Legendre polynomials and Jacobsthal sums

Transcription

1 Legendre olynomials and Jacobsthal sums Zhi-Hong Sun( Huaiyin Normal University( htt:// Notation: Z the set of integers, N the set of ositive integers, [x] the greatest integer not exceeding x, ( a the Legendre symbol, R the set of rational numbers whose denominator is corime to. Keywords: congruence, Legendre olynomial, Jacobsthal sum, ellitic curve 1

2 Main references: [S1] Z.H. Sun, Congruences concerning Legendre olynomials, Proc. Amer. Math. Soc. 139(011, [S] Z.H. Sun, Congruences concerning Legendre olynomials II, J. Number Theory 133(013, [S3] Z.H. Sun, Congruences involving ( J. Number Theory 133(013, ( 3, [S4] Z.H. Sun, Congruences concerning Legendre olynomials III, Int. J. Number Theory 9(013, [S5] Z.H. Sun, Legendre olynomials and suercongruences, Acta Arith. 159(013,

3 1. Cubic Jacobsthal sums Let be an odd rime and m, n Z. The sums ( x 3 + mx and are called cubic Jacobsthal sums. ( x 3 + n Let A, B, C Z. The sum ( x 3 + Ax + Bx + C is called general cubic Jacobsthal sum. Hasse s theorem: Let > 3 be a rime, A, B, C R, and let D be the discriminant of x 3 +Ax + Bx + C. If D 0 (mod, then 1 ( x 3 + Ax + Bx + C. 4

4 For a rime > 3 and A, B, C R let #E (y x 3 + Ax + Bx + C be the number of oints on the curve E : y x 3 + Ax + Bx + C over the field F of elements. For A, B, C R, it is easily seen that #E (y x 3 + Ax + Bx + C ( x3 +Ax +Bx+c ( x3 +Ax +Bx+c 1 ( x3 +Ax +Bx+c ( x 3 + Ax + Bx + C ( x3 +Ax +Bx+c

5 Let K Q( d be an imaginary quadratic field and the curve y x 3 + mx + n has comlex multilication by an order in K. By Deuring s theorem, we have #E (x 3 + mx + n { + 1 if is inert in K, + 1 π π if π π in K, where π is in an order in K and π is the conjugate number of π. If 4 u + dv with u, v Z, we may tae π 1 (u + v d. Thus, (1.0 ( x 3 + mx + n { ±u if 4 u + dv with u, v Z, 0 otherwise. In [Gr, Gross(198], [JM, Joux and Morain(1995] and [PV, Padma and Venataraman(1996] the sign of u was determined for those imaginary quadratic fields K with class number 1. In [LM, Lerévost and Morain (1997] and [I, Ishii(004] the sign of u was determined for imaginary quadratic fields K with class number. 6

6 From (1.0 and some nown results we derive that (1.1 ( x 3 11x + 14 { +3 ( 1 4 a if 4 1, a + b, 4 a 1, 0 if 3 (mod 4. (1. ( x 3 15x + { A if 3 1, A + 3B and 3 A 1, 0 if (mod 3. 7

7 (1.3 ( x 3 30x + 56 ( ( 3 c if 8 1, c + d, 4 c 1, ( ( 3 c if 8 3, c + d, 4 c 1, 0 if 5, 7 (mod 8. (1.4 ( x 3 35x + 98 ( 1 +1 ( C 7 C if C + 7D 1,, 4 (mod 7, 0 if 3, 5, 6 (mod 7. 8

8 (1.5 ( x 3 595x ( 1 +1 C( C 7 if C + 7D 1,, 4 (mod 7, 0 if 3, 5, 6(mod 7. (1.6 ( x 3 10x ( L if 3 1, 4 L + 7M, 3 L 1, 0 if (mod 3. I conjectured (1.6 in 010 and roved it in 011 by using Legendre olynomials and Jacobsthal sums. 9

9 It is also nown that (J.C. Parnami and A.R. Rajwade, 198 (1.7 ( x x u if ( 11 1 and 4 u + 11v, ( ( u 11 0 if ( 11 1 and that ([RPR,1984],[JM,1995],[PV,1996] 10

10 (1.8 ( x x + 19 ( ( u 19 0 if ( 19 1, ( x x u if ( 19 1 and 4 u + 19v, ( ( u 43 0 if ( 43 1, ( x x u if ( 43 1 and 4 u + 43v, ( ( u u if ( and 4 u + 67v, 0 if ( 67 1, ( x x ( ( u u if ( and 4 u + 163v, 0 if (

11 . Congruences for 1 (x3 +mx+n (mod Main roblem: For a given odd rime find a simle exression for 1 (x3 +mx+n (mod without Legendre symbols. Many mathematicians attaced this roblem, but their formulas are comlicated. In 006, using the theory of ellitic curves and differential equations P. Morton (JNT 10(006, obtained a general and exlicit formula for 1 (x3 +mx+n (mod in terms of certain Jacobi olynomials. 1

12 Morton (006: Let > 3 be a rime, m, n R and 4m 3 + 7n 0 (mod. Then ( x 3 + mx + n 1 ( 3 ( 48m (864n 1 ( 1 ( 16(4m 3 + 7n [ 1 ] 178 [ 1 ] P ( 1 3 ( 3, 1 ( 1 [ 1 ] ( 4m 3 + 7n 4m 3 + 7n (mod, where P (α,β (x is the Jacobi olynomial given by P (α,β (x 1 r0 ( + α ( + β (x 1 r (x+1 r. r r 13

13 The Legendre olynomials {P n (x} are given by P n (x P (0,0 n (x 1 n n! 1 [n/] n 0 ( n d n dx n(x 1 n ( 1 ( n x n. n A famous formula due to Murhy states that P n (x n 0 ( n ( n + ( x 1. We have P n ( x ( 1 n P n (x and P n (0 0 if n, 1 ( m ( 4 m if n m. m 14

14 Theorem.1 ([S5 (Z.H. Sun, Acta Arith. 159(013, , Theorem.1]. Let > 3 be a rime and m, n R with m 0 (mod. Then ( x 3 + mx + n ( 3m 1 4 P [ ( 3m+1 4 3m 3m 6 ](3n m (mod if 4 1, 3m 6 ](3n m (mod if 4 3. P [ I first deduced Theorem.1 by using Morton s wor, and later found an elementary and straightforward (? roof. One can easily rove the following lemma. 15

15 Lemma.1. Let be an odd rime. Then (i ( 1 1 (ii ( 1 ( 1 (mod for ( 4, ( 6 3 ( 3 4 ( (iii ( [ 3 ] + 1 ( 7 (mod for [ 6 ( 3 [ ] (mod for 3 ( 3. ], Using Lemma.1 one can easily rove: ( [ 6 ] Lemma.. Let > 3 be a rime and {0, 1,..., [ 1 ]}. Then ( [ 6 ] [ 6 ] ( 1 [ 6 ] ( 3 4 [ 6 ] ( 1 [ 3 ] ( ( ( 3 (mod. 16

16 Proof of Theorem.1. For any ositive integer it is well nown (see [IR, Lemma,.35] that x B { 1 (mod if 1, 0 (mod if 1. For, r Z with 0 r 1 we have 0 + r 3( 1. Thus, x +r and therefore { 1 (mod if 1 r, 0 (mod if 1 r 17

17 (.1 (x 3 + mx + n 1 1 ( 1/ 0 ( 1/ 0 r0 r 1 1 r0 ( ( 1/ (x 3 + mx n 1 ( ( 1/ ( ( 1/ ( 1 1 r 1 4 r 1 3 ( r r0 ( r x 3r (mx r n 1 m r n 1 ( 1 r r ( 1 ( 1 r 1 r r m 1 3r n r 1 (mod. x +r m 1 3r n r 1 18

18 If n 0 (mod, from the above we deduce that ( x 3 + mx + n ( (x 3 + mx + n 1 m 1 4 (mod if 4 1, 0 (mod if 4 3. Thus alying Lemma. (with [ 1 ] we get P [ 6 ](0 1 ( [ ( 4 [ 6 ] 1 ] [ ( 1 [ 1 ] ( ( ] ( (mod if 4 1, 0 if 4 3. Hence the result is true for n 0 (mod. Now we assume n 0 (mod. From (.1 we see that 19

19 ( x 3 + mx + n (x 3 + mx + n 1 ( 1 m 1 n 1 ( n ( n 1 4 r 1 3 [ 1 ] r 1 3 ( 1 [ 3 ] ( ( ( 1/ r ( 1 ( 1 r 1 r r ( 1 ( ( 3 r 1 3r ( n m 3 ( n m 3 r n r m 3r [ 3 ] (mod. 0

20 On the other hand, 3m P [ 6 ](3n m [ 6 ] [ [ 6 ] [ 1 ] 0 1 ] 0 ( [ 6 ] ( [ 6 ] ( 3n 3m ( 1 [ m 3m ( 1 ( [ 6 ] ( 3n [ 6 ] m ( 1 ( [ 6 ] [ 6 ] 1 ( 1 1 ] [ [ 6 ] ( 3n 3m m δ(m, 1 ( n [ 1 ] 0 ( [ 6 ] 1 ] 0 ( [ 6 ] 1 ( 1 ( 3 ( 7n [ 1 ] 4m 3 ( [ 6 ] [ 6 ] ( 7n [ 4m 3 ( 1 [ 1 ] [ 6 ] 3 ] ( 1 4 ( [ 6 ] ( 7n [ 3 ] [ 6 ] 4m 3 (mod, 1 [ 6 ]

21 where δ(m, ( 3m 1 4 if 1 (mod 4, ( 3m +1 4 if 3 (mod 4. 3m Hence, by the above and Lemma. we get 3m δ(m, P [ 6 ](3n m ( [ 1 ] 0 ( 1 ( 1 [ 6 ] ( 3 ( [ 6 ] ( ( 7n [ 3 ] ( 3 4m 3 ( 3 ( 1 [ 1 ] [ 6 ] (mod. ( n [ 3 ]

22 Since ( 3 ( 1 [ 1 ] [ 6 ] ( 1 [ 6 ] ( 3 ( 1 [ ( 1 [ 1 ]+[ 6 ]( 3 1 ]+[ 6 ]( 3 4 [ 6 ] ( 7 4 [ 6 ] [ 3 ] 3 3[ 3 ]+(1 ( 3 / ( 1 [ 1 ]+[ 6 ] ( 3 ( 1 6 ( 1 [+1 4 ] ( 1 [ 1 ] 1 (mod, from the above we deduce that ( 3n 3m δ(m, P [ 6 ] m ( [ n 1 ] ( 1 ( ( [ 3 ] ( 3 0 ( x 3 + mx + n This comletes the roof. (mod. ( n m 3 [ 3 ] [ 3 ] 3

23 For a rime > 3 and {0, 1,..., 1} it is easily seen that ( [ 6 ]( [ 6 ] + ( [ 6 ] + ( (mod. ( 43 Therefore ( ( for 6 < <, and [/6] (. P [ 6 ](t ( ( 3 ( 6 3 ( 3 ( 1 t 864 (mod. Corollary.1. Let > 3 be a rime and m, n R with m 0 (mod. Then P [ 6 ] ( n m 3 [/6] 0 ( 6 ( 3m 3 ( 3 ( m 3 n 1 3 m 3 ( x 3 3m x + n (mod. 4

24 Corollary.. Let > 3 be a rime, and let c(n be given by q 1 Then (1 q (1 q 11 c( P [ 6 ] ( 19 8 ( 1 1 n1 [/6] 0 c(nq n ( q < 1. ( ( (mod. Proof. It is easy to see that the result holds for 11. Now assume 11. By the well nown result of Eichler (see [KKS, Theorem 1.], we have {(x, y F F : y + y x 3 x } c(. Since 5

25 {(x, y F F : y + y x 3 x } ( 1 { (x, y F F : y + x 3 x { (x, y F F : y x 3 x + 1 } 4 ( x 3 x ( 1 6 we obtain ( (x (x ( x x ( ( x x (.3 c( 108 ( x 3 1x + 38 ( 6, ( x 3 1x }

26 Using Corollary.1 we see that Therefore, ( 19 P [ 6 ] c( 8 ( 6 P [ 6 ] ( 19 8 ( 1 [ 6 ] P [ 6 ] ( 1 1 ( 1 1 ( x 3 1x + 38 (mod. ( [/6] 0 [/6] 0 Thus the result follows ( 6 3 ( 3 ( ( ( / (mod. Remar.1 Set q e πiz and f(z q 1 (1 q (1 q 11. It is nown that f(z is the unique weight modular form of level 11.

27 For ositive integers a 1, a, a 3, a 4 let q 1 n1 (1 q a 1 (1 q a (1 q a 3 (1 q a 4 c(a 1, a, a 3, a 4 ; nq n ( q < 1. For (a 1, a, a 3, a 4 (1, 1, 11, 11, (,, 10, 10, (1, 3, 5, 15, (1,, 7, 14 and (4, 4, 8, 8 it is nown that (see [MO, Theorem 1] f(z n1 c(a 1, a, a 3, a 4 ; nq n (q e πiz are weight newforms. Conjecture.1 Let > 3 be a rime. Then c(,, 10, 10; c(, 4, 6, 1; c(1, 3, 5, 15; ( 3 ( 1 3 ( 1 3 ( x 3 1x 11 ( x 3 39x 70 ( x 3 3x 3 7,,

28 and c(1,, 7, 14; ( 3 ( x 3 75x 506 Is Conjecture.1 nown according to Wiles wor on Fermat s last theorem? Wiles revealed the connection between ellitic curves and modular forms. Theorem. ([S5, Theorem.6]. Let > 3 be a rime and m, n R with m 0 (mod. Then ( x 3 + mx + n ( 3m 1 4 [ 1 ] 0 3n ( 3m+1 4 m ( [ 1 ] [ 1 ] 0 ( [ 5 1 ] ( 4m 3 + 7n 4m 3. (mod if 4 1, ( [ 1 ] ( [ 5 1 ] ( 4m 3 + 7n 4m 3 (mod if

29 Proof. It is nown (see [AAR,.315] that P n (x P (0, 1 n (x 1, P n+1 (x xp (0,1 n (x 1. From [H. Bateman, Higher Transcendental Functions (vol II, 1953,.170] we now that P (0,β n (x n 0 ( n ( n β 1 ( 1 x. If 1 (mod 4, then [ 6 ] [ 1 ] and so ( 3n 3m P [ 6 ] [ 1 ] m ( [ 0 [ 1 ] ( [ 1 ] 0 [ 1 ] ( [ 0 1 ] 1 ] P (0, 1 ( [ 1 ] 7n 4m 3 1 ( 1 [ ( 1 ( [ 5 1 ] [ 1 ] 1 ] ( 1 + 7n 4m 3 ( 4m 3 + 7n 4m 3 ( 4m 3 + 7n (mod ; 4m 3 9

30 P [ 6 ] if 3 (mod 4, then [ 6 ] [ 1 ] + 1 and so ( 3n 3m m 1 1 ] 3n 3m m P (0, [ 3n 3m m 3n 3m m 3n 3m m [ 1 ] 0 [ 1 ] ( [ 0 [ 1 ] 0 ( 7n 4m 3 1 ( [ 1 ] 1 ] ( [ 1 ] ( 3 [ ( 3 ( [ 5 1 ] [ 1 ] 1 ] ( 1 + 7n 4m 3 ( 4m 3 + 7n 4m 3 ( 4m 3 + 7n (mod. 4m 3 Now combining the above with Theorem.1 we deduce the result. 30

31 Theorem.3 (Z.H. Sun [S1-S5] For any rime > 3 and t R, P 1 (t ( 6 1 ( x3 3(t + 3x + t(t 9 (mod ( t 1 P 1 ( t ( x 3 tx + tx (mod, P [ 1 4 ](t (6 ( x3 3(3t+5 x + 9t + 7 (mod, P [ 3 ](t ( 3 ( x3 + 3(4t 5x + (t 14t + 11 (mod, P [ 1 6 ](t (3 ( x3 3x + t (mod. 31

32 3. A congruence for ( 1 (x3 +mx+n (mod Lemma 3.1. we have n ( 0 n 0 For any nonnegative integer n ( 3 ( 6 ( ( 3 ( n ( 43 n ( 3(n ( 6(n n 3(n Lemma 3.1 can be roved by using WZ method. Using Lemma 3.1 we deduce: Theorem 3.1 ([S5, Theorem 3.1]. Let be an odd rime and let x be a variable. Then ( 0 ( 0 ( 3 ( 6 ( 3 3 ( 6 3 (x(1 43x (mod x. Theorem 3.1 can be viewed as a analogue of Clausen s identity.. 3

33 ( Theorem 3. ([S5, Theorem 4.]. Let > 3 be a rime and m, n R with m 0 (mod. Then ( x 3 + mx + n ( 3m 1 0 ( Moreover, if ( ( 3 ( 6 ( 3 ( 6 ( 4m 3 + 7n (mod m 3 (x3 +mx+n 0, then ( 4m 3 + 7n 0 (mod m 3. Proof. We first assume that 4m 3 + 7n 0 (mod. Clearly 3m ( m 9n (mod and so ( 3m 1. As x3 + mx + n (x 3n m (x + 33

34 3n m (mod we see that ( x 3 + mx + n ( (x 3n m (x + 3n x 3n t0 m ( t (mod m ( x 3n m ( 3n m 3n m ( mn.

35 Since m 0 (mod we have n 0 (mod and so 1 (x3 +mx+n ( mn ±1. Thus the result holds in this case. Now we assume that 4m 3 + 7n 0 (mod. Set t 3n 3m m and m m3 4m 3 +7n. Then t 1 178/m 1. From Theorems.1 and 3.1 we have ( ( x 3 + mx + n ( 3m 1 P [ ( 3m Recall that 1 (x3 +mx+n 6 ](t ( 1/ 0 ( 3 ( ( 6 3 ( 3 m 1 ( 6 3 for 6 (mod. < <. If 0, we must have P [ 6 ](t 0 (mod. Thus, alying Theorem 3.1 we see that ( 1/ 0 ( ( 3 m 1 ( (mod. 34

36 4. A theorem concerning eta roducts and binary quadratic forms Theorem 4.1 (Z.H. Sun, Adv. in Al. Math. 48(01, For a, b, n N let λ(a, b; n Z be given by q 1 (1 q a 3 (1 q b 3 n1 λ(a, b; nq n ( q < 1. (1 Suose ab and is an odd rime such that a, b, ab + 1 and ax + by with x, y Z. Let n ((ab + 1 a b/8. Then ( 1 a+b x+b+1 (4ax λ(a, b; n + 1. ( Let a, b N with (a, b 1. Let be an odd rime such that ab, ab+1 and x +aby with x, y Z. Let n (a + b( 1/8. If a, b, 8 b and 8 1, then ( 1 y (4x λ(a, b; n