Four Plus Minus Repeat

Let $\displaystyle \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-.....}}}}=x$ .

Also, let $\displaystyle \sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4+.....}}}}=y$ .

Then we see by observation that:-

$\displaystyle \large \sqrt{4+y}=x$

$\displaystyle \large \sqrt{4-x}=y$

Now whole square both equations and subtract to get:-

$\displaystyle x^2-y^2=x+y$ , which, in turn, means $\displaystyle x-y=1$ . Put $\displaystyle y=x-1$ in square of equation $\displaystyle 2$ , to get the final equation- $\displaystyle x^2-x-3=0$

Apply Sridharacharya's Rule (the quadratic formula) to get $\displaystyle x=\frac{1+\sqrt{13}}{2}$ . Note that $\displaystyle x$ can't be the other root $\displaystyle \frac{1-\sqrt{13}}{2}$ because $\displaystyle x$ is positive.

Given is
$x=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-\sqrt{4\pm\ldots}}}}}$ so we can say
$\begin{aligned}x&=\sqrt{4+\sqrt{4-x}}\\ x^2&=4+\sqrt{4-x}\\ (x^2-4)^2&=4-x\\ 0 &= x^4-8x^2+x+12= (x^2-x-3)(x^2+x-4)\end{aligned}$
this leaves us with only positive solution $x=\frac{1+\sqrt{13}}{2}$