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 $e^{2\pi ik/p}$ 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 $mn$th root of unity, thus we need only consider prime powers. But then if $\alpha$ is a primitive $p$th root of unity, then $\sqrt[k-1]{\alpha }$ 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 $x^2-yx+1=0$ 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. Vandermonde did just that, and we outline his answer in preparation for later.

Set $\beta$ to a primitive $10$th root of unity (so it is the negation of one of the solutions for the $p=5$ case above).

Suppose $\zeta$ is a primitive $11$th root of unity. Notice that 2 is a generator of ${\mathbb{Z}}_{11}^{*}$ as the powers of 2 are $2,4,8,5,10,9,7,3,6,1$. Define $$t_1=\zeta +\beta {\zeta }^2+{\beta }^2{\zeta }^4+...+{\beta }^9{\zeta }^6$$ and for $t_2,...,t_{10}$ define $t_i$ equal to $t$ with $\beta$ replaced by ${\beta }^i$. We shall see that $t_i^{10}$ can be expressed in terms of $\beta$ easily for all $i$.

Then $\zeta =\frac{1}{10}(t_1+...+t_{10})=\frac{1}{10}(\sqrt[10]{t_1^{10}}+...+\sqrt[10]{t_{10}^{10}})$ where we need to choose the correct 10th roots to find $\zeta$. It turns out that $t_i{t}_1^{10-i}$ is also easily expressed in terms of $\beta$, hence we may pick any 10th root of $t_1$ and then derive the values of $t_2,...,t_{10}$.