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Let $p$ be an odd prime. Define $S(p)$ as the sum of all primitive roots modulo $p$ taken from $\left[-\frac{p-1}2,\frac{p-1}2\right]$.

Now here's the strange thing. If the primitive roots were 'random', you'd expect $S(p)$ to be negative about as often as it is positive. However, experiments suggest that, at the very least, $$\lim_{n\to\infty}\frac{\#\{p\le n:S(p)<0\}}{\#\{p\le n:S(p)>0\}}>2.$$ This is extremely counter-intuitive to me.

I don't expect one can compute the limit, or even prove it exists. But is there some intuition behind this phenomenon?

EDIT: If $p\equiv 1\pmod 4$, it is easy to see $S(p)=0$. Now consider $p\equiv 3\pmod 4$. We can go further and split into cases mod $8$. I've computed $S(p)$ for all primes $p<10^5$ with $p\equiv 3\pmod 4$.

The value of $p$ modulo $8$ turns out to matter a whole lot. For $p\equiv 3\pmod 8$, I found $$ \begin{align*} \#\{p\le 10^5:p\equiv 3\pmod 8,S(p)>0\}&=204;\\ \#\{p\le 10^5:p\equiv 3\pmod 8,S(p)=0\}&=10;\\ \#\{p\le 10^5:p\equiv 3\pmod 8,S(p)<0\}&=2196. \end{align*} $$ Meanwhile, the behavior for $p\equiv 7\pmod 8$ is completely different: $$ \begin{align*} \#\{p\le 10^5:p\equiv 7\pmod 8,S(p)>0\}&=1278;\\ \#\{p\le 10^5:p\equiv 7\pmod 8,S(p)=0\}&=39;\\ \#\{p\le 10^5:p\equiv 7\pmod 8,S(p)<0\}&=1083. \end{align*} $$ Refining even further: up to $10^5$ there are $1203$ primes $p$ satisfying $p\equiv 11\pmod {24}$. With the sole exception of the prime $65171$, these primes satisfy $S(p)<0$.

Mastrem
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    It is known that when $p\equiv3\pmod4$, there are more quadratic nonresidues in $[-\frac p2,0]$ than there are in $[0,\frac p2]$ (this question's answers give entry points into the literature). Being a quadratic nonresidue is necessary for being a primitive root, so I suspect the two phenomena are connected. – Greg Martin Sep 18 '21 at 21:56
  • If $a$ is a primitive root $-a$ isn't – Roddy MacPhee Sep 18 '21 at 22:19
  • @RoddyMacPhee That’s false unless you assume $p=4k+3$. For instance, the primitive roots of $5$ are $2$ and $-2$, so it’s completely backwards for this case. – Erick Wong Sep 18 '21 at 22:36
  • Well $a^n$ and $(-a)^n$ are additive inverses( doh if $n$ is odd), so where $-1$ shows up in one $1$ shows up for the other. – Roddy MacPhee Sep 18 '21 at 22:51
  • @RoddyMacPhee Yep, you spotted the problem with your logic. When the modulus is $1$ mod $4$ the order of the multiplicative group is divisible by $4$, so the value of $n$ for which $a^n = -1$ is even. – Erick Wong Sep 18 '21 at 23:00
  • My answer here:https://math.stackexchange.com/questions/1388722/distribution-of-primitive-elements-finite-fields-prime-order/4683979#4683979 shows that $S(p)$ is approximately $0$ with an error term of size at most $O(p^{29/15})$. Of course, the size of the error can go down if we are more careful. However, this result does not give any information toward the sign bias. – Sungjin Kim Apr 26 '23 at 23:26

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