Competition problem — Number theory

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Number theory Difficulty 6.0 AIME, harder Prove it

12. If the prime $p \equiv 3(\bmod 4)$ , then $g$ is a primitive root modulo $p$ if and only if
$\delta_{p}(-g)=(p-1) / 2$

Solution

12. Necessity follows from part (iii) of question 10. If $\delta_{p}(-g)=(p-1) / 2$ , then $g^{(p-1) / 2} \equiv-1$ $(\bmod p)$ . From this and $\delta_{p}(g) /\left(\delta_{p}(g), 2\right)=\delta_{p}\left(g^{2}\right)=\delta_{p}\left((-g)^{2}\right)=\delta_{p}(-g) /\left(\delta_{p}(-g), 2\right)=$ $(p-1) / 2$ , it follows that $g$ is a primitive root.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.