Ostrich's Nested Radical

Let $A$ be a positive real number such that $\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=A=\sqrt{y+\sqrt{y-\sqrt{y+\sqrt{y-\cdots}}}}$

$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=A \\ \implies x+A=A^2 \\ \implies x=A^2-A \\$

$\sqrt{y+\sqrt{y-\sqrt{y+\sqrt{y-\cdots}}}}=A \\ \implies y+\sqrt{y-A}=A^2 \\ \implies y-A=(A^2-y)^2 \\ =A^4-2A^2y+y^2 \\ \implies y^2-2A^2y-y=-A-A^4 \\ \implies y^2-(2A^2+1)y=-A-A^4 \\ \implies \left(y-\frac{2A^2+1}{2}\right)^2=-A-A^4+\left(\frac{2A^2+1}{2}\right)^2 \\ =-A-A^4+A^4+A^2+\frac{1}{4} \\ =A^2-A+\frac{1}{4} \\ =\left(A-\frac{1}{2}\right)^2 \\$

Case 1: $y-\frac{2A^2+1}{2}=A-\frac{1}{2} \\ \implies y=A^2+A \\$

Case 2: $y-\frac{2A^2+1}{2}=\frac{1}{2}-A \\ \implies y=A^2-A+1 \\ =x+1 \\$

Let us first consider Case 1. Since $x=A^2-A$ and $y=A+A^2$ , $x+y=2A^2$ implies that $A$ must be an integer. We proceed to rewrite $x=A^2-A$ as $x+\frac{1}{4}=A^2-A+\frac{1}{4} \\ =\left(A-\frac{1}{2}\right) \\ A=\frac{1}{2}+\sqrt{x+\frac{1}{4}} \\ (\text{rej.} \quad \frac{1}{2}-\sqrt{x+\frac{1}{4}} \because A>0) \\$ For $A$ to be an integer, since $\sqrt{x+\frac{1}{4}}=\sqrt{\frac{4x+1}{4}}=\frac{\sqrt{4x+1}}{2},$ $\sqrt{4x+1}$ must be odd, i.e. $\sqrt{4x+1}=2n+1,$ where $n$ is a positive integer. We have $4x+1=(2n+1)^2 \\ =4n^2+4n+1 \\ \implies x=n(n+1) \\$ Substituting $n=1,2,3,4,\cdots,$ we have $x=2,6,12,20,\cdots$

We now proceed to consider Case 2. Since $y=x+1,$ we have $x=1,2,3,\cdots,20.$ Combining the results of both cases, the values of $x$ are $1,2,3,\cdots,20$ and the values of $y$ , consequently, would be $2,3,4,\cdots,21$ respectively. Hence our desired answer is $(1+2+3+\cdots+20)+(2+3+4+\cdots+21) \\ =\left(\frac{20\times 21}{2}\right)+\left(\frac{21\times 22}{2}-1\right) \\ =210+231-1 \\ =\boxed{440}, \\$ and we are done. $_{\square}$