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%I #82 Mar 19 2023 01:05:22
%S 11,13,23,37,47,59,61,71,73,83,97,107,109,131,157,167,179,181,191,193,
%T 227,229,239,241,251,263,277,311,313,337,347,349,359,373,383,397,409,
%U 419,421,431,433,443,457,467,479,491,503,541,563,577,587,599,601,613
%N Primes p that divide 3^((p-1)/2) - 1.
%C Rational primes that decompose in the field Q[sqrt(3)]. - _N. J. A. Sloane_, Dec 26 2017
%C 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.
%C Apart from initial terms, this and A038874 are the same. - _N. J. A. Sloane_, May 31 2009
%C Primes in A091998. - _Reinhard Zumkeller_, Jan 07 2012
%C Also, primes congruent to 1 or 11 (mod 12). - _Vincenzo Librandi_, Mar 23 2013
%C 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
%C 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
%C Comment from _Richard R. Forberg_, Feb 07 2023: (Start)
%C Conjecture: These are the exclusive prime factors of the set of integers d > 1 such that there exist primitive Heronian triangles with sides {b, b+d, b+2d} for one or more integers b.
%C Also b is always > d. For d=11 the b values begin {15, 17, 65, 75, 267, 305, 1025, ...}. For d=1 (not prime, thus not listed) the b values are given by A016064. (End)
%H Reinhard Zumkeller, <a href="/A097933/b097933.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Pri#primes_decomp_of">Index to sequences related to decomposition of primes in quadratic fields</a>
%e For p = 5, 3^2 - 1 = 8 <> 3*k for any integer k, so 5 is not in this sequence.
%e For p = 11, 3^5 - 1 = 242 = 11*22, so 11 is in this sequence.
%t Select[Prime[Range[300]], MemberQ[{1, 11, 13, 23}, Mod[#, 24]]&] (* _Vincenzo Librandi_, Mar 23 2013 *)
%t Select[Prime[Range[2,200]],PowerMod[3,(#-1)/2,#]==1&] (* _Harvey P. Dale_, Jun 02 2020 *)
%o (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","))) }
%o (PARI) {a(n)= local(m, c); if(n<1, 0, c=0; m=0; while( c<n, m++; if( isprime(m)& kronecker(3,m)==1, c++)); m)} /* _Michael Somos_, Aug 28 2006 */
%o (Haskell)
%o a097933 n = a097933_list !! (n-1)
%o a097933_list = [x | x <- a091998_list, a010051 x == 1]
%o -- _Reinhard Zumkeller_, Jan 07 2012
%o (Magma) [p: p in PrimesUpTo(1000) | p mod 24 in [1, 11, 13, 23]]; // _Vincenzo Librandi_, Mar 23 2013
%Y Cf. A003630, A010051, A038874, A091998.
%K nonn
%O 1,1
%A _Cino Hilliard_, Sep 04 2004
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