Exercise 2.8.23* (Elements of order $3$)

Question

Prove that if $a$ belongs to the exponent $3$ modulo a prime $p$, then $1+a+a^2 \equiv 0 \pmod p$, and $1+a$ belongs to the exponent $6$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

\begin{proof} 
Since $a$ has order $3$ modulo $p$,
$$a^3 \equiv 1 \pmod p,\qquad a 
ot \equiv 1 \pmod p.$$
Then
$$0\equiv a^3 - 1 = (a-1)(1+a+a^2)\pmod p.$$
Since $p$ is prime, and $a-1 
ot \equiv 0 \pmod p$, we obtain
$$1 + a + a^2 \equiv 0 \pmod p.$$

Moreover, using $p
e 2$,
\begin{align*}
(1 +a)^6 &\equiv (-a^2)^6 =(a^3)^4 \equiv 1 \pmod p,\
(1+a)^3 & \equiv 1+ 3a+3a^2+a^3\
&\equiv 1 + 3a -3(1+a) +1\
&\equiv -1 
ot \equiv 1 \pmod p\
(1+a)^2 &= 1 + 2a + a^2\
&\equiv 1+2a - (1+a)\
&\equiv a 
ot \equiv 1 \pmod p.
\end{align*}
This shows that $1+a$ has order $6$ modulo $p$.
\end{proof}

Note: $\overline{a}$ plays a similar role in $\Z/p\Z$ than $\omega = e^{2i\pi/3}$ in $\C^*$: $\omega$ has order $3$, and $1+ \omega = - \omega^2 =  e^{i\pi/3}$ has order $6$.

Exercise 2.8.23* (Elements of order $3$)

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Exercise 2.8.23* (Elements of order $3$)

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