A001479 -id:A001479 - cp's OEIS frontend

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

Wolfdieter Lang, Apr 21 2016

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.

			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)
...
----------------------------------------------------------
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)).

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)

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

Cf. A002476, A001479, A001480, A271228.

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).

A271228 P-defects p - N(p) of the elliptic curve y^2 = x^3 + 17 for primes p, where N(p) is the number of solutions modulo prime p.

Original entry on oeis.org

0, 0, 0, -5, 0, -7, 0, -7, 0, 0, -11, -11, 0, -13, 0, 0, 0, 13, 5, 0, 10, 4, 0, 0, -5, 0, -7, 0, -2, 0, -19, 0, 0, 7, 0, -19, -25, -8, 0, 0, 0, 7, 0, -23, 0, 28, 13, -28, 0, -22

Offset: 1

Wolfdieter Lang, Apr 21 2016

See A271227 for details and the conjecture for a(n) if prime(n) == 1 (mod 3).

a(n) is negative for the 1 (mod 3) primes 7, 13, 19, 31, 37, 43, 97, 103, 109, 127, 151, 157, 163, 193, 223, 229, 241, 271, 277, 307, 313, 331, ... and positive for the primes 61, 67, 73, 79, 139, 181, 199, 211, 283, 337, 349, ... See A271227 for a comment on the conjectured three types I, II, and III of 1 (mod 3) primes. All three types appear for primes with negative as well as positive a(n) values.

			n = 4, prime(4) = 7, A271227(4) = 12 (see the example in A271227 for the solutions), a(4) = 7 - 12 = -5. Prime 7 is of type II.
n = 25, prime(25) = 97, A271227(25) = 102, a(25) = -5. Prime 97 is of type III.
n = 29, prime(29) = 109, A271227(29) = 111, a(29) = -2. Prime 109 is of type I.
n = 18, prime(18) = 61, A271227(18) = 48, a(18) = +13. Prime 61 is of type II.
n = 19, prime(19) = 67, A271227(19) = 62, a(19) = +5. Prime 67 is of type III.
n = 21, prime(21) = 73, A271227(21) = 63, a(21) = +10. Prime 73 is of type I.

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)

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

Cf. A271227.

a(n) = prime(n) - A271227(n), where  A271227(n) is the number of solutions of the congruence y^2 = x^3 + 17 (mod prime(n)).
a(n) = 0 precisely for prime(n) == 0, or 2 (mod 3). See the Silverman reference, pp. 400 - 402 for the proof. (The case 0 (mod 3) is trivial.)
Conjecture [WL]: For prime(n) = A002476(m) (a prime  == 1 (mod 3)) one has  a(n) = + or - sqrt(4*prime(n)) - 3*q(m)^2), with three alternative cases for q(m)^2, namely (2*B(m))^2, (A(m) - B(m))^2 and (A(m) + B(m))^2, where A(m) = A001479(m+1) and B(m) = A001480(m+1), m >= 1.

cp's OEIS Frontend

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

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