

If $x$ is rational, then the LHS is also rational. However the RHS is irrational, contradiction, so $x$ is irrational.

Same argument as here and here. It should be put in the FAQ :-)

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.