If $y =\sqrt{5+\sqrt{5-\sqrt{5+ \cdots}}}$, what is the value of $y^2-y$? If $$y =\sqrt{5+\sqrt{5-\sqrt{5 + \cdots}}},$$
what is the value of $y^2-y$ ?
I am unable to get the clue due to these alternative signs of plus and minus please help on this thanks...
 A: Hint: 
$$\begin{align}
(y^2 -5)^2 - 5 &= -y \\
y^4 - 10y^2 + 25 - 5 + y &= 0 \\
y^4 -10y^2 + y + 20 &= 0.
\end{align}
$$
Edit: To elaborate a bit on the comments given below: you have produced a depressed quartic equation. As explained on the Wikipedia article you can try to factor this into quadratics
$$
y^4 -10y^2 + y + 20 = (y^2 +py + q)(y^2 + ry + s).
$$
All you have to do is find $p,q,r$, ands $s$. You get
$$\begin{align}
y^4 -10y^2 + y + 20 &= (y^2 +py + q)(y^2 + ry + s)\\
&=  y^4 + (p + r)y^3 + (q + s + pr)y^2 + (ps + qr)y + qs.
\end{align}
$$
Now put coefficients equal to each other and solve ...
A: Indeed, there is another way:
Let $x =\sqrt{5-\sqrt{5+\sqrt{5 - \cdots}}}$,
then it's easy to see 
$y^2 = 5 + x$, $x^2 = 5 - y$ 
then 
 $y^2 - x^2 = x+y$, therefore $y-x = 1$ 
Thus,
$$y^2 - y = y^2 - 1 - x = 5 + x - (1+x) =4$$
A: Consider $z_{-+} = -y = -\sqrt {5 + \sqrt{5 - \ldots}}$.
$z_{-+}$ is clearly a solution to the equation $(z^2 -5)^2-5 = z$,
and the other three solutions of this degree $4$ polynomial equation, should be
$z_{+-} = +\sqrt {5 - \sqrt{5 + \ldots}}, z_{++} = +\sqrt {5 + \sqrt{5 + \ldots}}$, and $z_{--} = -\sqrt {5 - \sqrt{5 - \ldots}}$ .
But $z_{++}$ and $z_{--}$ are in fact solution to a simpler polynomial equation, which is $z^2-5 = z$.
So $z^2-z-5$ has to be a factor of $(z^2-5)^2-z-5$, and doing the division, you end up with $(z^2-5)^2-z-5 = z^4-10z^2-z+20 = (z^2-z-5)(z^2+z-4)$, hence $z_{-+}^2 + z_{-+} - 4 = 0$.
From there you conclude that $y^2-y = z_{-+}^2 + z_{-+} = 4$.
