Find all possible integers $n$ such that $\sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$ is an integer. 
Find all possible integers $n$ such that $m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$ is an integer.

Guess what? This problem is adapted from a recent competition. There have been a solution below for you to check out. I am aware of the fact that there are other solutions that are more practical and suitable in test setting.
 A: We have that $$m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}} \ (m \in \mathbb N)$$
$$\iff m - \sqrt{n + 2} = \sqrt{n + \sqrt{n + 2}} \iff (m - \sqrt{n + 2})^2 = n + \sqrt{n + 2}$$
$$\iff m^2 - (2m - \sqrt{n + 2})\sqrt{n + 2} = (\sqrt{n + 2} + 1)\sqrt{n + 2} - 2$$
$$\iff m^2 + 2 = (2m + 1)\sqrt{n + 2} \iff \sqrt{n + 2} = \frac{m^2 + 2}{2m + 1}$$
As an addition, $\dfrac{m^2 + 2}{2m + 1} \in \mathbb Q^+, \forall m \in \mathbb N \implies \sqrt{n + 2} \in \mathbb Q^+$
$\implies \sqrt{n + 2} \in \mathbb N \implies \dfrac{m^2 + 2}{2m + 1} \in \mathbb N \iff \dfrac{4(m^2 + 2) - (2m + 1)(2m - 1)}{2m + 1} \in \mathbb N$
$\iff \dfrac{9}{2m + 1} \in \mathbb N \iff 2m + 1\mid 9 \iff 2m + 1 \in \{1, 3, 9\} \iff m \in \{0, 1, 4\}$
We can set up a table for different value of $m$ and $\sqrt{n + 2}$.
$$\begin{matrix} m&& 0&& 1&& 4\\ \sqrt{n + 2} = \dfrac{m^2 + 2}{2m + 1}&& 2&& 1&& 2 \end{matrix}$$
$\iff n \in \{-1, 2\}$.
Plugging $n \in \{-1, 2\}$ in $m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$, we have that $(m,n) = (1, -1)$ and $(m, n) = (4, 2)$ is the correct answer.
A: We have that $n+2 = a^2, n + a = b^2$, where $a$ and $b$ are non-negative integers.
If $a > 2$, then $$a^2  = n+2 < n+a = a^2 + a - 2 < a^2 + 2a + 1.$$ 
This means that $n+a$ cannot be a perfect square as it's bounded between 2 consecutive perfect squares, which is a contradiction. Hence $ 0 \leq a \leq 2$.
If $ a= 0$, then $n = -2$ but $n+a = -2$ is not a perfect square. No solution.
If $a = 1$, then $n = - 1$ and $n+a = 0$ is a pefect square. This gives $m=1$.
If $a=2$, then $n=2$ and $n+a = 4$ is a perfect square. This gives $m=4$.
A: A shorter approach:
Generally for numbers $m=\sqrt{n+a}+\sqrt{n+\sqrt{n+a}}$ we must have:
$n+a=b^2$
$b^2+b-a=c^2$
If this system of equations, with given a, has integer solutions for n then m is integer.For example $a=3$; $m=\sqrt{n+3}+\sqrt{n+\sqrt{n+3}}$ we have:
$n+3=b^2$ 
$b^2+b-3=c^2$
It can be seen that with $c=3$ we have:
$b^2+b-3=3^2=9$; or $b^2+b-12=0$ which gives $b=3$ and $b=-4$. With $b=3$ we get $n=6$ and $m=6$, with $b=-4$ we get $n+3=16$ or $n=13$ which gives $m=4+\sqrt{17}$  which is not integer. 
In question $a=2$ and  we can search and see that reasonable value of c can only be zero. Other values give irrational numbers for one of terms $\sqrt{n+2}$ or $\sqrt{n+\sqrt{n+2}}$. Hence we have:  
$$b^2+b-2=0$$
Which gives:
$b=1$; $a= 1$; $n=-1$; $m= 1$
$b=-2$;$a= 4$; $n=2$; $m=4$
