1
$\begingroup$

Let $p$ be a prime that is congruent to 1 mod 4. Show that the sum of the primitive roots mod $p$ is zero mod $p$.

Kind of at a loss here of how to do this. Can also reformulate the question as being if $S$ is a complete set of residues $mod\ p$, then the sum of the primitive roots in $S$ is $0\ mod\ p$.

$\endgroup$
1
$\begingroup$

If $p \equiv 1 \bmod 4$ and $g$ is a primitive root mod $p$, then so is $-g$.

Indeed, let $m$ be the order of $-g$. Then of course $m \le p-1$.

If $m$ is even, then $1=(-g)^m=g^m$ and so $m \ge p-1$. Hence, $m=p-1$.

If $m$ is odd, then $1=(-g)^m=-g^m$ and so $g^m=-1=g^{\frac{p-1}{2}}$. Therefore, $m \equiv \frac{p-1}{2} \bmod p-1$ and so $m=\frac{p-1}{2}$, because $m \le p-1$. But this cannot happen because $m$ is odd and $\frac{p-1}{2}$ is even.

If $g$ is a primitive root, then $g\ne -g$.

Indeed, $g=-g$ would imply $2g=0$ and so $g=0$.

The sum of all primitive roots is zero.

Both $g$ and $-g$ appear in the sum.

$\endgroup$
  • $\begingroup$ Hm, from here I guess you could group all the primitive roots $g$ with primitive roots $-g$ and get the sum to $0$ but how do you know this fact? $\endgroup$ – SS' Apr 10 '18 at 2:02
  • $\begingroup$ Hm so $(-g)^{\frac{p-1}{2}} = (-1)^{\frac{p-1}{2}} \cdot g^{\frac{p-1}{2}} = -1$ if $p \equiv 1\ mod 4$ $\endgroup$ – SS' Apr 10 '18 at 2:08
  • $\begingroup$ I think I follow the rest of the proof from here, but just to make sure, is it correct to say that since for every primitive root $g$, there is a primitive root $-g$, the sum therefore $= 0$? Thanks! $\endgroup$ – SS' Apr 10 '18 at 2:12
  • $\begingroup$ Tried to think about why $-g \neq g$ but the best I got is that they're elements from a complete set of residues, so hence incongruent. $\endgroup$ – SS' Apr 10 '18 at 2:21
  • $\begingroup$ @SS, see the details in my edited answer. Perhaps we can clear the comments? $\endgroup$ – lhf Apr 10 '18 at 2:27
1
$\begingroup$

An advanced way is to think of the Cyclotomic polynomial. $\Phi_{p-1}(x),$ which has primitive $p-1$th roots of unity as roots.

But you can show that in general: $$\Phi_{4m}(x)=\Phi_{2m}(x^2)$$

This can be seen by using the rule:

$$\Phi_n(x)=\prod_{d\mid n}\left(x^d-1\right)^{\mu(n/d)}$$

But when $n/d$ is divisible by $4$, $\mu(n/d)=0$ so:

$$\Phi_{4m}(x)=\prod_{d\mid 2m}\left(x^{2d}-1\right)^{\mu(2m/d)}=\Phi_{2m}(x^2)$$

This means that $\Phi_{4m}(x)$ has, as the sum of roots, zero.

This same argument shows that:

$$\Phi_{mq^2}(x)=\Phi_{mq}(x^q)$$

So for a prime $p\equiv 1\pmod{q^2}$ for any $q>1$ then the sum of the generators modulo $p$ is zero.

So it is true for $p=19,$ for example, which has generators $2,2^5=13,2^7\equiv14,2^{11}\equiv15,2^{13}\equiv3,2^{17}\equiv10$, and $$2+13+14+15+3+10=57$$ is divisible by $19.$


This leads to two conjectures:

The sum of the generators modulo $p$ is $0$ if and only if $\mu(p-1)=0,$ that is , if $p-1$ is divisible by a prime squared.

and

Define $Q$ to be the set or primes whose generators do not add up to $0.$ Then: $$q(x)=\left|\{p\in Q\mid p\leq x \}\right|$$ then $$\frac{q(x)}{\pi(x)}\to \prod_p\left(1-\frac{1}{p(p-1)}\right)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.