Proof that any number is equal to $1$ Before I embark on this bizzare proof, I will quickly evaluate the following infinite square root; this will aid us in future calculations and working:
Consider $$x=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}$$
$$x^2-2=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}=x \implies x^2-x-2=0\implies x=2$$
as $x>0$. Now for the proof:
I was attempting some different infinite expansions/square roots when trying to solve another question of mine (Evaluate $\sqrt{x+\sqrt{{x^2}+\sqrt{{x^3}+\sqrt{{x^4}...}}}}$ ) and I came across this:
$$x+\frac{1}{x}=\sqrt{(x+\frac{1}{x})^2}=\sqrt{2+x^2+\frac{1}{x^2}}=\sqrt{2+\sqrt{(x^2+\frac{1}{x^2}}})^2=\sqrt{2+\sqrt{2+x^4+\frac{1}{x^4}}}=\sqrt{2+\sqrt{2+\sqrt{(x^4+\frac{1}{x^4})^2}}}=\sqrt{2+\sqrt{2+\sqrt{2+x^8+\frac{1}{x^8}}}}=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}=2$$
if you keep on applying this and using the result found at the start of the question. So we have that for any real number $x$ that
$$x+\frac{1}{x}=2\implies x^2-2x+1=0\implies (x-1)^2=0$$
so we finally have:
$$x=1$$
Where have I gone wrong, for surely this cannot be correct?

Although they look similar at first glance, there is no reason for the two sequences
$$ \sqrt{2+2},\ \sqrt{2+\sqrt{2+2}},\ \sqrt{2+\sqrt{2+\sqrt{2+2}}}, \ldots $$
and
$$\sqrt{2+x^2+1/x^2},\ \sqrt{2+\sqrt{2+x^4+1/x^4}},\ \sqrt{2+\sqrt{2+\sqrt{2+x^8+1/x^8}}}, \ldots $$
to have the same limit unless $x=1$.
