Exercise 2.8.31* (Least period of the decimal expansion of $1/p$)

Question

Show that the decimal expansion of $1/p$ has period $p-1$ if and only if $10$ is a primitive root of $p$. (It is conjectured that if $g$ is not a square, and if $g 
e -1$, then there are infinitely many primes of which $g$ is a primitive root.)

Richard Ganaye · Accepted Answer

\begin{proof} 
If $p = 2$ or $p = 5$, then $10 \equiv 0 \pmod p$, so $10$ is not a primitive root modulo $p$. Now we can suppose that $p\wedge 10 = 1$. By Problem 30, the order $k$ of $10$ modulo $p$ is equal to the least period of the decimal expansion of $1/p$. Thus this least period is $p-1$ if and only if the order of $10$ is $p-1$, if and only if $10$ is a primitive root modulo $p$.

For instance the least period of the decimal expansion of $1/9967$ is $9966$. 
\end{proof}

Exercise 2.8.31* (Least period of the decimal expansion of $1/p$)

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Exercise 2.8.31* (Least period of the decimal expansion of $1/p$)

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