Show that there is no modulus with an odd prime number of primitive roots. I'm not really sure how to approach this at all. I was thinking of using the property that there exists a primitive root modulo $m$ if and only if $m = 2, 4$ or $m$ is of the form $p^k$ or $2p^k$, but I'm not really sure what to do next.

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    $\begingroup$ If $a$ is a primitive root is $-a$ ? $\endgroup$ Aug 1, 2021 at 18:38
  • $\begingroup$ @RoddyMacPhee: did you mean $a^{-1}$? $\endgroup$ Aug 1, 2021 at 18:42
  • $\begingroup$ No but that's also a good consideration @J.W.Tanner $\endgroup$ Aug 1, 2021 at 18:47
  • $\begingroup$ @RoddyMacPhee But $5$ is a primitive root $\pmod 7$ while $2$ isn't, for example. Or, more simply, $2$ is a primitive root $\pmod 3$ while $1$ is not. $\endgroup$
    – lulu
    Aug 1, 2021 at 18:48
  • $\begingroup$ Yes and what I was pointing out was that if $-1$ as a residue shows up in one line, $1$ is the residue of the line of it's additive inverse at that same exponent. For multiplicative inverses they end up at multiplicative inverses. $\endgroup$ Aug 1, 2021 at 18:51

1 Answer 1


If there are any primitive roots modulo $m$ at all, then there are $\varphi(\varphi(m))$ of them. Being a totient, it could never be an odd number greater than $1$, regardless of primality.


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