Exercise 2.8.15 (If $a$ has order $h$, then $a^{h/2} \equiv -1 \pmod p$)

Question

Prove that if $a$ belongs to the exponent $h$ modulo a prime $p$, and if $h$ is even, then $a^{h/2} \equiv -1 \pmod p$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Since $h$ is the order of $a$ modulo $p$,
$$a^{h} \equiv 1 \pmod p,\qquad a^{h/2} 
ot \equiv 1 \pmod p.$$
Then $(a^{h/2})^2 = a^h \equiv 1 \pmod p$, thus $(a^{h/2} - 1)(a^{h/2} + 1 )\equiv 0 \pmod p$. But $a^{h/2} - 1 
ot \equiv 0 \pmod p$, and $p$ is prime, therefore
$a^{h/2} + 1 \equiv 0 \pmod p,$
so
$$a^{h/2} \equiv -1 \pmod p.$$
\end{proof}

Exercise 2.8.15 (If $a$ has order $h$, then $a^{h/2} \equiv -1 \pmod p$)

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Exercise 2.8.15 (If $a$ has order $h$, then $a^{h/2} \equiv -1 \pmod p$)

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