# [Math] F ind the seven solutions to $x^{7} \equiv 1 \pmod{29}$

elementary-number-theorymodular arithmeticnumber theory

Use the fact that $$2$$ is a primitive root modulo 29 to find the seven solutions to $$x^{7} \equiv 1 \pmod{29}$$

As $$2$$ is primitive root modulo $$29$$ then $$2^{28} \equiv 1 \pmod{29}$$ $$2^{4*7} \equiv 1 \pmod{29}$$ $$16^{7} \equiv 1 \pmod{29}$$ where 16 is a solution to the equation. From this solution as I can get 6 missing solutions?

We know that $2^{28} \equiv 1 \bmod 29$, i.e. $16^7 \equiv 1 \bmod 29$. However we can also conclude from this that:
$16^{14},16^{21},16^{28},16^{35},16^{42},16^{49} \equiv 1 \bmod 29$.
So $16^2, 16^3, 16^4, 16^5, 16^6, 16^7$ are the other $6$ solutions.
Alternatively: we know every $x\in\mathbb{Z}/29\mathbb{Z}$ can be written as $x=2^k$ for some $k$.
To satisfy $x^7 \equiv 1 \bmod 29$ we require $2^{7k} \equiv 1 \bmod 29$. But since $2$ is a primitive root mod $29$ this means that $7k \equiv 0 \bmod 28$, i.e. $k \equiv 0 \bmod 4$ so we may take $k=0,4,8,12,16,20,24$ (after this the powers of $2$ repeat).