Number Plus One divides Power Plus One iff Odd

Theorem
Let $q, n \in \Z_{>0}$.

Then:
 * $\left({q + 1}\right) \mathrel \backslash \left({q^n + 1}\right)$

$n$ is odd.

In the above, $\backslash$ denotes divisibility.

Proof
Let $n$ be odd.

Then from Sum of Odd Positive Powers:
 * $\displaystyle q^n + 1 = \left({q + 1}\right) \sum_{k \mathop = 1}^n \left({-1}\right)^k q^{k - 1}$

Let $n$ be even.

Consider the equation:
 * $q^n + 1 = 0$

By elementary complex analysis:
 * $q \in \left\{{e^{\left({2 k + 1}\right) i \pi / n}: 0 \le k < n}\right\}$

from which it is apparent that $-1$ is not among these roots.

Thus $q + 1$ is not a divisor of $q^n + 1$ when $n$ is even.