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?
 A: 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, 
$$\sqrt{n}+\sqrt{m} = \sqrt{3}+\sqrt{12} =3\sqrt 3 \not \in\Bbb Q$$
A: Squaring we get, $m=a^2+n-2a\sqrt n\implies \sqrt n=\frac{a^2+n-m}{2a}$ which is rational
A: 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.  
A: 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 :-)
A: It's easy to show that if the result is rational, it has to be natural.
Hint:
$$m=(\lfloor\sqrt{m}\rfloor+\epsilon)^{2}=\lfloor\sqrt m \rfloor ^2+2\epsilon \lfloor\sqrt m \rfloor + \epsilon^2$$
$$n=(\lceil \sqrt n \rceil-\epsilon)^2=\lceil \sqrt n \rceil^2-2\epsilon\lceil \sqrt n \rceil+\epsilon^2$$
Subtract these two from each other and show that $\epsilon$ has to be rational.
