We determine whether 4 is a primitive root modulo 5.\\ We have
\begin{align*}
4^0\bmod 5 \amp =1\bmod 5=1 \\
4^1\bmod 5 \amp =4\bmod 5=4 \\
4^2\bmod 5 \amp =16\bmod 5=1 \\
4^3\bmod 5 \amp =(4\cdot 4^2) \bmod 5 = (4\cdot 1)\bmod 5=4 \\
4^4\bmod 5 \amp =(4\cdot 4^3) \bmod 5 = (4\cdot 4)\bmod 5=1
\end{align*}
We observe that the powers of \(4\) modulo 5 alternate between \(1\) and \(4\text{.}\)
But for
\(4\) to be a primitive root modulo
\(5\text{,}\) all elements of
\(\ZZ_5-\{0\}=\{1,2,3,4\}\) would have to be powers of
\(4\text{.}\) For example,
\(2\) is not a power of
\(4\text{.}\) So
\(4\) is a not primitive root modulo
\(5\text{.}\)
Thus we have found that
\(2\) and
\(3\) are primitive roots modulo
\(5\) while
\(1\) and
\(4\) are not primitive roots modulo
\(5\text{.}\)
