Expressing $\sqrt{n +m\sqrt{k}}$ Following this answer, is there a simple rule for determining when:
$$\sqrt{n +m\sqrt{k}}$$
Where $n,m,k \in \mathbb{N}$, can be expressed as:
$$a + b\sqrt{k}$$ 
For some natural $a,b$?
This boils down to asking for what $n,m,k \in \mathbb{N}$ there exist $a,b\in \mathbb{N}$ such that:
$$2ab= m,\ \ \text{and}\ \ a^2+b^2k = n$$

Hint: Please see this link. To translate it to your problem, note that $ b = m^2k $. From here, one of the square roots on the right hand side of their expression must reduce. 
In the case that $ a + \sqrt{a^2 - b} = 2p^2, p \in \mathbb{N} $, $ p^2 = \frac{a + \sqrt{a^2 - b}}{2} $. For the second case $ q^2 = \frac{a - \sqrt{a^2 - b}}{2} $. Hence, if $ \frac{a \pm \sqrt{a^2 - b}}{2} $ is a perfect square then the nested square root is reducible. 
For example, let $ a = 30, b = 896 $. In this case, $ \frac{a + \sqrt{a^2 - b}}{2} = 16 $ so $ \sqrt{30 + \sqrt{896}} $ is reducible, being equal to $ 4 + \sqrt{14} $. Also, note that if $ a = 30, b = 756 $, it is reducible because $ \frac{a + \sqrt{a^2 - b}}{2} = 9 $, being equal to $ 3 + \sqrt{21} $. 
