Competition problem — Number theory

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Number theory Difficulty 5.4 AIME, harder Find the answer

4. Try to find a $g$ , for all $\alpha \geqslant 1$ , it is a primitive root of $p^{\alpha}, 2 p^{\alpha}$ :
$p=11,13,17,19,31,37,53,71$

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

4. In sequence: $7,7,3,3,3,-5,3,7$ .

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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.