Given that the following polynomial\begin{equation*}f(x)=x^8-x^4+1\end{equation*}
is a cyclotomic polynomial $\Phi_n$ for some $n\in \mathbb N$. are there some basic tools to determine $n$?
I know that since $f(x)$ has degree 8, $n$ must have 8 relative primes smaller then $n$. for example, $n$ can be 16 or 24 since both have 8 relative primes smaller then them. I know that $n\not= 16$ since it is easier to determine that \begin{equation*}\Phi_{16}=x^8+1\end{equation*} since every root of the polynomial is 8-th root of $-1$ and hence 16-th root of $1$. but this trick does not seem to work for $f(x)$.
2026-03-26 22:52:37.1774565557
On
On
determining which cyclotomic polynomial is $x^8 -x^4+1$
167 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
4
There are 4 best solutions below
0
On
Just some trial and error here:
$$ (a^2 - a + 1)(a^2 + a + 1)(a^2 -1) = (a^4 + a^2 + 1)(a^2 -1) = a^6 - 1$$
In your case $a = x^4$, so the equation is $x^{\color{blue}{24}}-1$. Indeed $\phi(24) = \phi(3)\phi(8) = 2\cdot 4 = 8$.
0
On
Solving for $x^4,$
$$x^4=\dfrac{1\pm\sqrt{1-4}}2=\dfrac{1\pm\sqrt3i}2=e^{(2n\pi\pm\pi/3)i}$$ where $n$ is any integer
$$\implies x=e^{(2n\pm1)\pi i/12}$$ where $n\equiv0,1,2,3\pmod4$
$$\implies x^r=e^{(2n\pm1)r\pi i/12}$$
We need $(2n\pm1)r\pi /12=m\pi$ where $m$ is any integer
$\iff(2n\pm1)r=12m\implies12|r $ if $12|(2n\pm1)r\forall n$ in $[0,3]$
The $n$-th cyclotomic polynomial has degree $\varphi(n)$, hence we have to solve $\varphi(n)=8$.
Suppose $n=2^r\prod_\limits{p_i\:\text{odd}}p_i^{r_i}$. Then $$\varphi(n)=2^{r-1}\prod_{p_i\:\text{odd}}p_i^{r_i-1}(p_i-1),$$ If some $r_i>0$, $\varphi(n)$ is divisible by $p_i-1$, so $\varphi(n)$ is a power of $2$ only if $r_i=0$ or $1$, and for each $r_i>0$, the corresponding $p_i-1$ is a power of $2$, in other words $p_i$ is a Fermat prime. As it must be less than $8$, we have $4$ possibilities:
Thus the solution is $\Phi_{24}$.