LegendreIIJNTFinal.dvi

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1 J Number Theory , CONGRUENCES CONCERNING LEGENDRE POLYNOMIALS II ZHI-Hong Sun School of Mathematical Sciences, Huaiyin Normal University, Huaian, Jiangsu 3001, PR China Homeage: htt://wwwhytceducn/xsjl/szh È 3/m 0 mod 1, 0 /m È /m mod 1 In articular, we show that 0 Abstract Let > 3 be a rime, and let m È be an integer with m In the aer we 1 solve some conjectures of ZW Sun concerning È 1 3 mod and 0 0 mod for 3, 5, 6 mod 7 Let {P n x} be the Legendre olynomials In the aer we also show that P [ ]t 6 È 1 mod, where t is a rational adic integer, [x] is the greatest integer not exceeding x and a is the Legendre symbol As consequences we determine P [ ]t mod in the cases t 5 3, 7 9, 65 and confirm many 63 conjectures of ZW Sun x0 x3 3 3t+5x+9t+7 MSC: Primary 11A07, Secondary 33C5, 11E5, 11L10, 05A10, 05A19 Keywords: Legendre olynomial; congruence; character sum; binary quadratic form; ellitic curve 1 Introduction Let be an odd rime Following [A] we define A, λ λ 1 0 It is easily seen that see [S, Lemmas, and 5] for {0, 1,, 1 }, mod, 1 + Thus, by Fermat s little theorem, for any rational -adic integer λ, 16 mod 1 A, λ λ mod 6 The author is suorted by the National Natural Sciences Foundation of China No

2 For ositive integers a, b and n, if n ax + by for some integers x and y, we briefly say that n ax + by In 1987, Beuers[B] conjectured a congruence for A, 1 mod equivalent to 13 1/ 0 3 { 0 mod 6 if 3 mod, x mod if x + y 1 mod This congruence was roved by several authors including Ishiawa[Is] 1 mod, van Hamme[H] 3 mod and Ahlgren[A] In 1998, by using the hyergeometric series 3F λ over the finite field F, Ono[O] obtained congruences for A, λ mod in the cases λ 1, 8, 1 8,, 1, 6, 1 6, see also [A] Hence from 1 we deduce congruences for m mod in the cases m 1, 8, 16, 6, 56, 51, 096 Recently the author s brother Zhi-Wei Sun[Su1,Su,Su3] conjectured corresonding congruences modulo In articular, he conjectured that { 0 mod if 3, 5, 6 mod 7, x mod if x + 7y 1,, mod 7 We note that for +1 1 In the aer, by using Legendre olynomials and the wor of Coster and van Hamme[CH] we comletely solve Zhi-Wei Sun s such conjectures Let {P n x} be the Legendre olynomials given by P 0 x 1, P 1 x x, n + 1P n+1 x n + 1xP n x np n 1 x n 1 It is well nown that see [MOS, 8-3], [G, ] 15 P n x 1 n [n/] 0 n n 1 x n 1 n n n! where [x] is the greatest integer not exceeding x From 15 we see that d n dx n x 1 n, 16 P n x 1 n P n x, P m and P m 0 1m m The first few Legendre olynomials are as follows: m m P 0 x 1, P 1 x x, P x 1 3x 1, P 3 x 1 5x3 3x, P x x 30x + 3, P 5 x x5 70x x Let Z be the set of integers For a rime, let Z be the set of those rational numbers whose denominator is not divisible by, let Q denote the field of -adic numbers, and let a be the Legendre symbol On the basis of the wor of Ono[O] see also [A, Theorem ] and [LR], in [S, Theorem 11] the author showed that for any rime > 3 and t Z, 6 17 P 1 t 1 x 3 3t + 3x + tt 9 mod x0

3 In this aer, by using elementary arguments we rove that for any rime > 3 and t Z, 18 ] t 6 1 x 3 3 3t + 5x + 9t + 7 mod x0 From 18 we deduce 17 immediately As consequences of 18, we determine ] 5 3, ] 7 9, ] 65 mod and use them to solve ZW Sun s conjectures [Su1] on m For instance, for any rime > 7, mod and 1 0 m mod { C mod if 1,, mod 7 and so C + 7D, 81 0 mod if 3, 5, 6 mod 7 For any odd rime and m Z we also establish the following general congruences: m P mod for m 0 mod, m x1 6x 1 0 x mod Congruences for ] t mod Lemma 1 Let be an odd rime and {0, 1,, [ ]} Then i ii iii 1 1 mod, [ ] mod, mod Proof It is clear that 1 1! !! 1 3 1! mod This together with 11 yields i 3

4 Suose r 1 or 3 according as 1 or 3 Then clearly [ ] + r + r + 1 r + 1! r r r rr + + r 1! 1!!! mod 6 and ! ! ! !!!! 1 mod Thus ii and iii are true and the roof is comlete Lemma Let be an odd rime and let t be a variable Then ] t [/] 0 [/] 1 [ ] 0 1 t t 18 mod Proof It is nown that [G, 3135] 1 P n t Observe that n ] t n+ [/] 0 n+ [ ] + n n n + t 1 0 By 1 and Lemma 1ii we have t 1 [/] 0 By 16, ] t 1 [ ] ] t Thus the result follows t 1 mod 6

5 Theorem 1 Let be a rime greater than 3 and let t be a variable Then 1 ] t x 3 + x + 1 tx 1 x0 6 1 x0 x 3 3 3t + 5x + 9t mod Proof For any ositive integer it is well nown that see [IR, Lemma, 35] 1 { 1 mod if 1, x 0 mod if 1 x0 Suose that x 3 + x + 1 tx 1 3 1/ 1 a x From the above we deduce that 3 1/ a 1 1 a 1 x0 1 x x 3 + x + 1 tx 1 mod x0 On the other hand, x 3 + x + 1 tx 1 x 1 x x + x 1 1 t r0 r x r x 1 r 1 t Hence, alying 11 and Lemmas 1- we obtain a 1 [/] 0 [/] 0 [/] t t 1 t P[ ] t mod

6 Therefore, This yields the result 1 ] t a 1 x 3 + x + 1 tx 1 1 x0 1 x0 x0 x 3 1 x 3 x0 3 + x t x 3 x 3 3 3t + 5x t t + 5 x t + 7 x0 x 3 3 Corollary 1 Let > 3 be a rime and t Z Then 3t + 5x + 9t mod 1 x 3 33t+5 x + 9t x t 5 x0 x0 Proof Since ] t 1 [ ] ] t, by Theorem 1 we obtain 1 x 3 33t+5 x + 9t + 7 x0 1 x 1 [ ] 3 3 3t+5 x 9t + 7 x0 1 x t 5 x + 9t 7 mod x0 1 x + 9t 7 x t 5 x 9t + 7 For {5, 11, 13} it is easy to chec that the result is true for t 0, 1,, 1 Now suose 17 By Hasse s estimate [C, Theorem 11, 315], x0 1 x 3 3±3t+5 x + 9t ± 7 x0 Since <, from the above we deduce the result for 17 Theorem Let be a rime greater than 3 Then ] 7 [/] [/] [ ] { a mod if 1, a + b and a 1, 0 mod if 3 mod 6

7 Proof From [BEW, Theorem 69] or [S1, 15-16] we have 1 x 3 x { a if 1 mod, a + b and a 1 mod, 0 if 3 mod x0 Thus taing t 7/9 in Lemma and Theorem 1 we deduce the result Theorem 3 Let be a rime greater than 3 Then ] 5 [/] [/] 1 [ 3 8 ] { A mod if 3 1, A + 3B and 3 A 1, 0 mod if mod 3 3 Proof It is nown that see for examle [S1, 7-9] or [BEW, ] 1 x 3 8 x0 { A if 3 1, A + 3B and 3 A 1, 0 if mod 3 Thus, utting t 5/3 in Lemma and Theorem 1 we deduce the result Theorem Let, 3, 7 be a rime Then ] 65 [/] [/] 1 [ ] { C 3 C 7 mod if 1,, mod 7 and so C + 7D, 0 mod if 3, 5, 6 mod 7 Proof Putting t 65/63 in Lemma we get ] 65 [/] 1 [ Taing t 65/63 in Theorem 1 we see that ] From [R1,R] we have 1 x 3 + x + 56 x0 63 x [/] ] 1 x x 3 + 1x + 11x mod x0 1 x 3 + 1x + 11x x0 Thus the result follows mod 03 1 x3 + 1 x x { C C 7 if C + 7D 1,, mod 7, 0 if 3, 5, 6 mod 7 For related conjectures concerning Theorems -, see [Su1, Conjectures A7-A9] 7

8 Lemma 3 Let be an odd rime and let u be a variable Then u 1 P 1 [/] u 1 [ ] u 1 6 mod Proof From 15 we see that u 1 P 1 u 1 1 [/] Thus alying 11 and Lemma 1 we obtain u 1 P 1 Noting that u [/] [ ] we then obtain the result 1 1 u 1 1 u 1 mod Theorem 5 Let > 3 be a rime, u Z and u 0 mod Then u 1 P 1 u u ] u 1 1 x 3 ux + ux x0 6 1 ux 3 3u + 9x + 18 u mod x0 Proof By 16, Lemmas and 3 we have ] u 1 By Theorem 1, we have and u u 1 ] 1 u P 1 [/] 1 [ ] 0 u mod ] 1 /u 1 x 3 + x + u x 1 x u 3 + x u + u x u x0 x0 1 u 3 x 3 ux + ux u x0 ] u 1 Thus the theorem is roved 6 1 6u x u x 3 ux + ux mod x0 x 3 3u+3 x + 18 u u u ux 3 3u + 3x + 18 u mod x0 8

9 3 Congruences for m mod Lemma 31 For any nonnegative integers m and n we have m 0 n + m m 0 n + m m Proof Let m and n be nonnegative integers For {0, 1,, m} set n + F 1 m,, m n + m n + m F m, m m For {0, 1,, m + 1} set n + G 1 m, m +, m + n + m m G m, 3mn n + m + 3mn m + 6n n + m + 6n + m + n + m + 1 n + m + 1 m + 1 m + 1 For i 1, and {0, 1,, m}, using Male it is easy to chec that 31 m + 3 F i m +, + m + 3m n + 3m n + F i m + 1, + m + 1m + n + m nf i m, G i m, + 1 G i m, Set S i r r 0 F ir, for r 0, 1,, Then m + 3 S i m + F i m +, m + F i m +, m m + 3m n + 3m n + S i m + 1 F i m + 1, m m + 1m + n + m ns i m m m m + 3 F i m +, + m + 3m n + 3m n + F i m + 1, 0 + m + 1m + n + m n m F i m, 0 m G i m, + 1 G i m, G i m, m + 1 G i m, 0 G i m, m From the above we deduce that for i 1, and m 0, 1,,, 3 m + 3 S i m + + m + 3m n + 3m n + S i m m + 1m + n + m ns i m 0 G i m, m m + 3 F i m +, m + + F i m +, m m + 3m n + 3m n + F i m + 1, m

10 Since S S 0 and S 1 1 nn+1 S 1, from 3 we deduce S 1 r S r for all r 0, 1,, This comletes the roof Remar 31 We actually find 31 and rove Lemma 31 by using WZ method and Male For the WZ method, see [PWZ] Definition 31 Note that n n+ n+ For any nonnegative integer n we define S n x n 0 n n + x Lemma 3 Let n be a nonnegative integer Then Proof From 1 we have n P n x On the other hand, n 0 n + S n x P n x 1 + x and P n x 1 S n 0 n m0 n + x 1 n x 1 m m x 1 S n n 0 0 n 0 n + n + r0 r r m m x 1 n n + x 1 r0 x 1 m m n + m0 0 Hence, from the above and Lemma 31 we deduce x P n x 1 S n and so x n + r x 1 r r n + m m 1 + x 1 r x 1 m P n 1 + x S n x This roves the lemma Remar 3 Since S n 1 P n0 by Lemma 3, using 16 we deduce that n 0 n + 1 This is an identity due to Bell [G, 635] 10 { 0 if n is odd, 1 n n/ if n is even n r

11 Theorem 31 Let be an odd rime and m Z with m 0 mod Then m P mod m Proof By Definition 31 and 11 we have S 1 16 m On the other hand, by Lemma 3 we get S 1 16 P 1 m Thus the result follows 16 m m 1 P m mod m 6 Theorem 3 Let be an odd rime, m Z and m 0, 6 mod Then m Moreover, m 3 m mm 6 m + 6 P [ ] mod m 6 m mod ] 0 mod m 6 Proof Set u 1 6/m From Theorem 5 we now that u 1 P 1 u u P [ ] u 1 mod Thus, u P 1 u u 1 It then follows from Theorem 31 that m P 1 u ] u 1 mod P 1 u mm 6 m + 6 u P [ ] u 1 P [ mod ] m 6 From 15 we see that u m 0 mod P 1 u Z Thus, by Theorems 31 and 5, P 1 u 0 mod u 1 P 1 u 0 mod u 1 P 1 u 0 mod ] u 1 0 mod m + 6 ] 0 mod m 6 Thus the theorem is roved 11

12 Theorem 33 Let be an odd rime Then i ii iii iv mod for 3, 5, 6 mod 7, mod for 3 mod, mod for mod 3, 6 0 mod for 5, 7 mod 8 Proof Taing m 1, 096, 8, 51, 16, 56, 6 in Theorem 3 and then alying Theorems - and 16 we obtain the result Theorem 3 Let be an odd rime i If 1,, mod 7 and so C + 7D with C, D Z, then C mod ii If 1 mod and so a + b with a, b Z and a, then a mod iii If 1 mod 3 and so A + 3B with A, B Z, then A mod iv If 1, 3 mod 8 and so c + d with c, d Z, then c mod 1

13 Proof From [CH, 8, Tables II and III] we now that P x y 7 mod for x + 7y with x + y 1, P 1 x y 7 mod for x + 7y with x + y 1, 8 7 P x y 1 mod for x + y 1 mod with x 1, 3 P 1 P 1 P 1 1 y x y 1 mod for x + y 1 mod with x 1, x y 3 mod for x + 3y 1 mod 3 with x + y 1, 3 1 x y 3 mod 3 3 for x + 3y 1 mod 3 with x + y 1, P 1 yx y mod for x + y 1, 3 mod 8 with x 1, where d Q and x y d 1 is the usual valuation on Q If d 1, x +dy and A Q with A x y d mod and x y d 1, then A x dy x mod and so A xa mod Hence A x mod and so xa A x +xa A xa+x x mod Therefore, A x x mod and so A x mod From all the above we deduce that P P 1 x mod for x + 7y, 8 P P 1 x mod for x + y 1 mod with x see also [S, Corollary 3 and Theorem 9], P P 1 x mod for x + 3y 1 mod 3, P 1 1 y x 1 1 x mod for x + y 1, 3 mod 8 Now taing m 1, 096, 8, 51, 16, 56, 6 in Theorem 31 and then alying the above we derive the result 13

14 Corollary 31 Let be a rime such that, 3, 7 Then 1/ 3 { C mod if 1,, mod 7 and so C + 7D, 8 1 mod if 3, 5, 6 mod 7 0 Proof By [Su, Theorem 13] we have 1/ mod 0 Thus alying Theorems 33i and 3i we conclude the result Remar 33 Theorems 33, 3 and Corollary 31 were conjectured by Zhi-Wei Sun in [Su1,Su3] Lemma 33 For any ositive integer n we have the following identities: n 1 n 1 n 1 n + n + n + 3 n n 1 n n 1 if n, if n, nn+1 n n n n n +n 1 n 1 3 n n 1 n n+1 3 n 1 n n if n, if n, if n, n n n+1 nn+1+1 n 1 15 n n 1 n n+1 n+1 15 n n n if n Proof We only rove the first identity The other identities can be roved similarly Let f x d dx fx be the derivative of fx By Lemma 3, S nx P n 1 + x If n, then P n 1 + x is a olynomial in 1 + x Thus and so From 15 we see that S nx P n 1 + x S n d dx P n 1 + x d 1 dx n d dx P n 1 + x 1 Pn 0 d dx P n 1 + x x 1 n n 1 0 n/ 0 n n x n n n n 1 n n 1 + x n 1 1

15 and therefore d dx P n 1 + x x 1 n n n 1 1 n 1 n 1 n n 1 nn + 1 n 1 n n/ 1 n Combining the above with 16 we get S n 1 By Definition 31, 1 n/ n 33 S n n 1 n n/ 1 n 1 Thus the first identity is true when n is even Now we assume n Set 1 nn + 1 n n 1 n/ n + 1 nn + 1 n n n/ A n x n 1 0 n n x n 1 n From Lemma 3 and 15 we have Thus, and therefore S n Pn 1 + x 1 S n x 1 + x 1 + x n 1 + xa nx S n x 1 n A nx x A n xa n x 1 A n n n + 1 n 1 n n 1 n n n 1 n n 1 n 1 n 1 n 1 n n 1 This together with 33 yields the first identity in the case n The roof is now comlete We remar that Lemma 33 can also be roved by using WZ method Lemma 3 Let be a rime of the form + 1 and so a + b with a, b Z and a Then a mod 15 n

16 Proof From [CDE] or [BEW] we now that for a 1 mod, 1 1 Set q 1 1/ We then have This roves the lemma a mod a a Theorem 35 Let > 5 be a rime Then { q 1 + q a a mod + 1 3/ 3/ mod if 3 mod, a mod if a + b 1 mod, { 1 3/ 3/ mod if 3 mod, 1 1 a mod if a + b 1 mod, { / 3/ mod if 3 mod, Proof By 11 and Lemma 33, 0 mod if 1 mod / 3/ mod if 3 mod, Set q 1 1/ For 3 mod we see that q 11 q mod if 1 mod q + 1 mod Now combining all the above with Lemma 3 we obtain the first congruence The remaining congruences can be roved similarly Remar 3 In [Su3,110], Zhi-Wei Sun obtained the congruence for 1 He also conjectured that mod 6 0 mod for any rime 1 mod with 16

17 A general congruence modulo Lemma 1 For any nonnegative integer n, we have n n n n 6 n n n n 0 Proof Let m be a nonnegative integer For {0, 1,, m} set F 1 m, 6 m, m m m F m, m m For {0, 1,, m + 1} set G 1 m, 6 m + 6 m+1, m + 0 G m, 16m 16m + 55m m + m + 1 m + 1 m + 1 m + 1 For i 1, and {0, 1,, m}, using Male it is easy to chec that 1 m + 3 F i m +, 8m + 38m + m + 19F i m + 1, + 10m + 1m + 1m + 3F i m, G i m, + 1 G i m, Set S i n n 0 F in, for n 0, 1,, Then m + 3 S i m + F i m +, m + F i m +, m + 1 8m + 38m + m + 19S i m + 1 F i m + 1, m m + 1m + 1m + 3S i m m m m + 3 F i m +, 8m + 38m + m + 19 F i m + 1, m + 1m + 1m + 3 m F i m, 0 m G i m, + 1 G i m, G i m, m + 1 G i m, 0 G i m, m From the above we deduce that for i 1, and m 0, 1,,, 0 m + 3 S i m + 8m + 38m + m + 19S i m m + 1m + 1m + 3S i m G i m, m m + 3 F i m +, m + + F i m +, m + 1 8m + 38m + m + 19F i m + 1, m Since S S 0 and S 1 1 S 1, from we deduce S 1 n S n for all n 0, 1,, This comletes the roof 17

18 Theorem 1 Let be an odd rime and let x be a variable Then x1 6x Proof It is clear that 1 x1 6x m0 x min{m, 1} x m 0 r x r r x mod 6 m m Suose m and 0 1 If >, then and so If <, then m > and so m 0 Thus, from the above and Lemma 1 we deduce that m0 1 m x m m 0 x m m 0 x1 6x 0 x 1 6 m m m m m m m 1 x r0 r r 1 x 0 r r m m m m x r x 1 r x m r r r r x r mod Now suose 0 1 and r 1 If 3, then!, 3! and so!!! 0 mod If <, then r > 3 and so r r r r r! r!r! 0 mod If < <, then r >, r r and If < < 3, then r >, and r r r r Hence we always have r r r r 0 mod and so 1 1 r r x x r 0 mod r r r Now combining all the above we obtain the result 18

19 Corollary 1 Let be an odd rime and m Z with m 0 mod Then 1 0 m /m mod 18 Proof Taing x /m 18 in Theorem 1 we deduce the result Theorem Let be an odd rime, m Z, m 0 mod and t 1 56/m Then 1 0 m ] t 1 x 3 + x + 1 tx 1 mod Moreover, if ] t 0 mod or 1 x0 x3 + x + 1 tx 1 0 mod, then 1 0 x0 0 mod m Proof For < <,!! 0 mod For < <,!! 0 mod Thus, for < < Now combining Lemma, Theorem 1 with Corollary 1 gives the result Theorem 3 Let 1, 3 mod 8 be a rime and so c + d with c, d Z and c 1 mod Then [ 8 1 ]+ c mod c Proof By Lemma, Theorem 1 and [BE, Theorems 51 and 517], 1 0 Set 1 0 [/] P 18 [ 1 x 3 + x + x ] 0 x0 1 x 3 + x + x 1 x0 1 [ 1 8 ]+ c mod 1 [ ]+ c + q Then 1 [ 18 x 3 x + x 1 x0 1 8 ]+ c + q c + 1 [ 1 8 ]+ 19 cq mod

20 Taing x 1 18 in Theorem 1 we get mod 18 From [M] and [Su] we have 1 0 /56 c mod Thus c 1 0 and hence q 1 [ 1 8 ]+ 1 c c [ 1 8 ]+ mod So the theorem is roved cq mod We note that Theorem 3 was conjectured by Zhi-Wei Sun in [Su1] 5 Congruences for 1 0 / m Theorem 51 Let, 3, 7 be a rime Then { 0 mod if 3 mod, 68 a mod if a + b 1 mod, { 0 mod 1 if mod 3, A mod if A + 3B 1 mod 3, { 0 mod 3969 if 3, 5, 6 mod 7, C mod if C + 7D 1,, mod 7 Proof Taing m 68, 1, 3969 in Theorem and then alying Theorems - and 16 we deduce the result We mention that Theorem 51 was conjectured by the author in [S] Lemma 51 [S3, Lemma 1] Let be an odd rime and let a, m, n be algebraic numbers which are integral for Then 1 1 x 3 + a mx + a 3 n 1 a 1 x 3 + mx + n 1 mod x0 Moreover, if a, m, n are congruent to rational integers modulo, then x0 x0 1 x 3 + a mx + a 3 n a 1 x 3 + mx + n x0 0

21 Theorem 5 Let, 3, 7 be a rime Then 5 7 { ] C 7 C mod if 1,, mod 7 and so C + 7D, 0 mod if 3, 5, 6 mod 7 and 1 0 { C mod if 1,, mod 7 and so C + 7D, 81 0 mod if 3, 5, 6 mod 7 Proof By Theorem 1, 5 7 ] x0 x x mod Since by the above and Lemma 51 we have 1 7 and , ] x 3 35x 98 x0 mod As x 3 + 1x + 11x x x , by we get 51 1 x 3 35x 98 x0 { C C 7 if C + 7D 1,, mod 7, 0 if 3, 5, 6 mod 7 For 1,, mod 7 we see that mod Thus, from the above we deduce the congruence for with m 81 we obtain the remaining result ] mod Alying Theorem Let > 3 be a rime and let F be the field of elements For m, n F let #E x 3 + mx + n be the number of oints on the curve E: y x 3 + mx + n over the field F It is well nown that see for examle [S1, 1-] 1 x 5 #E x mx + n + mx + n x0 1

22 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 [C, Theorem 116],[PV],[I], we have { + 1 if is inert in K, 53 #E x 3 + mx + n + 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 u + dv with u, v Z, we may tae π 1 u + v d Thus, 5 1 x 3 + mx + n { ±u if u + dv with u, v Z, 0 otherwise x0 In [JM] and [PV] the sign of u in 5 was determined for those imaginary quadratic fields K with class number 1 In [LM] and [I] the sign of u in 5 was determined for imaginary quadratic fields K with class number Theorem 53 Let be a rime such that ±1 mod 1 Then { ] x mod if x + 9y 1 mod 1 with 3 x 1, 1 0 mod if 11 mod 1 and 1 0 { x 188 mod if x + 9y 1 mod 1, 0 mod if 11 mod 1 Proof From [I, 133] we now that the ellitic curve defined by the equation y x x has comlex multilication by the order of discriminant 36 Thus, by 5 and [I, Theorem 31] we have By Theorem 1, 1 n n n0 { x 1+ 3 if x + 9y 1 mod 1 with 3 x 1, 0 if 11 mod ] n0 6 1 n0 n n n 8 n n n mod n0

23 Now combining all the above we obtain the congruence for ] mod Alying Theorem with m 188 and t we deduce the remaining result Remar 51 In [Su1, Conjecture A], ZW Sun conjectured that for any rime > 3, [ x 6 ] x mod if x + y 1 mod 1 and x 1, xy 3 xy mod if x + y 5 mod 1 and x 1, 0 mod if 3 mod Theorem 5 Let be an odd rime such that ±1 mod 8 Then { 1 1 ] x 3 x mod if x + 6y 1, 7 mod, 3 0 mod if 17, 3 mod and 1 0 { x mod if x + 6y 1, 7 mod, 8 0 mod if 17, 3 mod Proof From [I, 133] we now that the ellitic curve defined by the equation y x x 8 + has comlex multilication by the order of discriminant Thus, by 5 and [I, Theorem 31] we have By Theorem 1, 1 n n 8 + n0 { x x 3 1+ if 1, 7 mod and so x + 6y, 0 if 17, 3 mod 6 ] 3 1 n0 n n mod Since / by Lemma 51 and the above we have and , n n 8 + ] 3 n0 { 6 xx 3 mod if x + 6y 1, 7 mod, 0 mod if 17, 3 mod 3

24 ] This yields the result for 3 mod Taing m 8 and t 3 in Theorem and alying the above we deduce the remaining result Let b {3, 5, 11, 9} and fb 8, 1, 158, 396 according as b 3, 5, 11, 9 For any odd rime with bfb, ZW Sun conjectured that [Su1, Conjectures A1, A16, A18 and A1] fb x mod if x + by, 8x mod if x + by, 0 mod if b 1 For m {5, 13, 37} let if m 5, fm 89 if m 13, 10 1 if m 37 and tm 1 56 fm 5 if m 5, if m 13, if m 37 Suose that is an odd rime such that mfm In [Su1], ZW Sun conjectured that 57 1 x mod if m 1 1 and so x + my, x mod if m fm and so x + my, 0 mod if m 1 Theorem 55 Let m {5, 13, 37} and let be an odd rime such that m 1 and fm Then 1 0 fm { x mod if 1 mod and so x + my, 0 mod if 3 mod Proof Let tm be given in 56 From [LM, Table II] we now that the ellitic curve defined by the equation y x 3 + x + tmx has comlex multilication by the order of discriminant m Thus, by 5 we have 1 n 3 + n + tmn { x if 1 mod and so x + my, 0 if 3 mod, n0 for m 5 see also [LM, Theorem 11] Now alying the above and Theorem we deduce the result Remar 5 Let 1 mod 0 be a rime and hence a + b x + 5y with a, b, x, y Z A result of Cauchy [BEW, 91] states that { x mod if 5 a, x mod if 5 a

25 Using the arguments similar to the roof of Theorem 55, [LM, Table II] or [I] and 5 we see that for any odd rime, 58 { 1 n 3 + n n x if 5 1 and so x + 10y, n0 0 if 5 1 and 1, { x if 11 1 and so x + y, 1 n 3 + n n 0 if 11 n0 1 and 1, { 1 n 3 + n n x if 9 1 and so x + 58y, 0 if n0 1 and 9 1, 1 n 3 + n n { x if 1, 19 mod and so x + 18y, 0 if 5, 3 mod, n0 1 n 3 + n n { x if 1, 9 mod 0 and so x + 5y, 0 if 11, 19 mod 0 n0 From 58 and Theorem we deduce the following results: { x mod if 5 1 and so x + 10y, 1 0 mod if 5 1 and 1, { x mod if 11 1 and so x + y, mod if 11 1 and 1, { x mod if 9 1 and so x + 58y, mod if 1 and 9 1, { x mod if 1, 19 mod and so x + 18y, 8 0 mod if 5, 3 mod, { x mod if 1, 9 mod 0 and so x + 5y, 0 mod if 11, 19 mod 0 We remar that all the congruences in 59 were conjectured by Zhi-Wei Sun in [Su1] Acnowledgements The author thans Professor Qing-Hu Hou for his hel in finding 31 and 1, and the referee for his/her helful comments References [A] S Ahlgren, Gaussian hyergeometric series and combinatorial congruences, in: Symbolic comutation, number theory, secial functions, hysics and combinatorics Gainesville, FI, 1999, 1-1, Dev Math, Vol, Kluwer, Dordrecht, 001 [BE] B C Berndt and R J Evans, Sums of Gauss, Eisenstein, Jacobi, Jacobsthal, and Brewer, Illinois J Math , [BEW] BC Berndt, RJ Evans and KS Williams, Gauss and Jacobi Sums, Wiley, New Yor,

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