Exercise 2.8.1 (Primitive roots modulo 3, 5, 7, 11, 13)

Question

Find a primitive root of the prime $3$; the prime $5$; the prime $7$; the prime 11; the prime $13$.

Richard Ganaye · Accepted Answer

\begin{proof} 
\begin{itemize}
\item $2^2 \equiv 1 \pmod 3$, and $2^1 
ot \equiv 1 \pmod 3$, thus

$2$ is is primitive root modulo $3$.

\item $2^4 = 16 \equiv 1 \pmod 5$, and $2^2 \equiv -1 
ot \equiv 1 \pmod 5$, thus

$2$ is is primitive root modulo $5$.

\item $3^6 \equiv 1 \pmod 7$ (Fermat's theorem), and $3^2 \equiv 2 
ot \equiv 1, 3^3 \equiv -1 
ot \equiv 1 \pmod 7$, thus

$3$ is is primitive root modulo $7$.

\item $2^{10} \equiv 1 \pmod {11}$ (Fermat), and $2^5 \equiv -1, 2^2 \equiv 4$, thus

$2$ is a primitive root modulo $11$.

\item $2^{12} \equiv 1 \pmod {13}$ (Fermat), and $2^6 \equiv -1, 2^4 \equiv 3 \pmod {13}$, thus

$2$ is a primitive root modulo $13$.
\end{itemize}
\end{proof}

With Sage
\begin{verbatim}
for p in range(14):
....:     if is_prime(p):
....:         Fp = GF(p)
....:         print(p, '=>', Fp.primitive_element())

(2, '=>', 1)
(3, '=>', 2)
(5, '=>', 2)
(7, '=>', 3)
(11, '=>', 2)
(13, '=>', 2)

a = Mod(2, 7)
a.multiplicative_order()
3
\end{verbatim}

Exercise 2.8.1 (Primitive roots modulo 3, 5, 7, 11, 13)

Answers

Comments

Exercise 2.8.1 (Primitive roots modulo 3, 5, 7, 11, 13)

Answers

Comments

Add answer