I read a math article on the net that stated:
For the nested radical equation
$$\pm \sqrt{n\pm \sqrt{n\pm\sqrt{n\pm \sqrt{n+x}} }}=x$$
By repeated squaring we get
$$\left( \left( \left( x^2-n \right)^2-n \right)^2-n \right)^2-x-n=0$$
The 16 roots of the polynomial are the solutions for the nested radical equation.
This polynomial can be decomposed into 4 quartics. First one is
$$\left( x^2-n \right)^2-x-n=0$$
The other three quartics had coefficients in the cubic
$$y^3+3y=4(1+ny)$$
The 12 roots are
$$x=-\frac{y-z}{4}\pm \frac{1}{2}\sqrt{\frac{(y-2)(y+z)z}{2y}},\quad z=\pm \sqrt{y^2+4}$$
3 choices of y and 4 combinations of $\pm$ sign make 12 solutions.
My question is : how to derive the coefficients of the cubic?
I tried to get the product of the 3 quartics by calculating
$$\frac{\left( \left( \left( x^2-n \right)^2-n \right)^2-n \right)^2-x-n}{\left( x^2-n \right)^2-x-n}=x^{12}-6n\cdot x^{10}+x^9+\mathcal{O}(x^8)$$
Suppose the 3 quaritcs are
$$x^4+a_1x^3+b_1x^2+c_1x+d_1=0 \\ x^4+a_2x^3+b_2x^2+c_2x+d_2=0 \\ x^4+a_3x^3+b_3x^2+c_3x+d_3=0 $$
and $a_1, a_2, a_3$ are 3 roots of the coefficients' cubic.
The product of the 3 quartics give the polynomial
$$x^{12}+(a_1+a_2+a_3)x^{11}+(a_1a_2+a_2a_3+a_1a_3+b_1+b_2+b_3)x^{10}+\\ (a_1a_2a_3+a_1b_2+a_1b_3+a_2b_1+a_2b_3+a_3b_1+a_3b_2+c_1+c_2+c_3)x^{9}+\mathcal{O}(x^8)$$
Comparing the coefficients with the original 12th degree polynomial above, apparently $a_1+a_2+a_3=0$.
We need to derive the following equalities to match with the original polynomial.
$$ a_1a_2+a_2a_3+a_1a_3=-4n+3 \\ b_1+b_2+b_3=-2n-3 \\ a_1a_2a_3=4 \\ a_1b_2+a_1b_3+a_2b_1+a_2b_3+a_3b_1+a_3b_2=-4n-3 \\ c_1+c_2+c_3=4n $$
How to derive these equalities to obtain the coefficients' cubic?
Applying the infinite iterations, we have:
$$\begin{align}\pm \sqrt{n\pm \sqrt{n\pm\sqrt{n\pm \sqrt{n\color{#c00}{\pm \sqrt{n\pm \sqrt{n\pm\sqrt{n\pm \sqrt{n±\sqrt{\cdots}}}}} }}}}}=x\end{align}$$
This tells us that, if the original equation has a real root, then it also has a root of the following equation:
$$ \begin{align}&±\sqrt {n+x}=x\\ \implies &n+x=x^2\\ \implies &x^2-x-n=0\end{align} $$
Now, we will shows that the real roots of the quadratic are equivalent to the original equation. Indeed, using a finite number of iterations and reversing the steps, we have:
$$\begin{align}&±\sqrt {n+x}=x\\ \implies &±\sqrt {n±\sqrt {n+x}}=x\\ \implies &±\sqrt {n±\sqrt {n±\sqrt {n+x}}}=x\\ \implies &±\sqrt{n±\sqrt{n±\sqrt{n± \sqrt{n+x}} }}=x.\end{align} $$
Therefore, we only need to solve the quadratic:
$$ \begin{align}&x^2-x-n=0\\ \implies &x_{1,2}=\frac {1\color{#c00}{±}\sqrt {1+4n}}{2}\end{align} $$
If all the signs are $«+»$ in the iteration, then the real root equals to:
$$x=\frac {1\color{#c00}{+}\sqrt {1+4n}}{2}$$
If all the signs are $«-»$, in the iteration, then the answer is:
$$x=\frac {1\color{#c00}{-}\sqrt {1+4n}}{2}$$
where $n\geqslant-\frac 14$.