# Number of roots of a particular polynomial in $F_{121}$

Question: Determine the number of roots of $f(x)=x^{12}+x^8+x^4+1$ in $\mathbb{F}_{121}$.

I was told that a solution to this problem could, or perhaps should, use the fact that $\mathbb{F}_{121}^\times$ is cyclic, but I don't currently see how this helps. My attempt at a solution is written below, but I don't think it's the right approach. How does the fact that $\mathbb{F}_{121}^\times$ is cyclic play a part?

My thoughts: First, notice that if $g(x)=x^3+x^2+x+1=(x^2+1)(x+1)$ then $g(x^4)=f(x),$ so $\beta\in \mathbb{F}_{121}$ is a root of $f$ if and only if $\beta^4=\alpha$ for some $\alpha\in \mathbb{F}_{121}$ that is a root of $g$. In particular, if $\alpha$ is a root of $g$ and $\alpha\in (\mathbb{F}_{121})^{\times 4}$ then $\alpha$ will correspond to a root of $f$ (the 'fourth root' of $\alpha$ in $\mathbb{F}_{121}$). Since $y^2+1\in\mathbb{F}_{11}[y]$ is irreducible, we know that $$\mathbb{F}_{121}\cong \frac{\mathbb{F}_{11}[y]}{(y^2+1)}\cong \mathbb{F}_{11}(i)$$ where $i^2=-1$. With this notation, we find that $g(x)=(x+1)(x+i)(x-i)\in \mathbb{F}_{11}(i)$. So we need to check whether $-1,\pm i$ are in $(\mathbb{F}_{121}^\times)^4$. At this point, I am not so sure how to proceed beyond checking for solutions to $(a+bi)^4=c$ for $a,b\in \mathbb{F}_{11}$ and $c\in \{-1,\pm i\}$, but I think there must be a better way... As a side note, I did a computation in PARI/GP and found that $x^{12}+x^8+x^4+1$ has $4$ roots, so we know how many there should be, but I'm at a loss as to how this can be computed nicely by hand.

• The roots of $f$ are the 16-th roots of unity excluding the 4-th roots of unity. Commented Feb 24, 2018 at 23:21
• Can you elaborate? Commented Feb 24, 2018 at 23:27
• That is mainly because $f(x) = (x^{16}-1)/(x^4-1)$. Commented Feb 24, 2018 at 23:28
• Ah, fantastic! I didn't catch that! So the number of roots of $f$ in $\mathbb{F}_{121}$ is just the number of 16th roots of unity in $\mathbb{F}_{121}$ minus the number of 4th roots of unity. Great! Commented Feb 24, 2018 at 23:34

As you said $\mathbb F_{121}^\times$ is cyclic, so it is isomorphic to $\mathbb Z/120\mathbb Z$ as a group, and $120 = 2^3 \times 3\times 5$.
Also $f(x) = \frac{x^{16}-1}{x^4-1}$, so the roots of $f(x)$ are the 16-th roots of unity minus 4-th roots of unity. $\mathbb Z/120\mathbb Z$ has 8 elements of order 8, 4 of which has order 4, but none of order 16. The answer is 4.
On the other hand, your calculations $f(x) = (x^8 + 1)(x^4 + 1)$ also gives much information: If $x^8 + 1 = 0$, then $x^{16} = 1$, so such $x$'s are the primitive 16-th roots of unity, and there are none. If $x^4 + 1 = 0$, then $x^{8} = 1$, so such $x$'s are the primitive 8-th roots of unity, and there are 4 of them.