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This is the weirdest thing I have observed so far! Take the set of Primitive Roots Modulo p (link to definition here) of a prime number $p$, $Pr(p)$. For those primes $p \gt 61$ there is always a pair of primitive roots $r1$ and $r2$ in $Pr(p)$ whose sum is the next prime to the current prime $p$, I will call it $\mathcal{N}(p)$.

(1) $\forall p \in \Bbb P\ ,\ p\gt 61\ ,\ \exists\ r1,r2\ \in Pr(p)\ /\ r1+r2 = \mathcal{N}(p)$

I have tested this with Python, in the interval $[62,10000]$ being always true.

E.g.:

$p=67\ ,\ Pr(67)=\{2,7,11,12,13,18,20,28,31,32,34,41,44,46,48,50,51,57,61,63\}$

$r_1=20$ and $r_2=51\ $,$\ r_1 + r_2 = 20+51=\mathcal{N}(67) = 71$

$p=499\ ,\ Pr(499)=\{7, 10, 11, 15, 17, 19, 23, 28, 35, 40, 41, 42, 44, 50, 53, 58, 60, 61, 63, 65, 66, 68, 71, 75, 76, 79, 85, 86, 87, 89, 90, 92, 94, 95, 98, 99, 102, 112, 113, 114, 129, 135, 138, 141, 146, 147, 153, 157, 160, 163, 164, 168, 171, 173, 176, 179, 182, 185, 193, 200, 202, 205, 206, 207, 210, 212, 214, 217, 218, 219, 223, 229, 232, 238, 240, 241, 242, 244, 246, 252, 260, 262, 264, 266, 271, 272, 273, 274, 275, 278, 284, 286, 295, 300, 301, 302, 303, 304, 309, 310, 311, 315, 316, 318, 319, 321, 325, 327, 329, 340, 341, 344, 347, 348, 349, 356, 357, 362, 363, 366, 367, 368, 369, 373, 376, 377, 378, 379, 380, 383, 390, 392, 393, 394, 396, 398, 399, 408, 411, 415, 417, 419, 426, 429, 430, 442, 443, 448, 450, 452, 453, 454, 456, 461, 465, 466, 469, 470, 474, 477, 478, 479, 485, 494\}$

$r_1=42$ and $r_2=461\ $,$\ r_1 + r_2 = 42+461=\mathcal{N}(499) = 503$

For $p \in [1,61]$ there are only five counterexamples: $\{2,3,5,7,61\}$ and for the rest of primes the observation is true as well.

I would like to share with you the following questions:

1 Does it make sense this kind of property having in mind the definition of primitive root modulo p?

2 It is just coincidental (probabilistic) because the size of the set of roots Pr(p) gets big enough to contain at least a pair of primitive roots complying with (1)? Or it is due to the Strong law of small numbers (I just tested $p \le 10000$)?

Thank you!

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  • $\begingroup$ @AlexeyBurdin hi! well, from that link my test verified up to the $gap=33$ for $p=1327$ because I have tested all the primes up to $p\le 7000$. My computer is very slow. I will try to test up to the next big $gap=35$ at $p=9551$... hopefully it will take one hour or two only... if I am able to run the test and my computer does not crash I will write here the results. :) $\endgroup$ Commented May 20, 2015 at 4:35
  • $\begingroup$ @AlexeyBurdin good point! but there is something I did not understand due to my lack of knowledge:why are you worried about big gaps between primes specifically?If I am not wrong,from the current prime p the next prime is in the interval [p,2p] and the primitive roots modulo p are distributed (depends on the case, but more or less half-half) in [1,p/2] and [p/2,p], so the probability of failing finding r1 and r2 would be the same independently of the gap to the next prime... math.stackexchange.com/questions/1186548/… $\endgroup$ Commented May 20, 2015 at 5:07
  • $\begingroup$ Oh I see. It's my lack of knowledge then. $\endgroup$ Commented May 20, 2015 at 5:10
  • $\begingroup$ @AlexeyBurdin not necessarily! :) I just observe things and ask about them, my explanation could be not totally correct, but in this case I think it should be as I said. I am testing now up to 10000, if my computer is able to finish I will put the info around here. Thanks for your comments! $\endgroup$ Commented May 20, 2015 at 5:15
  • $\begingroup$ Tested to 100k. Semi-OT, you may want to consider installing Pari/GP and learning it, as it will likely be a lot faster than SymPy. Perl/ntheory would be faster yet, especially for large nextprime/prevprime (not applicable to this question but to others of yours) but I imagine you don't want to switch. Or we could just improve SymPy. $\endgroup$ Commented May 20, 2015 at 5:58

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Yes, it's probabilistic (in other words, it doesn't have anything to do with primitive roots or the next prime).

The number of primitive roots modulo $p$ between $0$ and $p$ is $\phi(p-1) \gtrsim e^{-\gamma}p/\log\log p$ (using a known lower bound). In other words, the "probability" that a randomly chosen integer between $0$ and $p$ is a primitive root modulo $p$ is at worst something like $1/2\log\log p$. There are nearly $p/2$ ways to add two numbers less than $p$ and get the next prime after $p$, or indeed any nearby number. When $p$ is large, it is therefore overwhelmingly likely to get at least $p/8(\log\log p)^2$ pairs of primitive roots whose sum is the next prime.

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  • $\begingroup$ I suspected something like that, but the counterexample at $p=61$ was somehow making me doubt. Thank you for the explanation, I have learned a lot! $\endgroup$ Commented May 20, 2015 at 5:32
  • $\begingroup$ sorry, may I abuse of your time and knowledge? I have a similar question that could be also a probabilistic case... if you have time and you are interested in the topic, I would appreciate very much your thoughts. math.stackexchange.com/questions/1287585/… $\endgroup$ Commented May 20, 2015 at 5:41
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    $\begingroup$ Yes, that too is likely to be true for all large $n$, by a similar probabilistic heuristic which I will let you discover. Of course, it's a stronger statement than the Goldbach conjecture, as you point out, so it's not likely to be proved any time soon (nor, in my opinion, to lead to fresh insight on the Goldbach conjecture). $\endgroup$ Commented May 20, 2015 at 16:57
  • $\begingroup$ thanks for reviewing it, in the case of the link the probabilities are lower, the set must contain two odd prime numbers and both must sum the even number, so that probability is smaller than in the case of primitive roots of this question. I am never able to distinguish which conjectures are stronger than others, so I appreciate very much the comments you did, thank you! $\endgroup$ Commented May 21, 2015 at 0:14

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