ETA: since commenters noted I missed a sign I am correcting that.
This is one of those Mathsolutionzz instagram problems. I was curious if I did this correctly, because I looked at the comments, and the solutions others came up with are rather different (and not integers).
The problem:
$$\sqrt{2+\sqrt{2-\sqrt{2+x}}} = x$$
So my first step was simple: square both sides and get rid of one of the radicals. We then get
$$2+\sqrt{2-\sqrt{2+x}} = x^2 \tag{1}$$ which I can rearrange to: $$\sqrt{2-\sqrt{2+x}} = x^2-2 \tag{2}$$
Then we repeat the process, square both sides again: $${2-\sqrt{2+x}} = (x^2-2)^2 \tag{3}$$
rearrange: $$-\sqrt{2+x} = (x^2-2)^2-2 \tag{4}$$
rinse, repeat: $$2+x = ((x^2-2)^2-2)^2 \tag{5}$$
and where I originally had a solution at $x=2$, because I messed up a sign, this time we get a polynomial: $$x^8-8x^6+20x^4-16x^2-x+2 \tag{6}$$
Which is the really ugly polynomial that @Lubin noted (thanks!) and has several roots.
All that said, were I trying to find the roots by hand, I know the roots can't be negative (since this problem doesn't seem to allow for imaginary solutions and we are sticking to real numbers).
So my next go-to would be factoring out the polynomial. When I originally did this I would try factoring one side of the equation with the opposite-sign binomial, like this:
$$2+x = ((x^2-2)^2-2)^2$$ $$2+x = (x^2-2)^4-4(x^2-2)^2+4$$ $$x-2 = (x^2-2)^4-4(x^2-2)^2$$ $$x-2 = [(x^2-2)^2-2(x^2-2)][(x^2-2)^2+2(x^2-2)]$$ $$x-2 = (x^2-2)[(x^2-2)-2][(x^2-2)+2]$$ $$x-2 = (x^2-2)(x^2-4)(x^2)$$ $$x-2 = (x^2-2)(x-2)(x+2)(x^2)$$
Which at least tells me that +1, +2, -1, -2 don't work, really except insofar as x=2 makes the whole thing zero. But I am curious if this is a bad method, or simply erroneous. (I do realize that ultimately I'd probably want to move the x-2 to the RHS and expand it leaving me with the degree 8 polynomial anyway, but I thought this worth a shot).