Does the equation $x^3-9=0$ have any solutions in $(\mathbb{Z}/31 \mathbb{Z})$? This is my first time "solving an equation" in a group so I feel like that may be the source of my troubles. Here's what I have so far, although I'm not sure how to progress:


  Assume $c$ is a solution to $x^3-9=0$ in $(\mathbb{Z}/31 \mathbb{Z})$.
  We then have $$c^3-9=0\implies c^3=9 \implies |c^3|=|9|$$ and since on
  the LHS we're multiplying elements in $(\mathbb{Z}/31 \mathbb{Z})$, we
  consider the order of $[9]_{31}$ in $(\mathbb{Z}/31 \mathbb{Z})^*$
  which is $15$ since $$9^{15} \equiv 1 \ \text{mod} \ 31.$$ Thus
  $|c^3|=15$.


I feel like I should be able to deduce $|c|$ from $|c^3|$ but it doesn't seem to be clear to me. Could someone assist me in figuring out how to get $|c|$? 

From there, my next steps would be to ensure it divides $31$. Since $(\mathbb{Z}/31 \mathbb{Z})^*$ is abelian and cyclic thus has an element of maximal finite order of $31$, and if $|c| \leq 31$ then $|c|$ must divide $31.$
There is no solution as

$$c^{30} = 9^{10} = 5$$

but $$c^{30} = 1 \ \ \forall c \not =0$$
Slightly more efficiently, we get that if $c^3\equiv 9\equiv 3^2\pmod{31}$, then $c^{15}\equiv 3^{10}\pmod{31}$.

$$3^{10}\equiv (27)^3\cdot 3$$

$$\equiv (-4)^3\cdot 3\equiv -64\cdot 3$$

$$\equiv -2\cdot 3\equiv -6\pmod{31},$$

Since $\left(c^{15}\right)^2\equiv 1\pmod{31}$ for all $c\in\mathbb Z$ such that $31\not\mid c$ by Fermat's little theorem,

we get $31\mid \left(c^{15}+1\right)\left(c^{15}-1\right)$,

so $31\mid c^{15}+1$ or $31\mid c^{15}-1$ by Euclid's lemma,

so $c^{15}\equiv \pm 1\pmod{31}$, contradiction.

((In fact, $a^{\frac{p-1}{2}}\equiv \left(\frac{a}{p}\right)\pmod{p}$ for all $a\in\mathbb Z$, odd primes $p$ by Euler's criterion))

((and also, in this case, $c^{15}\equiv 1\pmod{31}$, because

if $c^{15}\equiv -1\pmod{31}$,

then $\left(3^5\right)^2\equiv -1\pmod{31}$,

contradiction, because $-1$ isn't a quadratic residue mod $31=4\cdot 7+3$ by Quadratic Reciprocity))