Help understanding some solution for a module

Question

I'm trying to solve the following

Determine if $a^3=-9$ is solvable in $\mod 31$ and in $\mod 11$.

So I did the following, for $a^3-9$ we have that $$a^3=-9=22\mod31$$

from there I'm not quite sure how to continue. I know that for that it must happen that $$a^3=22+31k$$ with $k\in\mathbb{Z}$, but I'm lost from here.

So my question is, what I need to do in order to find the solution or that there is no solution for $a$?

Is your question $a^3=9$ or $a^3=-9$? You seem to have used both here — lioness99a, Mar 23 '17 at 11:42
Not my area of expertise. I think it can be proven that since $11\equiv-1\pmod3$, everything should be a cube $\pmod{11}$. Not sure how to handle $31$ though. — Mike, Mar 23 '17 at 12:04

score 1 · Accepted Answer · answered Mar 23 '17 at 14:23

The answer appears to be no. It appears that $3$ generates the group $\mathbb Z_{31}^*$. This means that

$$a^3=-3^2\equiv3^{17+30k}\pmod{31}$$

which is impossible since the exponent is never a multiple of $3$.

The logic for why there is a solution for $a^3\equiv-9\pmod{11}$ is similar. Let $b$ be a generator of the group $\mathbb Z_{11}^*$. For some integers $k_1$ and $k_2$

$$b^{10k_1+k_2}\equiv-9\pmod{11}$$

Choosing the correct value of $k_1$ will make the exponent a multiple of $3$.

Help understanding some solution for a module

1 Answers1