This infinitely nested root gives me two answers $ \sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{...}}}}}} $

I am trying to evaluate $$ \sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{...}}}}}} $$ where those numbers inside roots are $$ a_{n+1}=\frac{a_n^2}{2}$$ And I found two ways to solve it that give different answers. I believe one of those is not right, but I don't know which and why. Please help.

Method-1. $$x+1=\sqrt{x^2+2x+1}=\sqrt{x^2+x+\sqrt{x^2+2x+1}}\\ =\sqrt{x^2+x+\sqrt{x^2+x+\sqrt{x^2+2x+1}}}=...$$ $x=1\rightarrow$ $$2=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}}}$$ Therefore $$\begin{align} &4=2\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}}}\\ &=\sqrt{2^2\cdot2+2^2\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}}}\\ &=\sqrt{8+\sqrt{2^4\cdot2+2^4\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}}}\\ &=\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{...}}}}}\end{align}$$ Finally, $$\sqrt{4+4}=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{...}}}}}}=2\sqrt2$$

Method-2. $$\begin{align} &x+2=\sqrt{x^2+4x+4}=\sqrt{x^2+3x+\sqrt{x^2+8x+16}}\\ &=\sqrt{x^2+3x+\sqrt{x^2+7x+\sqrt{x^2+32x+256}}}\\ &=\sqrt{x^2+3x+\sqrt{x^2+7x+\sqrt{x^2+31x+\sqrt{x^2+512x+256^2}}}}... \end{align}$$ $x=1\rightarrow$ $$3=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{...}}}}}} $$

Alternative Method-2. $$\begin{align} &3=\sqrt9=\sqrt{4+5}=\sqrt{4+\sqrt{25}}=\sqrt{4+\sqrt{8+17}}\\ &=\sqrt{4+\sqrt{8+\sqrt{2\cdot16+16^2+1}}}\\ &=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{2\cdot16^2+16^4+1}}}}\\ &=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{2\cdot16^4+16^8+1}}}}}\\ &=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{2\cdot16^8+16^{16}+1}}}}}}=... \end{align}$$

So, I have two answers $2\sqrt2$ and $3$. Which one is correct and what's the problem in the other solution?? Thanks.

Now I think I understand. Thanks for all the answers. Let me post this method-3 just to show that it could be any number $\geq2\sqrt2$ and conclude this topic.

Method-3.

$$\begin{align} &\sqrt{10}=\sqrt{4+6}=\sqrt{4+\sqrt{36}}=\sqrt{4+\sqrt{8+28}}\\ &=\sqrt{4+\sqrt{8+\sqrt{32+752}}}\\ &=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+564992}}}}\\ &=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{...}}}}}}=... \end{align}$$

Let $$x=\sqrt{4+\sqrt{8+\dots}}$$ Then $$x^2=4+\sqrt{8+\dots}=4+\sqrt{2}\cdot{x}.$$ So $x$ satisfies $x^2-\sqrt{2}\cdot x-4=0$, i.e. the first solution is correct. I am not sure off the top of my head what is wrong with the second one, but I will keep looking.

Suppose you have two integer sequences, $\{a_n\}$ and $\{b_n\}$. I believe the crux of the apparent paradox lies in the observation that if one defines

$$ x = \sqrt{a_1+\sqrt{a_2+\sqrt{a_3+\sqrt{a_4+\cdots}}}} $$

as the limit of the sequence

$$ l_1 = \sqrt{a_1} $$ $$ l_2 = \sqrt{a_1+\sqrt{a_2}} $$ $$ l_3 = \sqrt{a_1+\sqrt{a_2+\sqrt{a_3}}} $$

et cetera, one may end up with a different result than if one defines it as the limit of the sequence

$$ m_1 = \sqrt{a_1+b_1} $$ $$ m_2 = \sqrt{a_1+\sqrt{a_2+b_2}} $$ $$ m_3 = \sqrt{a_1+\sqrt{a_2+\sqrt{a_3+b_3}}} $$

et cetera. In particular, if the $\{b_n\}$ converge, or diverge gently, then the two sequences should have the same limit (if indeed it exists). My intuition is that with the nested square roots as they are, if the $\{b_n\}$ diverges in the limit as $2^{2^n}$, or faster, then the two limits (if they both exist) will be different, but I'm not confident of that.