Show that \(x^n + 1\) is divisible by \(x + 1\) when, and only when, \(n\) is an odd integer.
By the factor theorem, given a polynomial \(p(x)\) with real coefficients and some real number \(a\), we know that \(x-a\) is a factor of \(p(x)\) if and only if \(p(a) = 0\).
If we let \(p(x) = x^n + 1\) and \(a = -1\), by the factor theorem we have that \[\begin{align*} \text{$x^n + 1$ is divisible by $x + 1$} &\iff (-1)^n + 1 = 0 \\ &\iff (-1)^n = -1 \\ &\iff \text{$n$ is an odd integer}. \end{align*}\]
