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A276730 Number of solutions to y^2 == x^3 + 4*x (mod p) as p runs through the primes.

%I A276730 #37 Oct 24 2018 08:08:03
%S A276730 2,3,7,7,11,7,15,19,23,39,31,39,31,43,47,39,59,71,67,71,79,79,83,79,
%T A276730 79,103,103,107,103,127,127,131,159,139,135,151,135,163,167,199,179,
%U A276730 199,191,207,199,199,211,223,227,199,207,239,271,251,255,263,295,271,295,271
%N A276730 Number of solutions to y^2 == x^3 + 4*x (mod p) as p runs through the primes.
%C A276730 This elliptic curve corresponds to a weight 2 newform which is an eta-quotient, namely, (eta(4t)*eta(8t))^2, see Theorem 2 in Martin & Ono.
%C A276730 It appears that a(n) = prime(n) iff prime(n) == 2 or 3 (mod 4). - _Robert Israel_, Sep 28 2016 This is true due to the L-function of this elliptic curve. See A278720. - _Wolfdieter Lang_, Dec 22 2016
%C A276730 The rational solutions of y^2 = x^3  + 4*x are (x,y) = (0,0), (2,4), (2,-4). See the Keith Conrad link, Corollary 3.17., p. 9. - _Wolfdieter Lang_, Dec 01 2016
%C A276730 For the p-defects p - N(p) see A278720. - _Wolfdieter Lang_, Dec 22 2016
%H A276730 Seiichi Manyama, <a href="/A276730/b276730.txt">Table of n, a(n) for n = 1..10000</a>
%H A276730 Keith Conrad, Expository papers, <a href="http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf"> Proofs by Descent</a>.
%H A276730 Yves Martin and Ken Ono, <a href="http://dx.doi.org/10.1090/S0002-9939-97-03928-2">Eta-Quotients and Elliptic Curves</a>, Proc. Amer. Math. Soc. 125, No 11 (1997), 3169-3176.
%F A276730 a(n) is the number of solutions of the congruence y^2 == x^3 + 4*x (mod prime(n)), n >= 1.
%F A276730 a(n) is also the number
%F A276730 of solutions of the congruence y^2 == x^3 - x (mod prime(n)), n >= 1. - _Wolfdieter Lang_, Dec 22 2016 (See the Cremona link given in A278720).
%e A276730 The first nonnegative complete residue system {0, 1, ..., prime(n)-1} is used.
%e A276730 The solutions (x, y) of y^2 == x^3 + 4*x (mod prime(n)) begin:
%e A276730 n, prime(n), a(n)\  solutions (x, y)
%e A276730 1,   2,       2:   (0, 0), (1, 1)
%e A276730 2,   3,       3:   (0, 0), (2, 1), (2, 2)
%e A276730 3,   5,       7:   (0, 0), (1, 0), (2, 1),
%e A276730                    (2, 4), (3, 2), (3, 3),
%e A276730                    (4, 0)
%e A276730 4,   7,       7:   (0, 0), (2, 3), (2, 4),
%e A276730                    (3, 2), (3, 5), (6, 3),
%e A276730                    (6, 4)
%e A276730 ...
%e A276730 The solutions (x, y) of y^2 == x^3 - x (mod prime(n)) begin:
%e A276730 n, prime(n), a(n)\  solutions (x, y)
%e A276730 1,   2,       2:   (0, 0), (1, 0);
%e A276730 2,   3,       3:   (0, 0), (1, 0), (2, 0);
%e A276730 3,   5,       7:   (0, 0), (1, 0), (2, 1),
%e A276730                    (2, 4), (3, 2), (3, 3),
%e A276730                    (4, 0);
%e A276730 4,   7,       7:   (0, 0), (1, 0), (4, 2),
%e A276730                    (4, 5), (5, 1), (5, 6),
%e A276730                    (6, 0);
%e A276730 ... - _Wolfdieter Lang_, Dec 22 2016
%p A276730 seq(nops([msolve(y^2-x^3-4*x, ithprime(n))]),n=1..100); # _Robert Israel_, Sep 28 2016
%o A276730 (Ruby)
%o A276730 require 'prime'
%o A276730 def A(a3, a2, a4, a6, n)
%o A276730   ary = []
%o A276730   Prime.take(n).each{|p|
%o A276730     a = Array.new(p, 0)
%o A276730     (0..p - 1).each{|i| a[(i * i + a3 * i) % p] += 1}
%o A276730     ary << (0..p - 1).inject(0){|s, i| s + a[(i * i * i + a2 * i * i + a4 * i + a6) % p]}
%o A276730   }
%o A276730   ary
%o A276730 end
%o A276730 def A276730(n)
%o A276730   A(0, 0, 4, 0, n)
%o A276730 end
%Y A276730 Cf. A095978, A272207, A278720.
%K A276730 nonn
%O A276730 1,1
%A A276730 _Seiichi Manyama_, Sep 16 2016