Q:

Is this a legitimate proof? If not, how to prove?

Question: Determine all natural numbers $n$ such that: $7 \mid \left(3^n - 2\right)
\implies3^{n}\equiv 2\pmod{7}$
Multiply both sides by 7
$7 \cdot 3^{n}\equiv 7\cdot2\pmod{7}$
Divide both sides by seven, since $\gcd(7,7) = 7$, we have to divide modulus by $7$
$\implies3^{n}\equiv 2\pmod{7/7}$
$\implies3^{n}\equiv 2\pmod{1}$
Therefore $n$ is any natural number, since one divides everything. But I made a mistake somewhere since the original equation doesn't work for $n = 1$

A:

Noting that $n=1$ does not work, let $n \ge 2$.  Then as $3^2 \equiv 2 \pmod 7$, we have the equivalent statement
$$2\cdot 3^{n-2} \equiv 2 \pmod 7 \iff3^{n-2}\equiv 1 \pmod 7$$
Now that has solutions $n = 6k+2$ as $3^6$ is the smallest positive power of $3$ that is $\equiv 1 \pmod 7$, so the solution is for natural number s.t. $n = 2 \pmod 6$.

