Ramanujan's Nested Radicals: evaluating $\sqrt{4+\sqrt{16+\sqrt{64+\sqrt{\cdots}}}}$

Question

Find the exact value of $$\sqrt{4+\sqrt{16+\sqrt{64+\sqrt{\cdots}}}}$$

My approach:

Suppose

$$\sqrt{4+\sqrt{4^2+\sqrt{4^3+\sqrt{\cdots}}}} = p \tag{1}$$

By multiplying each side by $2$, we have $$2\sqrt{4+\sqrt{4^2+\sqrt{4^3+\sqrt{\cdots}}}} = 2p \tag{2}$$

and this equation is equivalent to $$\sqrt{4^2 + \sqrt{4^3 + \sqrt{4^4 + \sqrt{\cdots}}}} = 2p \tag{3}$$

Back to our original $p$. By squaring each side, we have $$4 + \sqrt{4^2 + \sqrt{4^3 + \sqrt{4^4 + \sqrt{\cdots}}}} = p^2 \tag{4}$$

It means, $4 + 2p = p^2$.

Solving this quadratic equation, I got $p = 1 + \sqrt5$, taking only the positive value. However, if I take account to scientific calculator approach, the answer tends to the surprisingly nice integer, that is $3$. How can it be? Did I do some mistakes?

Adding note: My friend said that it holds a theorem.

$$\sqrt{x + \sqrt{4x + \sqrt{4^2x + \sqrt{\cdots}}}} = 1 + \sqrt{x} \tag{5}$$

for every positive integer $x$.

Answer 1

The mistake is: $2\sqrt{4+\sqrt{4^2+\sqrt{4^3+\sqrt{\cdots}}}} \neq \sqrt{4^2 + \sqrt{4^3 + \sqrt{4^4 + \sqrt{\cdots}}}} $

$2\sqrt{4+\sqrt{4^2+\sqrt{4^3+\sqrt{\cdots}}}} = \sqrt{4^2 + \sqrt{4^4 + \sqrt{4^7 + \sqrt{\cdots}}}} $

I think that an a way is:

$$2^n+1=\sqrt{(2^n+1)^2}$$ $$2^n+1=\sqrt{4^n+2\cdot2^n+1}$$ $$2^n+1=\sqrt{4^n+\sqrt{(2\cdot2^n+1)^2}}$$

$$2^n+1=\sqrt{4^n+\sqrt{4^{n+1}+4\cdot2^n+1}}$$ $$2^n+1=\sqrt{4^n+\sqrt{4^{n+1}+\sqrt{(4\cdot2^n+1)^2}}}$$ $$2^n+1=\sqrt{4^n+\sqrt{4^{n+1}+\sqrt{4^{n+2}+8\cdot2^n+1}}}$$ $$2^n+1=\sqrt{4^n+\sqrt{4^{n+1}+\sqrt{4^{n+2}+\sqrt{(8\cdot2^n+1)^2}}}}$$ $$2^n+1=\sqrt{4^n+\sqrt{4^{n+1}+\sqrt{4^{n+2}+\sqrt{4^{n+3}+\sqrt{\cdots}}}}}$$

If $n=1$, Then $$2^1+1=\sqrt{4^1+\sqrt{4^2+\sqrt{4^3+\sqrt{4^4+\sqrt{\cdots}}}}}=3$$