# Show that there is no modulus with an odd prime number of primitive roots

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.

• If $a$ is a primitive root is $-a$ ? Aug 1, 2021 at 18:38
• @RoddyMacPhee: did you mean $a^{-1}$? Aug 1, 2021 at 18:42
• No but that's also a good consideration @J.W.Tanner Aug 1, 2021 at 18:47
• @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.
– lulu
Aug 1, 2021 at 18:48
• 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. Aug 1, 2021 at 18:51

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.