<s>[INST] 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. 
 [/INST] Like Find all solutions to $y^2 \equiv 5x^3 \pmod {7}$,
as $2$ is a primitive root $\pmod{37}$ as
$2^5\equiv-5\pmod{37}\implies2^{10}\equiv(-5)^2\equiv-12\implies2^{12}\equiv-12\cdot2^2\equiv26\not\equiv1$
and $2^{18}\equiv(-10)^3\equiv-1$
using Discrete Logarithm , $3$ind$_2x\equiv0\pmod{36}\iff$ind$_2x\equiv0\pmod{12}$
$\implies x\equiv2^{12k}\pmod{37}$ where $k\equiv0,1,2\pmod3$
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