The number of roots of a polynomial $p(x)=x^{12}+x^8+x^4+1$ in $\mathbb{F}_{11^2}$.

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 $\beta, \beta^3, \beta^5, \beta^7$.

Notice: $$p(x)= (x^4+1)(x^8+1)$$

Clearly no number divisible by 11 is a root. So suppose $c$ is a root, so $11\nmid c$. By Euler we have $$c^{110} \equiv_{121} 1$$ and $c^4\equiv_{121} -1$ or $c^8\equiv_{121} -1$

In first case we get $c^2\equiv_{121} -1$ and so $c^4\equiv_{121} 1$ which is impossibile.

In second case we get $c^6\equiv_{121} -1$ and so $c^2\equiv_{121} 1$ and so $c^8\equiv_{121} 1$ which is impossibile again.

So it has no root in $F_{121}$.