TITLE: The number of roots of a polynomial $p(x)=x^{12}+x^8+x^4+1$ in $\mathbb{F}_{11^2}$.
QUESTION [7 upvotes]: I am trying to count the number of roots of a polynomial $p(x)=x^{12}+x^8+x^4+1$ in $\mathbb{F}_{11^2}$. Of course, I can plug every element in $\mathbb{F}_{11^2}$ into $x$ in $p(x)$. But is there any simpler way to do this? Thanks in advance!

REPLY [3 votes]: Hint Since $(x^4 - 1) p(x) = x^{16} - 1$ (which has no repeated roots), the roots $x \in \Bbb F_{11^2}$ of $p$ are precisely the elements $x \in \Bbb F_{11^2}^* \cong \Bbb Z_{120}$ that satisfy $x^{16} = 1$ but not $x^4 = 1$.

Any element of $\Bbb Z_{120}$ has order dividing $120$ so any element of order dividing $16$ has order dividing $\operatorname{gcd}(16, 120) = 8$. Thus $p$ has four roots: If $\alpha$ is a generator of $\Bbb Z_{120}$, then $\beta := \alpha^{15}$ generates the subgroup of elements of order dividing $8$ and $\beta^2$ the subgroup of elements of order dividing $4$, so the roots are precisely $\beta, \beta^3, \beta^5, \beta^7$.