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 A097933 Primes such that p divides 3^((p-1)/2) - 1. 9
 11, 13, 23, 37, 47, 59, 61, 71, 73, 83, 97, 107, 109, 131, 157, 167, 179, 181, 191, 193, 227, 229, 239, 241, 251, 263, 277, 311, 313, 337, 347, 349, 359, 373, 383, 397, 409, 419, 421, 431, 433, 443, 457, 467, 479, 491, 503, 541, 563, 577, 587, 599, 601, 613 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Rational primes that decompose in the field Q[sqrt(3)]. - N. J. A. Sloane, Dec 26 2017 For all primes p > 2 and integers gcd(x, y, p) = 1, x^((p-1)/2) +- y^((p-1)/2) is divisible by p. This is because (x^((p-1)/2) - y^((p-1)/2))(x^((p-1)/2) + y^((p-1)/2)) = x^(p-1) - y^(p-1) is divisible by p according to Fermat's Little Theorem (FLT). This sequence lists p that divides 3^((p-1)/2) - 1^((p-1)/2), and A003630 lists the '+' case. Apart from initial terms, this and A038874 are the same. - N. J. A. Sloane, May 31 2009 Primes in A091998. - Reinhard Zumkeller, Jan 07 2012 Also, primes congruent to 1 or 11 (mod 12). - Vincenzo Librandi, Mar 23 2013 Conjecture: Let r(n) = (a(n) - 1)/(a(n) + 1) if a(n) mod 4 = 1, (a(n) + 1)/(a(n) - 1) otherwise; then Product_{n>=1} r(n) = (6/5) * (6/7) * (12/11) * (18/19) * ... = 2/sqrt(3). - Dimitris Valianatos, Mar 27 2017 Primes p such that Kronecker(12,p) = +1 (12 is the discriminant of Q[sqrt(3)]), that is, odd primes that have 3 as a quadratic residue. - Jianing Song, Nov 21 2018 LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 EXAMPLE For p = 5, 3^2 - 1 = 8 <> 3*k for any integer k, so 5 is not in this sequence. For p = 11, 3^5 - 1 = 242 = 11*22, so 11 is in this sequence. MATHEMATICA Select[Prime[Range], MemberQ[{1, 11, 13, 23}, Mod[#, 24]]&] (* Vincenzo Librandi, Mar 23 2013 *) 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", "))) } (PARI) {a(n)= local(m, c); if(n<1, 0, c=0; m=0; while( c

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Last modified December 6 04:14 EST 2019. Contains 329784 sequences. (Running on oeis4.)