I was given the following question:
Let $\mathbb{F}$ be a finite field with $991$ elements. Determine the number of solutions $x \in \mathbb{F}$ to the equation: $$x^{105} = 1$$
My attempt:
$\mathbb{F} \backslash \{0\}$ is an abelian group, $|\mathbb{F} \backslash \{0\}| = 990 = 2\cdot3^2\cdot5\cdot11$
Define the homomorphism: $$\phi: \mathbb{F} \backslash \{0\} \to \mathbb{F} \backslash \{0\} \\ \phi(x) = x^{105}$$
Now the question boils down to find $|K| = |\ker(\phi)|$. Since $K$ is a subgroup of $\mathbb{F} \backslash \{0\}$ (It is a Kernel which is always a subgroup), $|K| \mid 2\cdot3^2\cdot5\cdot11$.
Moreover, $x \in K \iff o(x) \mid 105 = 3\cdot5\cdot7$, and $\forall x :o(x) \mid |\mathbb{F} \backslash \{0\}|$, therefore $$x \in K \iff o(x) \mid 3\cdot5 \iff o(x) = 3,5,15$$
By theorems, if $p$ is a prime and $p \mid |K|$ then $\exists x\in K: o(x) = p$, so $2,11 \nmid |K| \Rightarrow |K| \mid 3^2\cdot5$.
Since $\exists x,y \in \mathbb{F} \backslash \{0\} : o(x) = 3, o(y) = 5 \Rightarrow x, y \in K \Rightarrow 3,5 \mid |K|$.
As conclusion from all of the above, we get $3\cdot5 \mid |K| \mid 3^2\cdot5$, and therefore either $|K| = 15$ or $|K| = 45$.
From here, I couldn't proceed.
One path I tried to follow was to try and determine whether $\exists x \in Im(\phi):o(x) = 3$. If I could prove it, or prove that it is not true, it will solve the problem since $\frac {\mathbb{F} \backslash \{0\}} K \cong Im(\phi)$. This is equivalent to the existence of an element of order $45$ in $\mathbb{F}$, which I also couldn't prove nor disprove.
Another thing that might solve it is to prove/disprove that $K$ is cyclic. (If $K$ is cyclic then $|K| = 15$, if it is not $|K| = 45$), but I couldn't do that as well.
At last, I'm quite bothered that I haven't used yet the fact that $\mathbb{F}$ is a field, and not just a group (of course it is abelian, but that is quite standard also, not nearly as developed as being a field). Maybe there is something about $\mathbb{F}$ being a field that I'm missing?
I feel that I'm terribly close to an answer... Could anyone just point me to the right direction? I will be very happy if so.
BTW: Much thanks if you read all this and got to here, I much appreciate it :)
Hint:
Let $G$ be a cyclic group of order $n$. Let $H \lhd G$ be the subgroup of all elements whose order divides $m$. By Lagrange's Theorem, $|H| \mid n$ as you've seen.
Now note that $H$ is cyclic because it is a subgroup of a cyclic group. Thus $H$ is generated by an element whose order divides $m$.