<s>[INST] How to solve this equation? $y^4-2y^2-\sqrt{y+1}=0$ How to solve this equation?   $y>0$ $$y^4-2y^2-\sqrt{y+1}=0$$  I found one solution that $y=\frac{1+ \sqrt5}{2} $ Can someone help me?
 [/INST] Let $f(y) = \sqrt{y+1} $, we have
$$ f(f(y)) = \sqrt{\sqrt{y+1}+1} = \sqrt{y^4-2y^2+1} = y^2 - 1 $$
and 
$$ f(f(f(y))) = \sqrt{f(f(y))+1} = \sqrt{(y^2-1)+1} = y $$
Because $f(y)$ is an increasing function, the only way to make $f(f(f(y))) = y$ is if  $f(y) = y$
Think about it. If $f(y) > y$ then that must mean $f(f(y) > f(y)$ and $f(f(f(y))) > f(f(y))$, which means $f(f(f(y)))) > y$. Conversely, if $f(y) < y$ then $f(f(f(y))) < y$. So $f(y) = y$
This means 
$$\sqrt{y+1} = y \\
y^2-y+1 = 0$$
Refer to this question also for a similar problem and explanation.
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