# Cyclotomic Equations

We try to solve the cyclotomic equation $$x^p - 1 = (x-1)(x^{p-1} + x^{p-2} + ... + 1) = 0$$ algebraically. (Transcendentally, 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.

Ben Lynn blynn@cs.stanford.edu 💡