Exercise 4.1

Proof. Let $p=29$. The integer $p$ is prime and $p-1=2^2\times 7$.

Note that

$2^4=16\not\equiv 1(\mathit{mod}29),$

$2^{14}=4^7=4\times 16^2=64\times 256\equiv 6\times (-34)=-204\equiv 86=3\times 29-1\equiv -1(\mathit{mod}29)$.

$2^{28}\equiv 1(\mathit{mod}29)$ and $2^d\not\equiv 1$ if $d\mid 28,d<28$, hence 2 is a primitive element modulo 29. □