AN INTERESTING RELATION

 

I just copy here an answer of mine of this post, just in case OP decided to delete it. 

Link: https://math.stackexchange.com/q/5005043/1231540

First we state two results:

 - **Proposition 1:** Let $n \ge 3$. The minimal polynomial of $\cos(2\pi/n)$ - denoted by $\psi_n$ - has degree $\phi(n)/2.$ Here $\phi$ is Euler totient function.

 - **Proposition 2:** The roots of $\psi_n$ are $\cos(2k\pi/n)$ where $\gcd(k,n)=1$ and $0 \le k \le \lfloor \frac{n}{2}\rfloor$.

The proofs of the above two propositions can be found in this paper: https://www.jstor.org/stable/2324301

***Proof of the first identity:***

It can be observed that $4\sin^2(\pi/n) = 2-4\cos(2\pi/n)$, so the polynomial $\Phi_n$ and $\psi_n$ have the same degree, which is $\phi(n)/2$. In particular, $\Phi_{2m}$ has degree $\phi(2m)/2$. 

If $m$ is odd, we clearly have $\phi(2m) = \phi(m)$, which means $\Phi_{2m}$ and $\Phi_m$ have the same degrees. By definition, $\Phi_{2m}(x)$ is irreducible and has a root $\alpha = 4\sin^2(\pi/2m)$, so it is sufficient to show that $\alpha$ is also a root of $\Phi_{m}(4-x)$. But this is clear as 

\begin{align}\Phi_m(4-\alpha) &= \Phi_m(4\cos^2(\pi/2m))\\ &=\Phi_m(4\sin^2(\pi/2-\pi/2m))\\&=\Phi_m(4\sin^2((m-1)\pi/2m)\end{align}

**Proposition 2** implies that $4\sin^2((m-1)\pi/2m)$ is a root of $\Phi_m$, thus $4-\alpha$ is also a root of $\Phi_m$. Since $\Phi_{2m}(x)$ and $\phi_m(4-x)$ have a common root and the same degree, we must have $$\Phi_{2m}(x) = c \cdot \Phi_m(4-x), $$

for some constant $c$. Comparing the leading coefficients of both sides yields $c = (-1)^{\phi(m)/2}$. 

**Proof of the second identity:**

We first prove an auxiliary identity. For $k \ge 2$, we claim that 

$$\Phi_{2^{k+1}}(x) = f(\Phi_{2^k}(x)) \quad (1)$$ 

where $f(x)=x^2-2$. 

It is easy to verify that the identity $(1)$ holds for $k=2$. Assume that it is also true for $k=n$, we will prove that it is true for $k =n+1$. Let's denote $f^{(n)} = \underbrace{f \circ f \circ \ldots \circ f}_\textrm{$n$-times}$. An inductive argument shows that for $n \ge 1$

$$ f^{(n)}(4\sin^2(\gamma)-2)=2- 4\sin^2(2^n\gamma) \quad (2)$$

Using the induction hypothesis and the identity $(2)$, we get

$$A(\gamma) = f\left(\Phi_{2^n} \right)(4\sin^2(\gamma))= f^{n-1}\left(\Phi_4(4\sin^2(\gamma))\right) = 2-4\sin^2\left(2^{n-1}\gamma)\right) $$

Let $\gamma = \pi/2^{n+1}$, it is immediate that $A(\gamma)=0$. In particular $4\sin^2(\pi/(2^{n+1})$ is a common root of two polynomials $\Phi_{2^{n+1}}$ and $f(\left(\Phi_{2^n})\right)$. Moreover, these two polynomials are of the same degree and leading coefficient, so they must be the same polynomial as one of them is irreducible. 

It is sufficient to show for $m \ge 3$ odd  and $k \ge 2$, the value $\gamma = 4\sin^2(\pi/2^{k}m)$ is also a root of $\Phi_m \circ \Phi^2_{2^k}$, as $\Phi_{2^k m}$ is irreducible and

$$ \deg(\Phi_m \circ \Phi^2_{2^k}) = \phi(2^k m)/2 = \deg(\Phi_{2^k m})$$

We proceed as follows: First apply the auxiliary identity we get

\begin{align}\Phi_{2^k}(\gamma) &= f^{k-2}(\Phi_4)(\gamma))\\&=f^{k-2}(\gamma-2)\\&=2-4\sin^2(2^{k-2}\pi/2^{k}m) \\&=2-4\sin^2(\pi/4m) \\ &= 2\cos(\pi/2m)\end{align}

Thus 

\begin{align}\Phi_m \circ \Phi^2_{2^k}(\gamma) &= \Phi_m(4\cos^2(\pi/2m)) \\&= (-1)^{\phi(m)/2}\Phi_{2m}(4\sin^2(\pi/2m))\\&=0 \end{align}

Here, the second equality comes from the very first identity we proved above, by applying for $x = 4\sin^2(\pi/2m)$. So we have proved $$\Phi_{2^k m}(x)=\Phi_m(\Phi_{2^k}(x)^2) $$

as desired.