Irish Mathematical Olympiad, 2014

Let $x = \sqrt{2}-1$, $t = m^\circ$. Then $x$ is irrational and $x^2 = 3-2\sqrt{2} = 1-2x$. We would like to establish that all the numbers $\cos(2nt)$ are distinct for $n \ge 1$. This will follow from the following claim.

Claim: The sequences of integers $(a_n)$ and $(b_n)$ given recursively by
$$\begin{align*} a_1 &= 1 & a_{k+1} &= 2a_k^2 + 2b_k^2 - 1 \\ b_1 &= 4 & b_{k+1} &= 4b_k(a_k + b_k) \end{align*}$$
are strictly increasing and, for each positive integer $n$, they satisfy
$$\cos(2nt) = a_n - b_n x.$$
Induction easily establishes that the two sequences are strictly increasing: Because $a_k \ge 1$ implies $2a_k^2 > a_k$, and $b_k > 1$ implies $2b_k^2 - 1 > 0$, we get $a_{k+1} = 2a_k^2 + 2b_k^2 - 1 > a_k$. Also, $4(a_k+b_k) > 1$, and so $b_{k+1} = 4b_k(a_k+b_k) > b_k$. We now establish the equality $\cos(2nt) = a_n - b_nx$ by induction on $n$. For $n=1$, we have $\cos(2t) = 2\cos^2(t) - 1 = 2x^2 - 1 = 2(1-2x) - 1 = 1-4x$, as required.

Assume $\cos(2kt) = a_k - b_kx$ for some $k \ge 1$. Then we have
$$\begin{align*} \cos(2^{k+1}t) &= 2 \cos^2(2^k t) - 1 = 2(a_k - b_k x)^2 - 1 \\ &= 2a_k^2 - 4a_k b_k x + 2b_k^2 x^2 - 1 \quad \text{(now use } x^2 = 1-2x) \\ &= (2a_k^2 + 2b_k^2 - 1) - (4a_k b_k + 4b_k^2) x \\ &= a_{k+1} - b_{k+1}x \end{align*}$$
and the claim is fully established.

It follows that all the numbers $\cos(2nt)$ are distinct for $n \ge 1$, since if $\cos(2pt) = \cos(2^q t)$, for positive integers $p < q$, then $a_p - xb_p = a_q - xb_q$, with $b_p \ne b_q$ and $x = (a_p - a_q)/(b_p - b_q)$ is rational, which is impossible.

However, note that $2^{12}-1 = (2^6+1)(2^6-1) = (2^2+1)(2^4-2^2+1)(2^3-1)(2^3+1)$ is divisible by $(2^2+1)(2^3+1) = 5 \cdot 9 = 45$. Therefore, for all integers $n \ge 3$ the number $2^{n+12} - 2^n$ is divisible by 360. So, if $m$ is an integer and $n \ge 3$, $2^{n+12}m - 2^m n$ is divisible by 360 and thus $\cos(2^{n+12}t) = \cos(2^m t)$, for $n \ge 3$. This contradiction shows that $m$ is not an integer.