IMC 1996 · Day 2 · P8

First we show that

(1) If $\cosh t$ is rational and $m \in \mathbb{N}$, then $\cosh mt$ is rational.

Since $\cosh 0 \cdot t = \cosh 0 = 1 \in \mathbb{Q}$ and $\cosh 1 \cdot t = \cosh t \in \mathbb{Q}$, (1) follows inductively from $$\cosh(m+1)t = 2 \cosh t \cdot \cosh mt - \cosh(m-1)t.$$ The statement of the problem is obvious for $k = 1$, so we consider $k \ge 2$. For any $m$ we have $$\tag{2} \begin{aligned} \cosh \theta &= \cosh((m+1)\theta - m\theta) = \\ &= \cosh(m+1)\theta \cdot \cosh m\theta - \sinh(m+1)\theta \cdot \sinh m\theta \\ &= \cosh(m+1)\theta \cdot \cosh m\theta - \sqrt{\cosh^2(m+1)\theta - 1} \cdot \sqrt{\cosh^2 m\theta - 1} \end{aligned}$$ Set $\cosh k\theta = a$, $\cosh(k+1)\theta = b$, $a, b \in \mathbb{Q}$. Then (2) with $m = k$ gives $$\cosh \theta = ab - \sqrt{a^2 - 1} \sqrt{b^2 - 1}$$ and then $$\tag{3} \begin{aligned} (a^2 - 1)(b^2 - 1) &= (ab - \cosh \theta)^2 \\ &= a^2 b^2 - 2ab \cosh \theta + \cosh^2 \theta. \end{aligned}$$ Set $\cosh(k^2 - 1)\theta = A$, $\cosh k^2 \theta = B$. From (1) with $m = k - 1$ and $t = (k+1)\theta$ we have $A \in \mathbb{Q}$. From (1) with $m = k$ and $t = k\theta$ we have $B \in \mathbb{Q}$. Moreover $k^2 - 1 > k$ implies $A > a$ and $B > b$. Thus $AB > ab$. From (2) with $m = k^2 - 1$ we have $$\tag{4} \begin{aligned} (A^2 - 1)(B^2 - 1) &= (AB - \cosh \theta)^2 \\ &= A^2 B^2 - 2AB \cosh \theta + \cosh^2 \theta. \end{aligned}$$ So after we cancel the $\cosh^2 \theta$ from (3) and (4) we have a non-trivial linear equation in $\cosh \theta$ with rational coefficients.