OFFSET
1,1
COMMENTS
Also 3 and primes p such that (p^2 - 1)/24 mod 10 = {0, 7}. - Richard R. Forberg, Aug 31 2013
Also primes p such that x^2 = 10 mod p has integer solutions, or Legendre(10, p) = 1. However, p could be irreducible but not prime in Z[sqrt(10)], especially if p = 3 or 7 mod 10. - Alonso del Arte, Dec 27 2015
Rational primes that decompose in the field Q(sqrt(10)). - N. J. A. Sloane, Dec 26 2017
From Jianing Song, Oct 13 2022: (Start)
Primes p such that kronecker(10,p) = 1 (or equivalently, kronecker(40,p) = 1).
Primes congruent to 1, 3, 9, 13, 27, 31, 37, 39 modulo 40. (End)
LINKS
EXAMPLE
For p = 13, 5^6 - 2^6 = 15561 is divisible by 13, so 13 is in the sequence.
MAPLE
select(p -> isprime(p) and 10 &^ ((p-1)/2) mod p = 1, [seq(i, i=3..1000, 2)]); # Robert Israel, Dec 28 2015
MATHEMATICA
Select[Prime[Range[100]], JacobiSymbol[10, #] == 1 &] (* Alonso del Arte, Dec 27 2015 *)
PROG
(PARI) \\ s = +-1, d=diff
ptopm1d2(n, x, d, s) = { forprime(p=3, n, p2=(p-1)/2; y=x^p2 + s*(x-d)^p2; if(y%p==0, print1(p", "))) }
ptopm1d2(1000, 5, 3, -1)
(PARI) isA097955(p) == isprime(p) && kronecker(10, p) == 1 \\ Jianing Song, Oct 13 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Cino Hilliard, Sep 06 2004
STATUS
approved