Exercise 4.22

Proof. If $a$ has order $3$ modulo $p$, then $0\equiv a^3-1=(a-1)(a^2+a+1)(\mathit{mod}p)$, with $a\not\equiv 1(\mathit{mod}p)$, thus $a^2+a+1\equiv 0(\mathit{mod}p)$. Thus

So ${(1+a)}^6\equiv 1(\mathit{mod}p)$.

${(1+a)}^2\equiv 1+2a+a^2=1+2a+(-1-a)\equiv a\not\equiv 1(\mathit{mod}p)$.

So ${(1+a)}^6\equiv 1,{(1+a)}^2\not\equiv 1,{(1+a)}^3\not\equiv 1(\mathit{mod}p)$, therefore the order of $1+a$ divides $6$, but doesn’t divides $2$ or $3$, thus $1+a$ has order $6$ modulo $p$. □