Magic of infinity

$\underline{\text{Solution 1}}:$

Ramanujan equation on nested radical is as follows:

$\small x+n+a = \sqrt{ax + (n+a)^2 + x\sqrt{a(x+n) + (n+a)^2 + (x+n)\sqrt{a(x+2n) + (n+a)^2 + (x+2n)\sqrt{...}}}}$

We note that: $y = \sqrt{a^2+a+\sqrt{a^2+a+\sqrt{a^2+a+\sqrt{...}}}}$ , $\Rightarrow x = 1$ , $n = 0$ and $a = a$ . Therefore, $y = 1+0+a = 101\quad \Rightarrow y^2 = 101^2 = \boxed{10201}$

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$\underline{\text{Solution 2}}:$

$\begin{aligned} y & = \sqrt{a^2+a+\sqrt{a^2+a+\sqrt{a^2+a+\sqrt{...}}}} \\ & = \sqrt{a^2+a+y} \\ \Rightarrow y^2 & = a^2 + a + y \\ y^2 - y - a^2 - a & = 0 \\ y^2 - y - 10100 & = 0 \\ (y-101)(y+100) & = 0 \\ \Rightarrow y & = 101 \quad \text{Since } y > 0 \\ \Rightarrow y^2 & = \boxed{10201} \end{aligned}$

Square both sides.

If a^2 = 10000, then a = 100

y^2 = 10000 + 100 + y

y^2 - y -10100 = 0

(y-101)*(y + 100) = 0

y = 101 or y = -100 but y must be positive because it is a square root.

y^2 = 101^2 = 10201