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?
 A: $y^4 - 2y^2 = \sqrt{y + 1}$, and this gives: $y^8 - 4y^6 + 4y^4 - y - 1 = 0$. Observe that $y^2 - y - 1$ is a factor of the left hand side of the equation. Using long division we can find all the zeroes.
A: This is actually solvable without guessing the root if we try two approaches to the problem:
The first is to get rid of the square root by squaring the expression like this:
$$y^4-2y^2=\sqrt{y+1}\Longrightarrow (y^4-2y^2)^2=y+1\Longleftrightarrow y^8-4y^6+4y^2-y-1=0$$
The second is to get rid of the square root by setting $x=\sqrt{y+1}$, so $y=x^2-1$ and $x>1$:
$$(x^2-1)^4-2(x^2-1)^2=x\Longleftrightarrow x^8-4x^6+6x^4-4x^2+1-2x^4+4x^2-2=x\Longleftrightarrow\\
x^8-4x^6+4x^4-x-1=0$$
There's something very suspicious about it. The polynomial is exactly the same as in the first case!
This means that if $x>1$ is a root, then $y=x^2-1$ is a root too! So what if the roots were actually the same? Then the root would be the solution of
$$x=x^2-1\Longleftrightarrow x^2-x-1=0$$
Trying to factor this out leads to success, indeed by polynomial division we get
$$x^8-4x^6+4x^4-x-1=(x^2-x-1)(x^6+x^5-2x^4-x^3+x^2+1)$$
Continuing the second approach, we see that $\varphi=\dfrac{1+\sqrt5}2$ which is the positive root of $x^2-x-1$ satisfies $\varphi>1$. Any other root would have to be the root of $x^6+x^5-2x^4-x^3+x^2+1$, which is not possible, because by AM-GM inequality $x^6+x^2\ge2\sqrt{x^6x^2}=2x^4$, so
$$x^6+x^5-2x^4-x^3+x^2+1\ge x^5-x^3+1>0$$
because $x^5>x^3$ for $x>1$ (it actually holds for $1\ge x\ge0$ too, because then $|x^3(x^2-1)|<1$)
So the only solution $y>0$ (and even $y\ge-1$) is $y=\varphi^2-1=\varphi=\dfrac{1+\sqrt5}2$.
A: 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.
