Exercise 2.8.5 (Elements of order $2$)

Question

Let $p$ be an odd prime. Prove that $a$ belongs to the exponent $2$ modulo $p$ if and only if $a \equiv -1 \pmod p$.

Richard Ganaye · Accepted Answer

\begin{proof} 
If the order of $a$ is $2$, then $a^2 \equiv 1 \pmod p$, and $a 
ot \equiv 1 \pmod p$. Therefore $p \mid (a^2 - 1) = (a-1)(a+1)$. Since $p$ is prime, $p \mid a-1$ or $p \mid a + 1$, so $a \equiv \pm 1 \pmod p$. Since $p 
ot \equiv 1 \pmod p$, $a \equiv -1 \pmod p$.

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Conversely, if $a \equiv -1 \pmod p$, then $a^2 \equiv 1 \pmod p$, and $a 
ot \equiv 1 \pmod p$, thus the order of $a$ modulo $p$ is $2$.

Exercise 2.8.5 (Elements of order $2$)

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Exercise 2.8.5 (Elements of order $2$)

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