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