## Cyclotomic Equations

We try to solve the cyclotomic equation $x^p - 1 = (x-1)(x^{p-1} + x^{p-2} + ... + 1) = 0$ algebraically. (Transcendentally, we know the roots are $e^{2\pi i k / p}$ for $k=0,...,p-1$.)

It can be easily shown that if $\gcd(m, n) = 1$, then a primitive $m$th root of unity times a primitive $n$th root of unity is a primitive $m n$th root of unity, thus we need only consider prime powers. But then if $\alpha$ is a primitive $p$th root of unity, then $\sqrt[k]{\alpha}$ is a primitive $p^k$th root of unity, so we need only consider the case where $p$ is prime.

In general we can use Gauss' method, but let us see how far elementary methods lead us.

$p = 3$: we merely solve the quadratic $x^2 + x + 1 = 0$ to obtain

$x = \frac{-1\pm i\sqrt{3}}{2}$

$p = 5$: we could solve the quartic $x^4 + x^3 + x^2 + x + 1 = 0$ but since it is palindromic we make the variable substitution $y = x + 1/x$, and solve

$y^2 + y - 1 = 0$

to find

$y = \frac{-1 \pm \sqrt{5}}{2}$

and $x^2 - y x + 1 = 0$ implies

$x = \frac{y \pm \sqrt{y^2 - 4}}{2}$

giving the four solutions

$x = \frac{\sqrt{5} - 1 \pm \sqrt{-2\sqrt{5}-10}}{4} , \frac{-\sqrt{5} - 1 \pm \sqrt{2\sqrt{5}-10}}{4}$

$p = 7$: the palindrome yields a cubic which can be solved for $x$.

$p = 11$: the palindrome yields a quintic. Now elementary methods fail us and we need resort to Gauss' method as Vandermonde did.