$(\zeta_n + 1 + \zeta_n^{-1})(\zeta_n^a + 1 + \zeta_n^{-a})=1$ where both $\zeta_n$ and $\zeta_n^a$ are different primitive $n$-th roots of unity such that the terms are distinct (i.e. $a\ne\pm1$). Also either $n$ is odd or $4$ does not divide $n$.
One solution is $n=10$, $a=3$. Are there any other solution such that $5\nmid n$?
Using Galois theory one can show that $a$ has order 4 in the group $(\mathbb{Z}/n\mathbb{Z})^\times$. So, there is a prime of the form $p\equiv 1 \pmod{4}$ (by decomposing the group into direct product of cyclic groups) which divides $n$. I am trying to show that that prime has to be $5$.