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.

As a solution to $x^2 \equiv 1 \mod 77$ you should have gotten
$x \equiv \pm 1 \mod 7$ and $x \equiv \pm 1 \mod 11$.  As this is FOUR systems of equations each system of equations should have yielded one unique solution.
System 1: $x \equiv 1\mod 7$ and $x \equiv 1 \mod 11$ yields $x \equiv 1 \mod 77$.
System 2: $x \equiv 1 \mod 7$ and $x \equiv -1 \mod 11$ yields $x \equiv 43 \mod 77$.  And $43^2 = 24*77 + 1$.
System 3: $x \equiv -1 \mod 7$ and $x \equiv 1 \mod 11$ yields $x \equiv 34 \mod 77$. And $34^2 =  15*77 + 1$.
System 4: $x \equiv -1 \mod 7$ and $x \equiv - 1 \mod 11$ yields $x \equiv 76\equiv -1 \mod 77$.
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For $x^3 = 1 \mod 77$.
To solve $x^3 \equiv 1 \mod 7$.  Note that if $x\not \equiv 0$ then $x^6 \equiv 1 \mod 7$ so $(x^2)^3 \equiv 1 \mod 7$ and so if $x \equiv (\pm 1)^2,(\pm 2)^2, (\pm 3)^2 \equiv 1, 4, 2 \mod 7$ then $x^3 \equiv 1 \mod 7$.
To solve $x^3 \equiv 1 \mod 11$, note that $\phi(11) = 10$.  $\gcd(3,10) = 1$.  For $x^k \equiv 1 \mod 11$ then $x\equiv y^j$ for some $j$ and $10|j*k$. That can only happen if $k = 10$ or $k < 10$ and $j=10$.  So $x \equiv 1 \mod 11$ is the only solution to $x^3 \equiv 1 \mod 11$.
So we have $3$ systems of equations.
$x \equiv 1 \mod 7$ and $x\equiv 1 \mod 11$ yields $x \equiv 1 \mod 77$.
$x \equiv 2 \mod 7$ and $x \equiv 1 \mod 11$ yields $x\equiv 23 \mod 77$.  And $23^3 = 12167 = 1 + 158*77$.
$x \equiv 4 \mod 7$ and $x \equiv 1 \mod 11$ yeilds $x \equiv 67 \mod 77$.  And $67^3 \equiv (-10)^3 \equiv -1000 + 770 \equiv -230 + 385 \equiv 155 - 154 \equiv 1 \mod 77$.
