<s>[INST] F ind the seven solutions to $x^{7} \equiv 1 \pmod{29}$ 
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?
 [/INST] 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).
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