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I was looking at the primitive roots $n \bmod p$ and $p^2$ to see how often we get primitive roots of a prime that are not primitive roots of the square of that prime. I'll call this a reluctant root of $p$.

It's really rare.

With $p$ limited to a little under $38000$, I didn't find any such cases at all for $n<10$. In particular I started out looking specifically for cases when $2$ is a reluctant root, and finding none I widened the search. The first case I found was $n=11, p=71$, when I was just looking at $n$ prime, but there is also a case for $n=10, p=487$. In these cases, as in most others, and taking only $n<p,$ this was the only such case despite many hundreds of primes for which $n$ was a primitive root.

Searching the other way around, looking for reluctant roots of various primes $p,$ I found that in the first $400$ primes, which have a total of $190111$ primitive roots, there are only $143$ reluctant roots across $122$ primes. Most primes have no reluctant roots; the first few primes that have reluctant roots are $29, 37, 43, 71, 103, 109, 113, 131\ldots$; even fewer ($18$ in this range) have more than $1$, with $653$ taking the golden cupcake (so far) with $4$ reluctant roots - $84, 120, 287, 410$.

Main question: what is the smallest prime that has $2$ as a reluctant root, or is there no such prime?

Other questions: why are reluctant roots so rare? and is there any more investigative work on these (perhaps under a less Harry-Potteresque name)?

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    $\begingroup$ I found the primes here are recorded under OEIS A060503, but no further information. They include $2$ in the set which I guess is technically true (the "primitive root" $\bmod 2$ is $1$, whereas $4$'s primitive root is $3$). $\endgroup$ – Joffan Apr 16 '18 at 1:51
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    $\begingroup$ Reluctant roots might also be called "jump the gun" roots in the sense that $r^{p-1}$, intended merely to be $\equiv 1 \bmod p$, turns out to be $\equiv 1 \bmod p^2$. For instance, $11^{70}\equiv 1 \bmod 5041$. Hitting this "jump the gun" condition is expected to be rare and to become more so as we go to larger primes. $\endgroup$ – Oscar Lanzi Apr 16 '18 at 2:03
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    $\begingroup$ en.wikipedia.org/wiki/Wieferich_prime $\endgroup$ – Lord Shark the Unknown Apr 16 '18 at 2:40
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    $\begingroup$ You might also be interested in oeis.org/A060520, primes that have at least 4 reluctant roots. $\endgroup$ – Gerry Myerson Apr 16 '18 at 3:09
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    $\begingroup$ See also mathoverflow.net/questions/27579/… $\endgroup$ – lhf Apr 16 '18 at 11:09
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If $g$ is a primitive root modulo $p$ (or even if it isn't), then $g^{p-1}\equiv1+kp\bmod{p^2}$ for some $k$, $0\le k\le p-1$. As Oscar Lanzi points out in the comments, $g$ is "reluctant" if and only if $k=0$, so there is (in some sense) only a 1-in-$p$ chance that $g$ is reluctant. That, I think, would explain why they are rare.

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  • $\begingroup$ Agreed, but the chances seem to be even lower than that, at least for primitive roots. $\endgroup$ – Joffan Apr 16 '18 at 3:11
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One word to search is Wieferich primes in base $a$ — for which $a^{p-1} = 1 (p^2)$, and they are (pretty obviously) related to periods of "decimal" expansions in different bases; they are really rare. For example, first (and only ones in range $\leq 10^{12}$) examples for which $2^{p-1} = 1 (p^2)$ are 1093 and 3511. So your reluctant roots are sort of $a$'s for generalized Wieferich primes. More scientific interest in these numbers is that if $p$ is Wieferich in $a$, then $\Bbb Z[a^{\frac 1 p}]$ is not full ring of integers of $\Bbb Q(a^{\frac 1 p})$; reluctant roots should be related to branching of analogous extensions.

On the reason of rarity: Dilcher and Pomerance proved in mid 90s that $(2^{p-1} - 1)/p$ is uniformly distributed mod $p$, and methods suggest that it should be close to uniform when you replace $2$ by something else. So when you have a "function", say $W$, defined on primes with values uniformly distributed on $\{0, \dots, p-1\}$ and nothing on composites, it'll have roughly $\ \log \log N$ zeroes in long run (by prime distribution theorem). And double logarithm is very slowly growing function. Even if you count primes for which $W(p)$ just divides $p-1$, you'll have not much, because divisor function is not so large.

I'd say that this business overall is not very popular (because it's pretty hard to prove something here).

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For odd primes $p$ and natural integers $a\neq 0,1$, the "Fermat quotients" $q_a(p):=\frac {a^{p-1}-1}{p}$ mod $p$ have been given much attention by Georges Gras and his collaborators (see ref. below). Let me sketch their results in two directions :

1) Concerning the "rarity" the "Harry Potteresque" reluctant roots mod $p$ : For fixed $a\ge 2$, it is suggested in [G] that the probability of nullity mod $p$ of the $q_p(a)$'s is less than $\frac {1}{p}$ for any arbitrary large prime $p$. To justify this the author proposes various heuristics, supported by numerical computations and analytical results, which may imply the finiteness of the number of the $q_p(a)$'s that are $0$ mod $p$, and the existence of integers $a$ such that $q_p(a) \neq 0$ mod $p$. He shows that the density of integers A such that $q_p(A)\neq 0$ mod $p$ is about $O(\frac {1}{log (x)})$ for all $p \le x$.

2) In relation with the works of Vandiver and Furtwängler, [GQ] lay the foundations of a new global cyclotomic approach to FLT. One significant results reads : Let $p>3$. If there exists a prime $l\neq p$ s.t. $q_p (l)\neq 0$ mod $p$ and s.t. for any prime ideal $\mathcal L$ above $l$ in $\mathbf Q(\mu _{l-1})$ one has the relation $\mathcal L^{1-c}=(\alpha)\mathcal A^p$, where $c$ denotes complex conjugation, $\mathcal A$ is an ideal in $\mathbf Q(\mu _{l-1})$ and $\alpha \in \mathbf Q(\mu _{l-1})$ with $\alpha\equiv 1$ mod $p^2$, then the first case of FLT holds; the second case holds for $p$ as soon as there exist infinitely many such primes $l$.

[G] G. Gras, "Etude probabiliste des $p$-quotients de Fermat", Functiones et Approximatio, Vol. 54.1 (2016), 1–26

[GQ] G. Gras, R. Quême, "Vandiver papers on cyclotomy revisited and FLT", Publ. Math. Besançon, Vol.2 (2012), 47-111

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