Isn't 100000 such a huge number?

$\sqrt{x+\sqrt {100000}}=\sqrt{(\sqrt a+\sqrt b)^2}$

$\sqrt{x+2\sqrt{25000}}=\sqrt{(a+b)+2\sqrt{ab}}$

Now, we have

$x=a+b$

$25000=ab$

We don't have to find each value of $x$ . What we have to do is add up all possible value of $a$ (or $b$ since they are symmetry). Since $a$ is a divisor of $25000=2^3\times 5^5$ , the sum of divisor is given by

$S(d)=(2^0+2^1+2^2+2^3)(5^0+5^1+5^2+5^3+5^4+5^5)$

The sum of $x$ is $58590$ .

For $\sqrt{x+\sqrt{100000}} = \sqrt{a} + \sqrt{b}$ , it implies that:

$\sqrt{x+\sqrt{100000}} = \sqrt{(\sqrt{a} + \sqrt{b})^2} = \sqrt{a} + \sqrt{b}$

$\Rightarrow x + \sqrt{100000} = (\sqrt{a} + \sqrt{b})^2 = a + 2\sqrt{ab} + b$

$\Rightarrow x = a+b, \quad 2\sqrt{ab} = \sqrt{100000}\quad \Rightarrow ab = 25000$

There are $12$ unordered pairs of $(a,b)$ that satisfy the equation $ab = 25000$ and the all the values of $x = a+b$ are as follows:

$\begin{matrix} 1×25000=25000 & 1+25000=25001 \\ 2×12500=25000 & 2+12500=12502 \\ 4×6250=25000 & 4+6250=6254 \\ 5×5000=25000 & 5+5000=5005 \\ 8×3125=25000 & 8+3125=3133 \\ 10×2500=25000 & 10+2500=2510 \\ 20×1250=25000 & 20+1250=1270 \\ 25×1000=25000 & 25+1000=1025 \\ 40×625=25000 & 40+625=665 \\ 50×500=25000 & 50+500=550 \\ 100×250=25000 & 100+250=350 \\ 125×200=25000 & 125+200=325 \end{matrix}$

And the sum of all the values of $x$ is:

$= 25001 + 12502 + 6254 + 5005 + 3133 + 2510 +$ $1270 + 1025 + 665 + 550 + 350 + 325 = \boxed{59580}$

First of all, this is a great question and what makes it great is that we have to find the sum of all the values and not the number of values.

One thing which is required to tackle this is -

$\sqrt{m + \sqrt{n}} =\sqrt { \frac { \sqrt { m\quad +\quad \sqrt { { m }^{ 2 }-n } } }{ 2 } } + \sqrt { \frac { \sqrt { m\quad -\quad \sqrt { { m }^{ 2 }-n } } }{ 2 } }$ [ I would like to have a good proof for this too ]

$\sqrt{x + \sqrt{100000}} = \sqrt{\frac{x \pm \sqrt{{x}^{2} - 100000}}{2}}$

So, for $a$ and $b$ to be positive integers,

$\sqrt { { x }^{ 2 }\quad -\quad 100000 } =\quad y$ for y being any positive integer

${x}^{2} - {y}^{2} = 100000$

$(x-y)(x+y) = 100000$

So, $100000$ can be 'split' into 2 parts $p, q$ such that

$x-y = p$ , $x+y = q$

Adding these 2 eqns, $2x = p+q$ . So, 100000 when split into 2 parts such that both of them are even, we will get a positive $x$ and $x$ will be $\frac{p+q}{2}$

Possible values of $p$ and $q$ can easily be fund now.

$(p,q) = (100, 1000), (10, 10000), (50, 2000), (500, 200), (5000, 20), (50000, 2), (250, 400), (2500, 40), (25000, 4), (1250, 80), (12500, 8), (6250, 16)$

After some calculations,

$x = 550, 5005, 1025, 350, 2510, 25001, 325, 1270, 12502, 665, 6254, 3133$

After adding and finally adding our calculations, the answer we get is $\boxed{58590}$