Find the smallest $n\in \mathbb Z^+$ that makes $\sqrt{100+\sqrt n}+\sqrt{100-\sqrt n}$ integer. Find the smallest $n\in \mathbb Z^+$ that makes $\sqrt{100+\sqrt n}+\sqrt{100-\sqrt n}$
Clearly if $n=0$ then we will have $20$ but I couldn't decide that how can I find the other integers. Any hint?
If I say that $x=\sqrt{100+\sqrt n}+\sqrt{100-\sqrt n}$ then we have $200+2\sqrt{10^4-n}=x^2$

Alternative approach:
You want
$$M^2 = \left(\sqrt{100 + \sqrt{n}} + \sqrt{100 - \sqrt{n}}\right)^2 = 200 + 2\sqrt{10000 - n}.$$
To minimize $n$, you need $M^2$ as large as possible, 
and still less than $200 + 2\sqrt{10000} = 400$.
The largest such square is 
$19^2 = 361 \implies 2\sqrt{10000 - n} = 161 \implies $
$\sqrt{10000 - n} = 80.5 \implies (10000 - n)$ is not an integer.
Therefore, the largest feasible square is 
$\displaystyle 18^2 = 324 \implies 2\sqrt{10000 - n} = 124 = 2\sqrt{(62)^2}.$
Therefore, $10000 - n = (62)^2 \implies n = 10000 - (62)^2 = 6156.$
