OFFSET
1,1
COMMENTS
Primes p such that ord((1+sqrt(5))/2,p) = (p-1)/5, where ord(z,p) is the smallest integer k > 0 such that (z^k-1)/p is an algebraic integer.
Let {T(n)} be a sequence defined by T(0) = 0, T(1) = 1, T(n) = k*T(n-1) + T(n-2), K be the quadratic field Q[sqrt(k^2+4)], O_K be the ring of integer of K, u = (k+sqrt(k^2+4))/2. For a prime p not dividing k^2 + 4, the Pisano period of {T(n)} modulo p (that is, the smallest m > 0 such that T(n+m) == T(n) (mod p) for all n) is ord(u,p); the entry point of {T(n)} modulo p (that is, the smallest m > 0 such that T(m) == 0 (mod p)) is ord(-u^2,p).
For an odd prime p:
(a) if p decomposes in K, then (O_K/pO_K)* (the multiplicative group of O_K modulo p) is congruent to C_(p-1) X C_(p-1), so the Pisano period of {T(n)} modulo p is equal to (p-1)/s, s = 1, 2, 3, 4, ...;
(b) if p is inert in K, then u^(p+1) == -1 (mod p), so the Pisano period of {T(n)} modulo p is equal to 2*(p+1)/r, r = 1, 3, 5, 7, ...
Here k = 1, and this sequence gives primes such that (a) holds and s = 5.
Number of terms below 10^N:
N | Number | Decomposing primes*
3 | 6 | 78
4 | 40 | 609
5 | 280 | 4777
6 | 2289 | 39210
7 | 18903 | 332136
8 | 163395 | 2880484
* Here "Decomposing primes" means primes such that Legendre(5,p) = 1, i.e., p == 1, 4 (mod 5).
MATHEMATICA
pn[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0 && Mod[Fibonacci[k + 1], n] == 1, Return[k]]];
Reap[For[p = 2, p < 11000, p = NextPrime[p], If[Mod[p, 5] == 1, If[pn[p] == (p - 1)/5, Print[p]; Sow[p]]]]][[2, 1]] (* Jean-François Alcover, Jul 05 2019 *)
PROG
(PARI) Pisano_for_decomposing_prime(p) = my(k=1, M=[k, 1; 1, 0], Id=[1, 0; 0, 1]); if(isprime(p)&&kronecker(k^2+4, p)==1, my(v=divisors(p-1)); for(d=1, #v, if(Mod(M, p)^v[d]==Id, return(v[d]))))
forprime(p=2, 11000, if(Pisano_for_decomposing_prime(p)==(p-1)/5, print1(p, ", ")))
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Jun 25 2019
STATUS
approved