# How to solve the congruence $x^{30} ≡ 81x^6 \pmod{269}$ using primitive roots(without indices)?

So I know that 3 is a primitive root of 269.

How can I solve $$x^{30} ≡ 81x^6 \pmod{269}$$

Even if I substitute $$x$$ with $$3^y$$, where $$y$$ lies between 0 and 267, I can’t get any solutions.

We can easily see that $$x=0$$ is a solution.
Since $$3$$ is a primite root modulo $$269$$, let $$x \equiv 3^t \pmod{269}$$. hence, we get the congruence:
$$3^{30t} \equiv 3^4 \cdot 3^{6t} \pmod{269}$$ We know that if $$g$$ is a primitive root modulo $$n$$, then $$g^r \equiv g^s \pmod{n} \iff r \equiv s \pmod{\phi(n)}$$ Since $$269$$ is a prime we get that $$\phi(269) = 268$$, and then we get the congruence: $$30t \equiv 4 \cdot 6t \pmod{268}$$ which gives $$2$$ solutions for $$t$$, $$\space$$ $$t\equiv 0 \pmod{268}$$ or $$t\equiv 134 \pmod{268}$$.
From the first solution we get that $$x\equiv 1 \pmod{269}$$ which means that $$\big\{ x = 1 +269k \mid k\in \mathbb{Z}\big\}$$ is one set of solutions.
From the second solution we get that $$x\equiv 268 \pmod{269}$$ which means that $$\big\{ x = 268 +269k \mid k\in \mathbb{Z}\big\}$$ is another set of solutions.