Factorization of $x^n + y^n$, what sort of coefficients show up?

Question

We know that$$a^2 + b^2 = (a + bi)(a - bi).$$ What are the complete factorizations of $a^3 + b^3$, $a^4 + b^4$, $\ldots$ , $a^k + b^k$, etc.? What sort of coefficients show up?

Answer 1

You could always write $a^n+b^n = a^n - c^n$ where $c = \xi_n b$, where $\xi_n = e^{\pi i/n}$ (so that $\xi_n^n = -1$). Then fixing $a$, the roots of $a^n - c^n$ are solutions to $a^n = c^n$ and so there are $n$ solutions for $c$; namely, the numbers in the form of $\zeta_n^k a$ where $0\leq k\leq n-1$ and $\zeta = e^{2\pi i/n}$ so that $\zeta_n^n = 1$. So the original polynomial factors completely over $\Bbb{C}$ as $$(c - a)(c - \zeta_na)(c - \zeta_n^2a)...(c - \zeta_n^{n-1}a)$$

Such $\zeta_n$ is called an $n^{th}$ root of unity. There are always $n$ $n^{th}$ roots of unity, namely, the distinct powers of $\zeta_n$.

Answer 2

The problem simplifies if you divide through by $a$ and lent $\frac{b}{a} = x$. Once you have the factors of $1+x^n$ it is easy to re-introduce $a$ and $b$.

The pattern in these is easy to see. For odd $n$,

$$a^3 + b^3 = (a+b) (a+b\mbox{ cis } (\frac{2\pi}{3})) (a+b\mbox{ cis }(\frac{4\pi}{3})) $$

$$a^5 + b^5 = (a+b) (a+b\mbox{ cis } (\frac{2\pi}{5})) (a+b\mbox{ cis }(\frac{4\pi}{5})) (a+b\mbox{ cis } (\frac{6\pi}{5})) (a+b\mbox{ cis }(\frac{8\pi}{5}))$$

and so forth. For even $n$, $$a^4 + b^4 = (a+b\mbox{ cis } (\frac{1\pi}{4})) (a+b\mbox{ cis } (\frac{3\pi}{4})) (a+b\mbox{ cis } (\frac{5\pi}{4})) (a+b\mbox{ cis } (\frac{7\pi}{4}))$$

and so forth.

You can in fact express the full factorization in terms of easy to swallow radicals for $n=2,3,4,5,6,8$. Also for $n=17$ for example; see Fermat primes.