Factor it 3

We know that the form $x^{n} - 1$ denotes a difference of two squares. The positive factor left over for the difference of two squares process is irreducible over the rationals, thus only the negative factors can break up into more factors. Let $k$ be the number of irreducible factors over the rationals. If we do some casework, $x^{2} - 1 = (x + 1)(x-1) \Longrightarrow k = 2$ $x^{4} - 1 = (x^{2} + 1)(x^{2}-1) = (x^{2} + 1)(x+1)(x-1) \Longrightarrow k = 3$ $x^{8} - 1 = (x^{4} + 1)(x^{4}-1) = (x^{4} + 1)(x^{2}+1)(x^{2}-1)= (x^{4} + 1)(x^{2}+1)(x+1)(x-1) \Longrightarrow k = 4$

Through induction, we can conclude that every $n$ that is a power of two equals $2^{k-1}$ . $n=2^{5-1} \Longrightarrow n= \boxed{16}$