Find all $4<n,m\in\mathbb N$ for which $\cos^2(\frac{2\pi}n)=\cos(\frac{2\pi}m).$

Question

Find all $4<n,m\in\mathbb N$ for which
$\cos^2\left(\frac{2\pi}n\right)=\cos\left(\frac{2\pi}m\right).$

I think it has one non-trivial solution, $n=8,m=6$. Intuitive idea for a proof: $\cos(x)\approx1-\frac{x^2}{2}$.

$\left(1-\left(\frac{2\pi}{n}\right)^2\right)^2\approx1-\left(\frac{2\pi}{m}\right)^2$.

$1-2\left(\frac{2\pi}{n}\right)^2+\left(\frac{2\pi}{n}\right)^4\approx1-\left(\frac{2\pi}{m}\right)^2$; $\sqrt2 m\approx n$.

On the other hand, $\mathbb Q\to\mathbb Q\left[\cos\left(\frac{2\pi}{m}\right)\right]\to \mathbb Q\left[\sqrt{\cos\left(\frac{2\pi}{m}\right)}\right]$ is of degree $2\varphi(m)$. So, $2\varphi(m)=\varphi(n)$.

From here, I think it is possible to formalize these arguments and arrive at a contradiction for large enough $m,n$.

Answer 1

Start by observing that $\cos(2 \pi/r)$ is an algebraic number and its conjugates are exactly the numbers $\cos(2 \pi k/r)$ for $(k,r) = 1$.

Since the conjugates of $\alpha^2$ are the squares of the conjugates of $\alpha$, it follows that the conjugates of the LHS are all positive real numbers.

OTOH, the conjugates of the RHS include $\cos(2 \pi k/m)$ for all $(m,k) = 1$. Hence, if there exists a $k$ with $(k,m)=1$ and

$$m/4 < k < 3m/4,$$

then $\cos(2 \pi k/m) < 0$ which would be a contradiction, implying that there cannot be any solutions to the original equation.

Define $k$ as follows:

1. If $m$ is odd, then take $k = (m+1)/2$.
2. If $2|m$ and $m/2$ is odd, then take $k = (m+4)/2$.
3. If $4|m$, then take $k = (m+2)/2$.

It's easy to verify that $(k,m)=1$, since $2k-m \in \{1,2,4\}$ so the only common factor could be $2$, but $k$ is odd when $m$ is even.

It's also easy to verify the required inequality holds unless $m=1$, $m=2$, or $m=6$. For $m=1$ one has $n=1,2$, for $m=2$ the RHS is negative, and for $m=6$ there is the solution $n=8$.

Answer 2

We note that if $k \ne 2,3,4,6, k \ge 2$ the real field $Q[\cos 2\pi/k]$ is not $Q$ since $\phi(k) >2$ and the algebraic degree of $\cos 2\pi/k$ is $\phi(k)/2$.

Now the formula for the discriminant of $Q[\cos 2\pi/k]$ immediately shows that the only two cases where $Q[\cos 2\pi/k]= Q[\cos 2\pi/l]$ with $k \ge l$ as above (meaning $Q[\cos 2\pi/k] \ne \mathbb Q$) are if $k=l$ or if $k=2l, l$ odd

(one implication is obvious as if $l$ odd, $\cos \pi/l$ is $-\cos m(2\pi/l)$ for some $m$ so the two fields are equal, but for the other, some AGNT may be needed)

Now we can solve the problem, noting that $\cos^2\left(\frac{2\pi}n\right)=\cos\left(\frac{2\pi}m\right)$ means $\cos 4\pi/n=2\cos 2\pi/m-1$ so if $n$ odd then $Q[\cos 2\pi/n]= Q[\cos 2\pi/m]$ since as above $\cos 4\pi/n$ generates $\cos 2\pi/n$ algebraically, so since $n \ge 5$ we must have $m=n$ or $m=2n$ and either is easily seen impossible since $n > m$ because $0<\cos 2\pi/n <1$

If $n=2k$ we get $Q[\cos 2\pi/k]= Q[\cos 2\pi/m]$ so if they are not $\mathbb Q$ we must have $k=m$ or $k=2m$ or $m=2k$ so $n=m,2m,4m, n >4$ and we can immediately exclude $n=m, n=2m$ since $\cos \pi/m \ne 0,1$ as $n>4$ means $m>2$, while if $n=4m$ we get $\cos \pi/m=2\cos 2\pi/m-1=4\cos^2 \pi/m-3$ and the equation $x=4x^2-3$ has roots $1, -3/4$ which are not $\cos \pi/m$ for any integer $m \ge 1$ since for $m \ge 2$ we have $\cos \pi/m \ge 0$

Hence the only case remains where $m,k$ are $2,3,4,6$ and by inspection the only solution for which $n>4$ is indeed $n=8, m=6$