elementary number theory - Prove that if $x^9 equiv 3 mod{61}$, then $x^{12}equiv 34 mod{61}$

Prove that if $x^9 \equiv 3 \mod{61}$, then $x^{12}\equiv 34 \mod{61}$.

The first thing I tried was the obvious:

$$(x^9)^{4/3} \equiv 3^{4/3} \mod{61}$$
$$x^{12} \equiv 3^{4/3} \mod{61}$$

But this doesn't get me anywhere.

I've also tried breaking it down by saying that there exists a $k_1 \in \mathbb{Z}$ s.t. $x^9 - 3 = 61k_1$ and trying to work it towards $x^{12} - 34 = 61k_2$, for some $k_2 \in \mathbb{Z}$, but again, I didn't get too far.

Answer

Recall that $x^{60} = 1$ modulo $61$. This is Fermats Little Theorem (of course $x$ is a not divisible by $61$ so it is applicable).

Thus, $3^7 = (x^9)^7 = x^{63}=x^3$ modulo $61$.
So you know $x^3$ and the rest is direct.

The general strategy would be to solve $9 k = 12$ modulo $60$, and then compute $x^{12}= (x^{9})^k = 3^k$, but I took a slight shortcut.