Roots of unity and a system of equations by Ramanujan

Question

Is it immediately apparent that the solution to the system of equations,

$$\begin{aligned} x_1^2 &= x_2+2\\ x_2^2 &= x_3+2\\ x_3^2 &= x_4+2\\ &\vdots\\ x_n^2 &= x_1+2\\ \end{aligned}\tag{1}$$

can be given by the roots of unity? Specifically,

$$x =\frac{y_k^2+1}{y_k}\tag{2a}$$

where the $y_k$ are,

$$\begin{aligned} y_k &= \exp\Big(\frac{2\pi i k}{2^n-1}\Big),\; k = 0\dots 2^{n-1}-1\\ y_k &= \exp\Big(\frac{2\pi i k}{2^n+1}\Big),\; k = 1\dots 2^{n-1}\\ \end{aligned}\tag{2b}$$

Example. Let $n=4$. Then (1) is equivalent to,

$$x = (((x^2-2)^2-2)^2-2)^2-2\tag{3}$$

Expanded out, (3) is a $2^4=16$-deg polynomial and its 16 roots are given by (2) where,

$$y_k = \exp\Big(\frac{2\pi i k}{15}\Big),\; k = 0\dots 7$$ $$y_k = \exp\Big(\frac{2\pi i k}{17}\Big),\; k = 1\dots 8$$

Ramanujan considered the system (1) for $n=3,4$ in the general case and also as nested radicals. For $x = (((x^2-a)^2-a)^2-a)^2-a$, see this related post. (Interestingly, $n=5$ in the general case is no longer completely solvable in radicals.)

Question:

I observed (2) empirically. How do we prove from first principles that this is indeed the solution?

Let $f(x)=x^2-2$. Then $f(2\cos u)=2\cos2u$. So, $f^n(2\cos u)=2\cos2^nu$. Then $f^n(x)=x$ gives $\cos2^nu=\cos u$. — Gerry Myerson, Jan 04 '14 at 03:15
Oh, I recognize that $\Re(e^{2\pi i k/p}) = \cos(m)$, but I can't follow the rest completely. Can you give a detailed version as an answer? — Tito Piezas III, Jan 04 '14 at 16:35

score 4 · Accepted Answer · answered Jan 04 '14 at 16:42

4

Let $f(x)=x^2-2$. Then $$f(2\cos u)=4\cos^2u-2=2(2\cos^2u-1)=2\cos2u$$ so $$f^n(2\cos u)=2\cos2^nu$$ Then $f^n(x)=x$ becomes $$\cos u=\cos2^nu$$ for which solutions are given by $$(2^n-1)u=2k\pi,k=0,\pm1,\pm2,\dots$$

answered Jan 04 '14 at 16:42

Gerry Myerson

179,216

For $n=4$, this gives $u = 2k\pi/15$. How do we account for the other half of solutions given by $u = 2k\pi/17$? – Tito Piezas III Jan 04 '14 at 16:46
$\cos x=\cos y$ has solutions other than $x-y$ a multiple of $2\pi$, namely, those with $x+y$ a multiple of $2\pi$. I Didn't claim my answer would give all the solutions, but I think that with this modification, it may. – Gerry Myerson Jan 04 '14 at 16:50
Ok, but it is certainly a good start. – Tito Piezas III Jan 04 '14 at 16:52

Roots of unity and a system of equations by Ramanujan

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