Worked solution: Dehn's invariant: the tetrahedron and cube are not scissors-congruent (1900)

Let $\theta = \arccos(1/3)$. By induction using the identity $\cos((n+1)\theta) = 2\cos\theta\cos(n\theta) - \cos((n-1)\theta)$, one shows $\cos(n\theta) = a_n/3^n$ for an integer $a_n$ with $3 \nmid a_n$, for every $n \ge 1$: the base case $a_1 = 1$ is immediate, and the recurrence $a_{n+1} = 2a_n - 9a_{n-1}$ gives $a_{n+1} \equiv 2a_n \pmod 3$, which is never $0$ when $a_n$ is not, since $2$ is invertible modulo $3$.

Now suppose, for contradiction, that $\theta/\pi = p/q$ for integers $p, q$ with $q \ge 1$. Then $q\theta = p\pi$, so $\cos(q\theta) = \cos(p\pi) = (-1)^p = \pm 1$. But the lemma gives $\cos(q\theta) = a_q/3^q$ with $3 \nmid a_q$, forcing $a_q = \pm 3^q$ — an integer visibly divisible by $3$ since $q \ge 1$, contradicting $3 \nmid a_q$. This contradiction shows $\theta/\pi$ is irrational, i.e. $[\theta] \ne 0$ in $\mathbb{R}/\pi\mathbb{Q}$.

This argument (essentially the case $\cos\theta = 1/3$ of what is now called Niven's theorem on rational values of trigonometric functions at rational multiples of $\pi$) is exactly the missing piece needed to compute $\operatorname{Dehn}$ of a regular tetrahedron in the next step.