Solve $a^6 \equiv 12 \pmod{13}$

Question

So I'm going over my notes for a test, and I can't read my horrible slop that I call handwriting. The question was

$a^6 \equiv 12 \pmod{13}$

I know that there is a primitive root $r$ and so that $a = r^x$

Then after a bunch of scribbles I end with up

$x \equiv 1 \pmod{2}$, but I can't figure out how to get there.

I know $12 = r^y$ for some $y$ so I can say that $6x \equiv y \pmod{12}$

So $12 \mid 6x - y$, but Im not sure if thats getting me anywhere.

Can someone explain to me the steps would would take to solve the original question?

Answer 1

Since $ 2 $ is a primitive root modulo $ 13 $, suppose $ a=2^x, 1\leq x\leq 12 $, then $ a^6\equiv 2^{6x}\equiv 12\equiv -1\mod 13 $, which implies $ 6x\equiv 6\mod 12 $, thus $ x=1, 3, 5, 7, 9, 11 $. Therefore $ a=2, 8, 6, 11, 5, 7 $ is the solutions of $ a^6\equiv 12\mod 13 $.

Answer 2

Note that $2$ is a primitve element modulo $12$. The powers are \begin{eqnarray*} \color{red}{2},4,\color{red}{8},3,\color{red}{6},12,\color{red}{11},9,\color{red}{5},10,\color{red}{7},1. \end{eqnarray*} Note that elements in red satisfy $a^6 \equiv 12 \pmod{13}$.

Answer 3

Note that

$$a^{12} \equiv 1 \pmod{13}\implies a^6 \equiv \pm 1 \pmod{13}$$

and since $\operatorname{ord}_{13}(2)=12$, $2$ is a generator $\mod {13}$ we can easily check that the solutions are

• $a=2$
• $a=2^3\equiv 8 \mod {13}$
• $a=2^5\equiv 6 \mod {13}$
• $a=2^7\equiv 11 \mod {13}$
• $a=2^9\equiv 5 \mod {13}$
• $a=2^{11}\equiv 7 \mod {13}$