How to solve $x^3 \equiv 1 \pmod{37}$ We are asked to solve $x^3 \equiv 1 \pmod{37}$. I know that the answer is $10$ since $27\cdot37 = 999$ and $10^3 = 1000$ but how do I show this rigorously? If it helps, we are given the primitive roots of $37$ which are $2, 5, 13, 15, 17, 18, 19, 20, 22, 24, 32$, and $35$. But I am not sure how this is useful. 

You know that $2^{36}\equiv 1$, since $2$ is a primitive root.  Hence you may conclude that $(2^{12})^3\equiv 1$, and also $(2^{24})^3\equiv (2^{36})^2\equiv 1^2$.  These correspond to $26$ and $10$ respectively.  Of course $2^{36}=1$.
Now, suppose that $(2^a)^3\equiv 1$.  Then $2^{3a}\equiv 1$, so $36|3a$, or $12|a$.  That doesn't leave many choices for $a$, since $1\le a\le 36$...
