Number Plus One divides Power Plus One iff Odd
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Theorem
Let $q, n \in \Z_{>0}$.
Then:
- $\paren {q + 1} \divides \paren {q^n + 1}$
if and only if $n$ is odd.
In the above, $\divides$ denotes divisibility.
Proof
Let $n$ be odd.
Then from Sum of Odd Positive Powers:
- $\ds q^n + 1 = \paren {q + 1} \sum_{k \mathop = 1}^n \paren {-1}^k q^{k - 1}$
Let $n$ be even.
Consider the equation:
- $q^n + 1 = 0$
We have:
- $\paren {-1}^n + 1 = 2$
So $-1$ is not a root of $q^n + 1 = 0$.
By Polynomial Factor Theorem, $q + 1$ is not a divisor of $q^n + 1$ when $n$ is even.
$\blacksquare$
Sources
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 2.4$: The rational numbers and some finite fields: Further Exercises $8$