Solve in Z :
$\displaystyle
x=20-\sqrt{20-\sqrt{x}}
$
Hllo : $\displaystyle x=u^{2}...(x\geq 0)$
I'have :
$\displaystyle
20-u^{2}=\sqrt{20-u}
$
$\displaystyle u^{4}-40u^{2}+u+380=0$
$\displaystyle (u-4)(u+5)(u^{2}-u-19)=0$
(u-4=0)or (u+5=0)or($\displaystyle u^{2}-u-19=0$)=0
conclusion :
$\displaystyle
u=4
$
but u=-5 is not solution ( $\displaystyle 20-u^{2}\geq 0$ )
the equation : $\displaystyle u^{2}-u-19=0 $ have not solution in Z.
$\displaystyle x=20-\sqrt{20-\sqrt{x}}$
$\displaystyle x - 20=-\sqrt{20-\sqrt{x}}$
$\displaystyle 20 - x=\sqrt{20-\sqrt{x}}$
$\displaystyle (20 - x)^2=20-\sqrt{x}$
$\displaystyle 400 - 40x + x^2=20-\sqrt{x}$
$\displaystyle 400 - 40x + x^2 - 20 + \sqrt{x}=0$
$\displaystyle 380 - 40x + x^2 + \sqrt{x}=0$
$\displaystyle x^2 - 40x + \sqrt{x} + 380=0$
Let $\displaystyle u = \sqrt{x}$ :
$\displaystyle u^4 - 40u^2 + u + 380=0$
Solve this fourth degree polynomial (I can't do this yet so just assume you solve this for $\displaystyle u$)
And you find that $\displaystyle u = -5$, $\displaystyle u = 4$, $\displaystyle u = -\frac{\sqrt{77}-1}{2}$, $\displaystyle u = \frac{\sqrt{77}+1}{2}$.
If $\displaystyle u < 0$ is negative, then $\displaystyle \sqrt{x} < 0$, which is not possible, so if $\displaystyle u < 0$, it is not a solution in $\displaystyle \mathbb{Z}$. The last solution is positive, but when squared does not give a relative number, so it is not a solution in $\displaystyle \mathbb{Z}$.
Substitute back the only potential solution into $\displaystyle x$ ($\displaystyle x = u^2$) to find out that the only solution in $\displaystyle \mathbb{Z}$ is :
$\displaystyle x = 4^2 = 16$
Conclusion : $\displaystyle x=20-\sqrt{20-\sqrt{x}}$ admits one solution in $\displaystyle \mathbb{Z}$ : $\displaystyle x_1 = 16$.
Another approach :
Note that the only perfect squares under $\displaystyle 20$ are $\displaystyle 16$, $\displaystyle 9$, $\displaystyle 4$ and $\displaystyle 1$. Denote $\displaystyle a^2$ any of these squares. If $\displaystyle 20 - \sqrt{a^2} = 20 - a = b^2$, $\displaystyle b \in \mathbb{Z}$, then $\displaystyle b^2$ is a solution for $\displaystyle x$.
$\displaystyle 20 - \sqrt{1} = 20 - 1 = 19$
$\displaystyle 20 - \sqrt{4} = 20 - 2 = 18$
$\displaystyle 20 - \sqrt{9} = 20 - 3 = 17$
$\displaystyle 20 - \sqrt{16} = 20 - 4 = 16 = 4^2$
If $\displaystyle a^2 = 16$, then we get a perfect square, thus the only solution in $\displaystyle \mathbb{Z}$ for your equation, is $\displaystyle x_1 = 16$.
This is far simpler than my other fully algebraïc method, but it is conceptually harder to grasp, so it needs to be written a little better. Any tips ?