Solving congruences like $3^p\equiv 1\pmod{\! p}$, $p$ prime [order computation]
Solution 1:
When $p\neq 3$, Fermat's little theorem gives:
$$3^{p-1}-1\equiv 0\pmod{p}$$
Thus $n=p-1$ is a solution. It follows that all multiples of $p-1$ are also solutions. Clearly for $p=3$ the only solution is $n=1$.
Solution 2:
You know from Fermat’s little theorem that $3^{p-1}\bmod p=1$ if $p$ is a prime greater than $3$. There may be smaller solutions: $3^5\bmod 11=1$, for instance. However, they must divide $p-1$, so there’s only a limited number to try. Once you find the minimum solution $m$, you have all solutions: they’re the positive integer multiples of $m$.