A271227 Number of solutions to y^2 == x^3 + 17 (mod p) as p runs through the primes.

2, 3, 5, 12, 11, 20, 17, 26, 23, 29, 42, 48, 41, 56, 47, 53, 59, 48, 62, 71, 63, 75, 83, 89, 102, 101, 110, 107, 111, 113, 146, 131, 137, 132, 149, 170, 182, 171, 167, 173

Offset

1,1

Comments

If prime(n) == 0 or 2 (mod 3) then a(n) = prime(n), i.e., the p-defect d(n) = prime(n) - a(n) = A271228(n) vanishes for these n. See A271228, and the Silverman reference, Theorem 45,2., p. 400. (The 0 (mod 3) case, i.e., prime(2) = 3, is trivial.)

If prime(n) == 1 (mod 3) = A002476(m) (for a unique m = m(n)) then prime(n) = A(m)^2 + 3*B(m)^2 with A(m) = A001479(m+1) and B(m) = A001480(m+1), m >= 1. In this case (4*prime(n) - d(n)^2)/3, with the p-defect d(n), seems to be a square, q(m)^2, if prime(n) = A002476(m). Three disjoint and exhaustive cases for these squares seem to apply: q(m)^2 = (2*B(m))^2, (A(m) - B(m))^2 and (A(m) + B(m))^2. See exercise 45.3, p. 404, of the Silverman reference, asking for a special form of 4*prime(n) - d(n)^2. These three cases (call them I, II and III) apply to the primes 73, 79, 109, 163, 199, 223, 229, 241, 307, 337, 349, 373, 397, ...; 7, 13, 19, 31, 37, 43, 61, 127, 157, 283, 313, 367, 409, ...; and 67, 97, 103, 139, 151, 181, 193, 211, 271, 277, 331, 379, 421, 433, ..., respectively. The shown numbers cover the first 40 primes 1 (mod 3).

The discriminant of the elliptic curve y^2 = x^3 + 17 is -3^3*17^2 = -7803. The bad primes (besides 2) are 3 and 17. See the Silverman reference p. 408.

References

J. H. Silverman, A Friendly Introduction to Number Theory, 3rd ed., Pearson Education, Inc, 2006, Table 45.5, Theorem 45.2, p. 400, Exercise 45.3, p. 404, p. 408 (4th ed., Pearson 2014, Table 5, Theorem 2, p. 366, Exercise 3, p. 370, p. 376)

Links

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

Formula

a(n) gives the number of solutions of the congruence y^2 == x^3 + 17 (mod prime(n)), n >= 1.

Proved [Silverman]: a(n) = prime(n) if prime(n) = 0 or 2 (mod 3).

Conjecture [WL]:

If prime(n) = 1 (mod 3), i.e., prime(n) = A002476(m), then a(n) = prime(n) + or -sqrt(4*prime(n) - 3*q(m)^2), with q(m)^2 of the form (2*B(m))^2 or (A(m) - B(m))^2 or (A(m) + B(m))^2 (exclusive or), with A(m) = A001479(m+1) and B(m) = A001480(m+1). See a comment above for the three cases applying to the first 40 primes 1 (mod 3). The +sqrt or -sqrt applies for negative or positive d(n) = A271228(n), respectively.

a(n) = prime(n) - A271228(n).

Example

Here P(n) stands for prime(n).
n,  P(n), a(n)\ Solutions (x, y) modulo P(n)
1,   2,    2:  (0, 1), (1, 0)
2,   3:    3:  (1, 0), (2, 1), (2, 2)
3,   5,    5:  (2, 0), (3, 2), (3, 3), (4, 1), (4, 4)
4,   7,   12:  (1, 2), (1, 5), (2, 2), (2, 5), (3, 3),
               (3, 4), (4, 2), (4, 5), (5, 3), (5,4),
               (6, 3), (6, 4)
5,  11,   11:  (2, 5), (2, 6), (3, 0), (4, 2), (4, 9),
               (8, 1), (8, 10), (9, 3), (9, 8), (10, 4),
               (10, 7)
...
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The conjecture is for example true for n=4: prime(4) = 7 == 1 (mod 3) = A002476(1). A(1) = 2 , B(1) = 1, q(1)^2 = 1 = (A(1) - B(1))^2 (case 2). a(4) = 7 + sqrt(4*7 - 3*1^2 ) = 7 + 5 = 12 (+sqrt is used here, because d(4) = A271228(4) = -5 (negative)).

Crossrefs

Cf. A002476, A001479, A001480, A271228.

Sequence in context: A187129 A341700 A158936 * A293696 A002139 A140489

Adjacent sequences: A271224 A271225 A271226 * A271228 A271229 A271230

Keyword

nonn,easy

Author

Wolfdieter Lang, Apr 21 2016

Status

approved