# 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.

• If $x=\sqrt{1+\cdots}$ then (you'll have to prove that) $x=\sqrt{1+x}$, and similarly for the other continued radical. – J. M. isn't a mathematician Dec 15 '16 at 11:21
• Evaluate $y$ for $y=\sqrt{x+y}$ and $x\in\{31;1\}$ . The result is $(a;b)=(6;5)$ . – user90369 Dec 15 '16 at 11:29
• Let me guess: from Brilliant.org? – Bart Michels Dec 17 '16 at 0:30

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.

Hint: To find

$$\sqrt{a+\sqrt{a+\cdots}}$$

solve the equation

$$x = \sqrt{a+x}$$

As for the answer, I get $11$.