Let $a$ be a primitive root for modulo $n$. Then, $a^{\frac{\phi(n)}{2}}\equiv-1\pmod{n}$.

I have a question for its converse. In general, its converse is false.

Is it possible to make(it means 'adding certain conditions')the converse is true?

Give some advice! Thank you!

  • 1
    $\begingroup$ $$\phi(n)=2^m$$ $\endgroup$ – lab bhattacharjee Mar 8 at 3:54
  • $\begingroup$ @labbhattacharjee, Thank you for comment. How to show that? I can't initiate $\endgroup$ – Primavera Mar 8 at 4:24

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