The sum of the solutions of $x^3 ≡ 8 \mod q$.

Question

Let $q$ be a prime equal to $1 \mod 12$ and $\lambda_1, \lambda_2, \ldots , \lambda_k$ denote the solutions of $x^3 ≡ 8 \mod q$.

How would I find the sum of these solutions $\mod q$? I'm completely stuck on this, is it something to do with Fermat's?

Answer 1

Your sum has the form $x_0 (\sum_{z \ in G} z)$, where $x_0$ is one solution, and $G=\{z^3=1\}$ is a subgroup of units. This is non trivial because $3 \mid q-1 = |\left ( \mathbb{Z}/q\mathbb{Z}\right )^* |$. Take a $1\neq \lambda \in G$, and convince yourself that $\lambda S =S \pmod q$. This yields $(\lambda -1)S= 0 \pmod q$, thus $S=0$ because of $\lambda\neq 1$.

Answer 2

In each case the number of solutions is three because the equation $x^3-8=0$ has three roots in some extension of the prime field $\mathbb F_q$. Furthermore $$x^3-2^3=(x-2)(x^2+2x+4)\Rightarrow\begin{cases}x_1=2\\x_2=-1+\sqrt{1-4}\\x_3=-1-\sqrt{1-4}\end{cases}$$ We see explicitement that the required sum is equal to zero.