Radical Relations

Let $\space z = \sqrt{y-\sqrt{y-\sqrt{y-...}}} = \sqrt{x+\sqrt{x+\sqrt{x+...}}}$

$\Rightarrow \begin{cases} z^2 = y - z & \Rightarrow y = z^2+ z \\ z^2 = x + z & \Rightarrow x = z^2 - z \end{cases}$

$\Rightarrow \begin{cases} y - x + 1 = z^2+ z - z^2+ z + 1 = 2z + 1 \\ y - x + 1 = z^2+ z - z^2+ z - 1 = 2z - 1 \end{cases}$

$\Rightarrow \begin{cases} z^2 = y - z & \Rightarrow z^2+ z - y = 0 & \Rightarrow z = \dfrac {-1+\sqrt{1+4y}}{2} \\ z^2 = x + z & \Rightarrow z^2 - z - x = 0 & \Rightarrow z = \dfrac {1+\sqrt{1+4x}}{2} \end{cases}$

$\Rightarrow \dfrac {y-z+1}{y-x-1} = \dfrac {2z+1}{2z-1} = \dfrac {-1+\sqrt{1+4y} +1} {1+\sqrt{1+4x} -1} = \dfrac {\sqrt{1+4y}} {\sqrt{1+4x}}$

$\Rightarrow a + b = 4 + 4 = \boxed{8}$

Let

$\displaystyle \sqrt{y-\sqrt{y-\sqrt{y}\dots}}=\sqrt{x+\sqrt{x+\sqrt{x}\dots}}=p$

Notice that p appears as the second term under the square roots, so we can write

$\displaystyle \sqrt{y-p}=p$

$\implies y=p^2+p\dots(1)$

and similarly

$x=p^2-p$

and so

$y-x=2p$

and the required numerator becomes

$2p+1=\sqrt{1+ay}$

$\implies 4p^2+4p+1=1+ay$

$\implies 4(p^2+p)=ay$

and now using (1) we can see that $a=4$

Analogous reasoning for the denominator yields $b=4$ and so

$a+b=\boxed{8}$