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.


Ben Lynn blynn@cs.stanford.edu 💡