Find all solutions to the congruence $x^3\equiv 1 {\pmod{77}}$ Find all solutions to the congruence $x^3\equiv 1 \pmod{77}$
I was able to do this for $x^2 \equiv 1\pmod{77}$ since $x$ would either be $1$ or $-1$, in each of $\bmod 7$ and $\bmod 11$. I don't know what $x$ would be in this case. 
I can do the Chinese remainder theorem to find the final answers.

Presumably you know that Fermat's little theorem (FLT) says that for prime modulus $p$ and $a$ coprime to $p$ we have $a^{p-1}\equiv 1 \bmod p$.
This also means that if we have $a^k \equiv 1 \bmod p$, and $k$ is the smallest positive integer for which this is true ($k$ is the order of $a \bmod p$), then $k$ divides $p{-}1$ - since otherwise FLT would not hold.
This means that we cannot have any such elements for $p=11$ and $k=3$, since $3 \nmid 10$. So $x^3\equiv 1 \bmod 11$ has only one solution, $x\equiv 1 \bmod 11$, since only $1$ divides both $3$ and $10$.
By constrast $3\mid (7-1) $ so there should be three solutions (including $1$), and $7$ is small enough to search case-by-case even if $7+1=8=2^3$ doesn't occur to you. The other cube root is $2^2 = 4$, which is apparent without further multiplying-out since $4^3=2^6\equiv 1^2\equiv 1 \bmod 7$.
