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 $200+2\sqrt{10^4-n}=x^2$
 A: If
$$
x=\sqrt{100-\sqrt n}+\sqrt{100+\sqrt n}\tag{1}
$$
then squaring both sides gives us
$$
x^2=200+2\sqrt{100^2-n}
$$
Rearranging and squaring again gives
$$
n=100x^2-\frac14x^4\tag{2}
$$
For this to be an integer greater than zero, $x$ must be non-zero and even.
From (1) we have that $\sqrt n\leq 100$. This gives us $x>\sqrt {200}\approx14.2$, so since $x$ is an integer, $x\geq 16$. Also from (1), $n>0$ implies that $x<20$, so again by the integer condition, $x\leq 18$.
There are only two possibilities, $x=16$ or $x=18$. (2) is decreasing for $x>10$, so we have $x=18$ as the solution, and $n=6156$.
A: Notice that $x$ must be even so that $n$ must be a multiple of $4$. Let $n=4m, x=2y$ then $$50+\sqrt{2500-m}=y^2\implies 2500-m=(y^2-50)^2,m=y^2(100-y^2)$$
Can you proceed?

$y^2 \geq 50$, so $m\geq 81\cdot 19= 1539$ and the minimum for $n$ is $4(1539)=6156$.

A: your thinking is absolutely right ,
$x=\sqrt{100+\sqrt n}+\sqrt{100-\sqrt n}$ then we have $200+2\sqrt{10^4-n}=x^2$
for the purposes of this question
$200+2\sqrt{10^4-n}$ has to be perfect square number . and we can also see that x is even
therefore let $g^2=10^4-n$ here $g\in\Bbb Z^+$.
we can also see that $g\leq100$ , therefore $x_{max}=20$
Since $x^2\geq200$
$\implies x\in {\{16,18,20\}}$
$\implies 200+2g \in {\{16^2,18^2,20^2\}}$
$\implies   g \in {\{28,62,100\}}$
$\implies n\in {\{9816,6156,0\}}$
There you go .
A: 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, $10000 - n = (62)^2 \implies n = 10000 - (62)^2 = 6156.$
