A134307 Primes p such that A^(p-1) == 1 (mod p^2) for some A in the range 2 <= A <= p-1.

11, 29, 37, 43, 59, 71, 79, 97, 103, 109, 113, 127, 131, 137, 151, 163, 181, 191, 197, 199, 211, 223, 229, 233, 241, 257, 263, 269, 281, 283, 293, 307, 313, 331, 347, 349, 353, 359, 367, 373, 379, 397, 401, 419, 421, 433, 439, 449, 461, 463, 487, 499, 509

Offset: 1

Author

Joerg Arndt, Aug 27 2008

Keywords

• nonn

Comments

It's worth observing that there are p-1 elements of order dividing p-1 modulo p^2 that are of the form r^(k*p) mod p^2 where r is a primitive element modulo p and k=0,1,...,p-2. Heuristically, one can expect that at least one of them belongs to the interval [2,p-1] with probability about 1 - (1 - 1/p)^(p-1) ~= 1 - 1/e.

Numerically, among the primes below 1000 (out of the total number pi(1000)=168) there are 103 terms of the sequence, and the ratio 103/168 = 0.613 which is already somewhat close to 1-1/e ~= 0.632.

If we replace p^2 with p^3, heuristically it is likely that the sequence is finite (since 1 - (1 - 1/p^2)^(p-1) tends to 0 as p grows). - Max Alekseyev, Jan 09 2009

Replacing p^2 with p^3 gives just the one term (113) for p < 10^6. - Joerg Arndt, Jan 07 2011

If furthermore the number A can be taken to be a primitive root modulo p, i.e., A is a generator of (Z/pZ)*, then that p belongs to A060503. - Jeppe Stig Nielsen, Jul 31 2015

The number of terms not exceeding prime(10^k), for k=1,2,..., are 2, 55, 652, 6303, 63219, ... - Amiram Eldar, May 08 2021

Examples

			Examples (pairs [p, A]):
[11, 3]
[11, 9]
[29, 14]
[37, 18]
[43, 19]
[59, 53]
[71, 11]
[71, 26]
[79, 31]
[97, 53]
		

References

• L. E. Dickson, History of the theory of numbers, vol. 1, p. 105.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)
• Wilfrid Keller and Jörg Richstein Fermat quotients that are divisible by p.

Crossrefs

Cf. A001220, A055578, A039678, A143548, A222184, A060503.

Programs

Mathematica

Select[ Prime[ Range[100]], Product[ (PowerMod[a, # - 1, #^2] - 1), {a, 2, # - 1}] == 0 &] (* Jonathan Sondow, Feb 11 2013 *)

PARI

{ forprime (p=2, 1000,
   for (a=2, p-1, p2 = p^2;
     if( Mod(a, p2)^(p-1) == Mod(1, p2), print1(p, ", ") ;break() );
  ); ); }