<s>[INST] If $\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$, find the value of $a+b$ 
$\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$ where $a,b$ are natural numbers. Find the value of $a+b$. 

I am not able to proceed with solving this question as I have no idea as to how I can calculate $\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}$. A small hint would do.
 [/INST] To find $\sqrt{a+\sqrt{a+\cdots}} $, solve the equation
$x = \sqrt{a+x}$
The solution of $\sqrt{31+\sqrt{31+\cdots}}$ gives us $x = \frac{1\pm 5\sqrt{5}}{2}$.  
The solution of $\sqrt{1+\sqrt{1+\cdots}}$ gives us $y = \frac{1\pm \sqrt{5}}{2}$.  
Thus, we have $$\frac{x}{y} = 6-\sqrt{5}$$ Hence, $a=6, b=5 \Rightarrow a+b = 11$. Hope it helps.
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