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!
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.