1. ## solve in Z

Solve in Z :
$

x=20-\sqrt{20-\sqrt{x}}
$

2. Originally Posted by dhiab
Solve in Z :
$

x=20-\sqrt{20-\sqrt{x}}
$
hi
you can put $x=u^2$.

3. Or just say $x=16$.

4. Hllo : $x=u^{2}...(x\geq 0)$
I'have :
$
20-u^{2}=\sqrt{20-u}
$

$u^{4}-40u^{2}+u+380=0$
$(u-4)(u+5)(u^{2}-u-19)=0$
(u-4=0)or (u+5=0)or( $u^{2}-u-19=0$)=0
conclusion :
$
u=4
$

but u=-5 is not solution ( $20-u^{2}\geq 0$ )
the equation : $u^{2}-u-19=0$ have not solution in Z.

5. $x=20-\sqrt{20-\sqrt{x}}$

$x - 20=-\sqrt{20-\sqrt{x}}$

$20 - x=\sqrt{20-\sqrt{x}}$

$(20 - x)^2=20-\sqrt{x}$

$400 - 40x + x^2=20-\sqrt{x}$

$400 - 40x + x^2 - 20 + \sqrt{x}=0$

$380 - 40x + x^2 + \sqrt{x}=0$

$x^2 - 40x + \sqrt{x} + 380=0$

Let $u = \sqrt{x}$ :

$u^4 - 40u^2 + u + 380=0$

Solve this fourth degree polynomial (I can't do this yet so just assume you solve this for $u$)

And you find that $u = -5$, $u = 4$, $u = -\frac{\sqrt{77}-1}{2}$, $u = \frac{\sqrt{77}+1}{2}$.

If $u < 0$ is negative, then $\sqrt{x} < 0$, which is not possible, so if $u < 0$, it is not a solution in $\mathbb{Z}$. The last solution is positive, but when squared does not give a relative number, so it is not a solution in $\mathbb{Z}$.

Substitute back the only potential solution into $x$ ( $x = u^2$) to find out that the only solution in $\mathbb{Z}$ is :

$x = 4^2 = 16$

Conclusion : $x=20-\sqrt{20-\sqrt{x}}$ admits one solution in $\mathbb{Z}$ : $x_1 = 16$.

6. Another approach :

Note that the only perfect squares under $20$ are $16$, $9$, $4$ and $1$. Denote $a^2$ any of these squares. If $20 - \sqrt{a^2} = 20 - a = b^2$, $b \in \mathbb{Z}$, then $b^2$ is a solution for $x$.

$20 - \sqrt{1} = 20 - 1 = 19$
$20 - \sqrt{4} = 20 - 2 = 18$
$20 - \sqrt{9} = 20 - 3 = 17$
$20 - \sqrt{16} = 20 - 4 = 16 = 4^2$

If $a^2 = 16$, then we get a perfect square, thus the only solution in $\mathbb{Z}$ for your equation, is $x_1 = 16$.

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 ?

7. Originally Posted by dhiab
Hllo : $x=u^{2}...(x\geq 0)$
I'have :
$
20-u^{2}=\sqrt{20-u}
$

$u^{4}-40u^{2}+u+380=0$
$(u-4)(u+5)(u^{2}-u-19)=0$
(u-4=0)or (u+5=0)or( $u^{2}-u-19=0$)=0
conclusion :
$
u=4
$

but u=-5 is not solution ( $20-u^{2}\geq 0$ )
the equation : $u^{2}-u-19=0$ have not solution in Z.
which means the only solution your equation has in $\mathbb{Z}$ is $x=16$.
you might also consider the way "Bacterius" solved your problem,it's a very good method.