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

2, 3, 7, 7, 11, 7, 15, 19, 23, 39, 31, 39, 31, 43, 47, 39, 59, 71, 67, 71, 79, 79, 83, 79, 79, 103, 103, 107, 103, 127, 127, 131, 159, 139, 135, 151, 135, 163, 167, 199, 179, 199, 191, 207, 199, 199, 211, 223, 227, 199, 207, 239, 271, 251, 255, 263, 295, 271, 295, 271

Offset

1,1

Comments

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.

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

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

For the p-defects p - N(p) see A278720. - Wolfdieter Lang, Dec 22 2016

Links

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Keith Conrad, Expository papers, Proofs by Descent.

Yves Martin and Ken Ono, Eta-Quotients and Elliptic Curves, Proc. Amer. Math. Soc. 125, No 11 (1997), 3169-3176.

Formula

a(n) is the number of solutions of the congruence y^2 == x^3 + 4*x (mod prime(n)), n >= 1.

a(n) is also the number

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).

Example

The first nonnegative complete residue system {0, 1, ..., prime(n)-1} is used.
The solutions (x, y) of y^2 == x^3 + 4*x (mod prime(n)) begin:
n, prime(n), a(n)\  solutions (x, y)
1,   2,       2:   (0, 0), (1, 1)
2,   3,       3:   (0, 0), (2, 1), (2, 2)
3,   5,       7:   (0, 0), (1, 0), (2, 1),
                   (2, 4), (3, 2), (3, 3),
                   (4, 0)
4,   7,       7:   (0, 0), (2, 3), (2, 4),
                   (3, 2), (3, 5), (6, 3),
                   (6, 4)
...
The solutions (x, y) of y^2 == x^3 - x (mod prime(n)) begin:
n, prime(n), a(n)\  solutions (x, y)
1,   2,       2:   (0, 0), (1, 0);
2,   3,       3:   (0, 0), (1, 0), (2, 0);
3,   5,       7:   (0, 0), (1, 0), (2, 1),
                   (2, 4), (3, 2), (3, 3),
                   (4, 0);
4,   7,       7:   (0, 0), (1, 0), (4, 2),
                   (4, 5), (5, 1), (5, 6),
                   (6, 0);
... - Wolfdieter Lang, Dec 22 2016

Maple

seq(nops([msolve(y^2-x^3-4*x, ithprime(n))]), n=1..100); # Robert Israel, Sep 28 2016

Prog

(Ruby)
require 'prime'
def A(a3, a2, a4, a6, n)
  ary = []
  Prime.take(n).each{|p|
    a = Array.new(p, 0)
    (0..p - 1).each{|i| a[(i * i + a3 * i) % p] += 1}
    ary << (0..p - 1).inject(0){|s, i| s + a[(i * i * i + a2 * i * i + a4 * i + a6) % p]}
  }
  ary
end
def A276730(n)
  A(0, 0, 4, 0, n)
end

Crossrefs

Cf. A095978, A272207, A278720.

Sequence in context: A011161 A305420 A171464 * A179894 A060215 A309666

Adjacent sequences: A276727 A276728 A276729 * A276731 A276732 A276733

Keyword

nonn

Author

Seiichi Manyama, Sep 16 2016

Status

approved