Modular exponentiation and two primes

Question

Given two primes $11$ and $5$, find all $\alpha> 1$ such that $$\alpha^{5} \equiv 1 \pmod{11}$$

What theorem will help me to find it out?

Answer 1

As $2^2\equiv4,2^5\equiv-1\pmod{11}$

$\implies2$ is a primitive root $\pmod{11}$

So taking Discrete Logarithm wrt $2$ on $$\alpha^5\equiv1\pmod{11}$$

$5$ind$_2\alpha\equiv0\pmod{\phi(11)}\implies$ind$_2\alpha\equiv0\pmod2$

$\implies\alpha\equiv2^{2n}\pmod{11}$ where $0\le2n<10$