Jounal of Numbe Theoy 101, 157-1595 CONGRUENCES INVOLVING ZHI-Hong Sun School of Mathematical Sciences, Huaiyin Nomal Univesity, Huaian, Jiangsu 001, PR China Email: zhihongsun@yahoocom Homeage: htt://wwwhytceducn/xsjl/szh Abstact Let > be a ime, and let m be an intege with m In this ae, based on the wo of Billhat and Moton, by using the wo of Ishii and Deuing s theoem fo ellitic cuves with comlex multilication we solve some conjectues of Zhi-Wei Sun concening È 1 0 Æ m mod MSC: Pimay 11A07, Seconday C5, 11E5, 11G07, 11L10, 05A10, 05A19 Keywods: Conguence; Legende olynomial; chaacte sum; ellitic cuve; binay quadatic fom 1 Intoduction Fo ositive integes a, b and n, if n ax + by fo some integes x and y, we biefly say that n ax + by Let > be a ime In 00, Rodiguez-Villegas[RV] osed some conjectues on sueconguences modulo One of his conjectues is equivalent to 1 0 { x mod if x + y 1 mod, 108 0 mod if mod This conjectue has been solved by Motenson[Mo] and Zhi-Wei Sun[Su] Let Z be the set of integes, and fo a ime let R be the set of ational numbes whose denominato is coime to Recently the autho s bothe Zhi-Wei Sun[Su1] osed many conjectues on 1 0 m mod, whee > is a ime and m Z with m Fo examle, he conjectued that see [Su1, Conjectue A1] 11 1 0 7 0 mod if 7, 11, 1, 1 mod 15, x mod if x + 15y 1, mod 15, 0x mod if 5x + y, 8 mod 15 Let {P n x} be the Legende olynomials given by see [MOS, 8-], [G, 1-1] 1 P n x 1 n [n/] 0 n n 1 x n 1 n n n! 1 d n dx n x 1 n,
whee [a] is the geatest intege not exceeding a Fom 1 we see that 1 P n x 1 n P n x Let > be a ime Then autho showed that 1 ] t [/] 0 1 t 5!! 0 mod fo 1 0 1 t 5 In the ae, using the wo of Billhat and Moton[BM] we ove that 15 ] t 1 < < In [S] the mod x + t 5x + t 1t + 11 mod, whee a is the Legende symbol Based on 15 and the wo of Ishii[I], we detemine ] t mod fo t 5, 5 11 1,,, 5, 9 17 5, 0, 5 5 1, 89 500 Fo instance, if 1, mod 5 is a ime, we ove that ] { x x 5 mod if x + 15y 1, mod 15, 0 mod if 11, 1 mod 15 Let > be a ime, m R, m 0 mod and t 1 108/m In the ae we show that 16 1 m 0 ] t mod and that 17 ] t 0 mod imlies 1 0 m 0 mod On the basis of 16 and 17, we ove some conguences fo 1 0 m in the cases m 8, 6, 16, 7, 19, 860, 1, 8, 00 Thus we atially solve some conjectues osed by Zhi-Wei Sun in [Su1,Su,Su] As a tyical examle, fo odd imes 11 we have 1 0 { x mod if 11 1 and so x + 11y, 6 0 mod if 11 1
A geneal conguence modulo We begin with a useful combinatoial identity Lemma 1 Fo any nonnegative intege n we have n 0 n 7 n n 0 n n n n Poof Let m be a nonnegative intege Fo {0, 1,, m} set Fo {0, 1,, m + 1} set F 1 m, 7 m, m m m F m, m m G 1 m, 7 m + m m + 1 + + 1 + m + G m, 9m 9m + 0m 1 + m + m + 1 m + 1 m + 1 m + 1 Fo i 1, and {0, 1,, m}, it is easy to chec that 7 m, 1 m + F i m +, m + 9m + 7m + F i m + 1, + 81m + 1m + m + F i m, G i m, + 1 G i m, Set S i n n 0 F in, fo n 0, 1,, Then m + S i m + F i m +, m + F i m +, m + 1 m + 9m + 7m + S i m + 1 F i m + 1, m + 1 + 81m + 1m + m + S i m m m m + F i m +, m + 9m + 7m + F i m + 1, 0 + 81m + 1m + m + 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 + 1 0 0
Obseve that and m + m + F 1 m + 1, m + 1, m + 1 m + 1 m + m + F 1 m +, m + 1 7m + 1 m + 1 m + 1 6m + m + m + 5 F 1 m +, m + m + m + m + F m + 1, m + 1, m + 1 m + 1 m + m + F m +, m + 1 6, m + 1 m + 1 m + m + 5 m + F m +, m + m + m + 1, m + m + 1 m + m + 1 m + m + 1 m + m + G 1 m, m + 1 7m + 1 m +, m + 1 m + 1 m + m + G m, m + 1 m + 1 7m + 10 m + 1 m + 1 Fom the above we deduce that fo i 1, and m 0, 1,,,, m + S i m + m + 9m + 7m + S i m + 1 + 81m + 1m + m + S i m G i m, m + 1 + m + F i m +, m + + F i m +, m + 1 m + 9m + 7m + F i m + 1, m + 1 0 Since S 1 0 1 S 0 and S 1 1 1 S 1, fom we deduce S 1 n S n fo all n 0, 1,, This comletes the oof Rema 1 We actually find 1 and ove Lemma 1 by using WZ method and Male The autho thans Pofesso Qing-Hu Hou fo his hel in finding 1 Fo the WZ method, see [PWZ] Theoem 1 Let be an odd ime and let x be a vaiable Then 1 0 x1 7x 1 0 x mod
Poof It is clea that 1 0 1 0 1 m0 x1 7x x min{m, 1} x m 0 0 7x m 7 m Suose m and 0 1 If >, then and so If <, then m > and so m 0 Thus, fom the above and Lemma 1 we deduce that 1 0 1 m0 1 m0 1 0 1 0 1 0 1 0 x m m 0 x m m 0 x1 7x x 1 m m m m m m 1 x 0 x 1 0 1 x 0 If 1, then! then 1 and so 1, then >, 7 m m m m m x x 1 1 x x m x x mod! 0 mod If 0 and 1,!! 0 mod If < < and!! 0 mod and!! 0 mod and so Hence, fo 0 1 and 1 we have 1 0 x 1 x 0 mod Thus the esult follows 5
Coollay 1 Let > be a ime and m R with m 0 mod Then 1 m 0 1 0 1 1 108/m mod 5 Poof Taing x 1 1 108/m 5 in Theoem 1 we deduce the esult A conguence fo /] t mod Let W n x be the Deuing olynomial given by 1 W n x It is nown that [G,1],[BM] n 0 n x 1 + x W n x 1 x n P n 1 x Let > be a ime, m, n R and m + 7n 0 mod Fom [Mo, Theoem ] we have 1 x + mx + n 8m 1 1 1 86n 16m + 7n [ 1 ] 8 m J mod, m + 7n whee J t is a cetain Jacobi olynomial given by J t 178 [ 1 ] P 1, 1 1 [ 1 ] 1 t 86 and P α,β x 1 + α 0 + β Theoem 1 Let > be a ime and t R Then x 1 x + 1 ] t 1 x + t 5x + t 1t + 11 mod Poof It is well nown that P n 1 1 Since ] 1 1 and 1 x x 1 x + 1 x 1 x 1 6,
the esult is tue fo t 1 mod As ] 1 1 [ ] ] 1 and 1 x 7x + 5 1 x 7 x + 5 1 x x 1, the esult is also tue fo t 1 mod Now we assume t ±1 mod Set W n x n n x 0 Fom [BM, Theoem 6] we now that 5 W [ 1 ] x u xx 7 [ xx 1 ] J mod, 7 x 7 whee J x is given by and 1 if 1 mod 1, x if 5 mod 1, u x x 6x + 16 if 7 mod 1, x x 6x + 16 if 11 mod 1 Taing x 5/t + 1 in 5 and alying the above we obtain 6 W [ ] t 1/t + 1 71 t 1+t [ 1 ] J 5 t 1 t1+t mod if 1 mod 1, 18t 5 t+1 71 t 1+t [ 1 ] J 5 t 1 t1+t mod if 5 mod 1, 108t 1t+11 t+1 71 t 1+t [ 1 ] J 5 t 1 t1+t mod if 7 mod 1, 19t 5t 1t+11 t+1 71 t 1+t [ 1 ] J 5 t 1 t1+t mod if 11 mod 1 Taing x t 1/t + 1 in we get 7 ] t t + 1 [ ] t 1 W [ ] t + 1 If mod and t 5 mod, fom the above we get On the othe hand, 5 ] t ] 5 + 1 [ ] W [ ] 5 1 5 + 1 0 mod 1 x + t 5x + t 1t + 11 1 x 7/ 1 y 7/ 0 7 y0
Thus the esult is tue when mod and t 5 mod If mod and t 1t + 11 0 mod, fom 6 and 7 we deduce that As ] t t + 1 [ ] t 1 W [ ] 0 mod t + 1 1 x + t 5x 1 x + t 5 x 1 x + t 5x, we see that 1 x + t 5x + t 1t + 11 1 x + t 5x 0 Thus the esult is tue when mod and t 1t + 11 0 mod Set m t 5 and n t 1t + 11 Then 8 m + 7n 1 t1 + t and so 8 m 5 t m + 7n 1 t1 + t By the above, we may assume that m 0 mod fo mod and n 0 mod fo mod Fom we see that 9 5 t 8 m J 1 t1 + t J m + 7n 8m 1 1 1 1 86n 16m + 7n [ 1 ] x + mx + n mod If 1 mod 1, fom 6-9 we deduce that ] t t + 1 [ ] t 1 1 + t W [ ] t + 1 1 71 t 1 + t 1 1 1 1 + t 1 t 1 8 1 t1 + t 1 1 1 x + t 5x + t 1t + 11 mod 8 1 1 J 5 t 1 t1 + t 1 x + mx + n
If 5 mod 1, fom 6-9 we deduce that ] t t + 1 [ ] t 1 t + 1 W [ ] t + 1 18t 5 t + 1 71 t 1 + t 5 + t 51 + t 5 1 t 5 1 1t 5 1 8 1 t1 + t 5 1 1 x + mx + n 1 x + t 5x + t 1t + 11 mod 5 1 J 5 t 1 t1 + t If 7 mod 1, fom 6-9 we deduce that ] t t + 1 t + 1 [ ] W [ 1 ] t 1 t + 1 108t 1t + 11 71 t t + 1 1 + t 7 1 J 5 t 1 t1 + t 7 +5 t 1t + 111 + t 7 1 t 7 1 6 t 1t + 11 1 8 1 t1 + t 7 1 1 x + mx + n 1 x + t 5x + t 1t + 11 mod If 11 mod 1, fom 6-9 we deduce that ] t t + 1 t + 1 [ ] W [ ] t 1 t + 1 19t 5t 1t + 11 71 t t + 1 1 + t 11 1 J 5 t 1 t1 + t 11 11 +5 t 5t 1t + 111 + t 11 1 t 11 1 8m 1 86n 1 8 1 t1 + t 11 1 1 x + mx + n 1 x + t 5x + t 1t + 11 mod This comletes the oof of the theoem 9
Coollay 1 Let > be a ime and let t be a vaiable Then [/] 0 ] t 1 t 5 1 x + t 5x + t 1t + 11 1 mod 1 t 5 mod By Poof Fom [S, Lemma ] we have ] t [/] 0 Theoem 1 and Eule s citeion, the esult is tue fo t 0, 1,, 1 Since both sides ae olynomials in t with degee at most 1, using Lagange s theoem we obtain the esult Coollay Let > be a ime and t R Then 1 x + t 5x + t 1t + 11 1 x t + 5x + t + 1t + 11 Poof Since ] t 1 [ ] ] t P[ ] t, by Theoem 1 we have 1 x + t 5x + t 1t + 11 1 x t + 5x + t + 1t + 11 mod By Hasse s estimate [C, Theoem 11, 15], 1 x ±t + 5x + t ± 1t + 11 Fo 17 we have <, fom the above we deduce the esult Fo {5, 11, 1} and t {0, 1,, 1} one can easily veify that the esult is also tue Thus the coollay is oved Coollay Let > be a ime Then 1 x 10x + 506 { L if 1, L + 7M and L 1, 0 if mod 10
Poof Taing t 5 in Coollay we find that 1 x 7 1 x 0x + 5 1 1 x 10x + 506 Fo mod it is clea that 1 the esult is tue when mod x 7 1 x 7 x 0 x + 5 1 x 0 Thus Now assume that 1 mod, A + B, L + 7M and A L 1 mod It is nown that 1 1 mod if and only if B When B we choose the sign of B so that B 1 mod By [S1, 1], 1/ 1 1 A mod B Fom [S1, 9-11] we deduce that Thus 1 x 7/ { A L if 1 1 mod, A + B L if 1 1 mod and B 1 mod 1 x 10x + 506 1 x 7/ This comletes the oof Theoem Let > be a ime i If mod, then [/] 0 16 [/] 0 L 5 P [ ] 0 mod ii If 1 mod and so L + 7M with L, M Z and L 1 mod, then [/] 0 16 [/] 0 Poof Putting t ± 5 5 ] [/] 0 This togethe with 1 yields 5 P [ ] L in Coollay 1 we get 16 mod and P [ ] 5 [/] 0 16 [/] 0 11 1 [ 1 ] mod [/] 0 mod mod
If mod, fom the oof of Theoem 1 we now that ] 5 0 mod Thus i is tue Now assume that 1 mod, L + 7M and L 1 mod By Theoem 1 and the oof of Coollay we have 5 1 ] x 7 On the othe hand, by the oof of Theoem 1, 5 ] By, Theefoe ] 5 9 8 1 5 + 1 71 5 1+ 5 [ ] W [ ] 5 9 8 1 108 5 1 5 +11 1 5 + 1 1 1 J 0 1 1 1 1 5 +1 71 5 1+ 5 7 1 J 0 L mod J 0 mod if 1 mod 1, 8 1 7 1 7 J 0 mod if 7 mod 1 J 0 178 [ 1 ] P 1, 1 1 [ 1 ] 1 [ 1 ] 178 [ 1 ] [ 1 ] 0 178 [ 1 ] [ 1 ] 1 [ 1 ] 1 178 [ 1 ] [ 1 ] 1 1 1 1 8 1 7 1 7 178 1 1 178 7 1 [ 1 ] 1 [ 1 ] 1 1 [ 1 ] 1 178 [ 1 ] 1 [ 1 ] mod 1 1 1 1 7 1 0 [ 1 ] 1 1 mod if 1 1, 1 mod if 1 7 1 7 1 Now utting all the above togethe we obtain the esult Rema 1 Fo any ime >, Zhi-Wei Sun conjectued that [Su1, Conjectue A6] 1 0!! 1 0! 16! 1/ 1/ mod if 1 mod, +1/ +1/ 1 mod if mod 1
Conguences fo 1 0 / m Let > be a ime and m Z with m In the section we atially solve ZW Sun s conjectues on 1 0 /m mod Theoem 1 Let > be a ime, m R, m 0 mod and t 1 108/m Then 1 1 ] t x + t 5x + t 1t + 11 1 mod 0 m Moeove, if ] t 0 mod o 1 x + t 5x + t 1t + 11 1 0 mod, then 1 0 / m 0 mod 1 Poof Since 1 t 1 t 5 1 7 5 1 m, by Theoem 1 we have 1 m 0 Obseve that 1 0 ] t 1 0 1 t 5 mod fo [ ] < < Fom Coollay 1 we see that 1 t 5 1 x + t 5x + t 1t + 11 1 mod This togethe with 1 yields the esult Theoem [Su1, Conjectue A8] Let > be a ime Then 1 0 { L 19 mod if 1 mod and so L + 7M, 0 mod if mod Poof Putting m 19 and t 5 in Theoem 1 and then alying Theoem we obtain the esult Lemma 1 Let be an odd ime and let a, m, n be algebaic numbes which ae integal fo Then 1 1 x + a mx + a n 1 a 1 x + mx + n 1 mod Moeove, if a, m, n ae conguent to ational integes modulo, then 1 x + a mx + a n a 1 x + mx + n 1
Poof Fo any ositive intege it is well nown that see [IR, Lemma, 5] Since 1 { 1 mod if 1, x 0 mod if 1 1 x + a mx + a n 1 1 1 1/ 0 1 1/ 0 1/ 0 1/ 1/ 1/ 1/ 0 a 1 1 1/ 1 1 x + a mx a n 1 0 x a mx a n 1 a m a n 1 1/ 1 1 1/ 1 1 1 x + a m 1 a n m 1 n 1 1 mod, we see that the conguence in Lemma 1 is tue Now suose that a, m, n ae conguent to ational integes modulo If a 0 mod, then 1 x + a mx + a n 1 x 1 x a 0 1 x + mx + n If a 0 mod, then 1 x + a mx + a n 1 ax + a max + a n a 1 x + mx + n Thus the lemma is oved Lemma Let be an odd ime Then 1 x 0x 56 { 1 [ 8 ]+1 c if 1, mod 8, c + d and c 1, 0 if 5, 7 mod 8 1
Poof Fom [BE, Theoems 51 and 517] we now that 1 x x + x { 1 [ 8 ]+1 c if c + d 1, mod 8 with c 1, 0 if 5, 7 mod 8 As 7x x + x x 0x 56, we see that Thus the esult follows 1 x x + x 1 x 0x 56 Lemma Let be an odd ime Then 1 n 15 + 0 n 8 + 70 1 n0 { + 1 [ 8 ]+1 c mod if c + d 1, mod 8 and c 1, 0 mod if 5, 7 mod 8 Poof It is easily seen that 151 + 1 0 and 8 + 70 1 56 Thus, by Lemmas 1 and we have 1 n0 n 15 + 0 n 8 + 70 1 1 1 1 n 0n 56 1 n0 + 1 1 n 0n 56 { + 1 [ 8 ]+1 c mod if c + d 1, mod 8 and c 1, 0 mod if 5, 7 mod 8 This oves the lemma Theoem Let > be a ime Then ] 5/ { 1 [ 8 ] c mod if c + d 1, mod 8 and c 1, 0 mod if 5, 7 mod 8 n0 and 1 0 { c mod if c + d 1, mod 8, 8 0 mod if 5, 7 mod 8 15
Poof Fom Coollay 1 with t 5/, Lemma and Theoem 1 with m 8 and t 5/ we deduce the esult Rema 1 Let be an odd ime Zhi-Wei Sun conjectued that [Su1, Conjectue A5] 1 0 { c mod if c + d 1, mod 8, 8 0 mod if 5, 7 mod 8 Lemma Let be an odd ime with 11 Then 1 x 11x + 1 11 { u 11 u if 11 1 and so u + 11v, 0 if 11 1 Poof It is nown that see [RP] and [JM] 1 x 96 11x + 11 11 { u 11 u if 11 1 and u + 11v, 0 if 11 1 Since x 96 11 x + 11 11 8x 11x + 1 11, we deduce the esult Lemma 5 Let 11 be an odd ime Then 1 n0 n + 5 + 11n + 7 1 11 11 { 11+ 11 u 11 u mod if 11 1 and so u + 11v, 0 mod if 11 1 Poof It is easily seen that 5 + 11 + 1 11 11 7 and 11 11 11 1 11 11 + 1 11 Thus, by Lemma 1 we have 1 n0 n + 5 + 11n + 7 1 11 11 11 + 1 11 1 11 + 11 11 1 1 x 11x + 1 11 1 1 x 11x + 1 11 mod Now alying Lemma we deduce the esult 16
Theoem Let 11 be an odd ime Then { 11 ] 11+ 11 u 11 u mod if 11 1 and so u + 11v, 0 mod if 11 1 and 1 0 6 { u mod if 11 1 and so u + 11v, 0 mod if 11 1 Poof Fom Coollay 1, Lemma 5 and Theoem 1 with m 6 and t 11 we deduce the esult Rema Let be an odd ime such that 11 Zhi-Wei Sun conjectued that [Su, Conjectue 5] 1 0 6 { u mod if 11 1 and so u + 11v, 0 mod if 11 1 Let > be a ime and let F be the field of elements Fo m, n F let #E x + mx + n be the numbe of oints on the cuve E: y x + mx + n ove the field F It is well nown that see fo examle [S1, 1-] 1 x #E x + mx + n + mx + n + 1 + Let K Q d be an imaginay quadatic field and the cuve y x + mx + n has comlex multilication by an ode in K By Deuing s theoem [C, Theoem 116],[PV],[I], we have { + 1 if is inet in K, #E x + mx + n + 1 π π if π π in K, whee π is in an ode in K and π is the conjugate numbe of π If u + dv with u, v Z, we may tae π 1 u + v d Thus, 1 x + mx + n { ±u if u + dv with u, v Z, 0 othewise In [G], [JM] and [PV] the sign of u in was detemined fo those imaginay quadatic fields K with class numbe 1 In [LM] and [I] the sign of u in was detemined fo imaginay quadatic fields K with class numbe Fo geneal esults on the sign of u in, see [M], [St], [RS] and the suvey [Si] 17
Lemma 6 Let be a ime with ±1 mod 8 Then 1 n + 15 + 6 n + 1 n0 { x x if 1, 7 mod and so x + 6y, 0 if 17, mod Poof Fom [I, 1] we now that the ellitic cuve defined by the equation y x + 1+1 x 8+ has comlex multilication by the ode of disciminant Since u + v imlies u and u + 6v, by and [I, Theoem 1] we have Obseve that 1 n + 1 + 1 n 8 + n0 { x x 1+ if 1, 7 mod and so x + 6y, 0 if 17, mod 15 6 1 + 1 1 + and + 1 8 + 1 + Using Coollay with t 1/ and Lemma 1 we see that 1 n + 15 + 6 n + 1 n0 1 n 15 + 6 n + + 1 n0 1 + 1 n0 n + 1 + 1 n 8 + Now utting all the above togethe we obtain the esult Theoem 5 Let be a ime such that 1, 7 mod 8 Then { x x ] mod if x + 6y 1, 7 mod, 0 mod if 17, mod and 1 0 { x mod if x + 6y 1, 7 mod, 16 0 mod if 17, mod 18
Poof Fom Theoem 1, Lemma 6 and Theoem 1 with m 16 and t / we deduce the esult Rema Fo any ime >, ZW Sun conjectued that [Su1, Conjectue A1] 1 0 x mod if x + 6y 1, 7 mod, 8x mod if x + y 5, 11 mod, 16 0 mod if 1, 17, 19, mod Lemma 7 Let be a ime with ±1 mod 5 Then 1 n + 15 + 1 5n + 8 5 n0 { x x if 1, mod 15 and so x + 15y, 0 if 11, 1 mod 15 Poof Fom [I, Poosition ] we now that the ellitic cuve defined by the equation y x + 105 + 8 5x 78 50 5 has comlex multilication by the ode of disciminant 15 Since u + 60v imlies u and u + 15v, by and [I, Theoem 1] we have 1 n + 105 + 8 5n 78 50 5 n0 { x x 1+ 5/ if 1, mod 15 and so x + 15y, 0 if 11, 1 mod 15 Obseve that 15 + 1 5 105 + 8 5 5 and 8 5 78 50 5 5 Using Lemma 1 we see that 1 n + 15 + 1 5n + 8 5 n0 5 1 n0 n + 105 + 8 5n 78 50 5 Note that 5 1+ 5 1 5 We then have 5 1+ 5/ Now utting all the above togethe we obtain the esult 19
Theoem 6 Let be a ime such that 1, mod 5 Then ] { x x 5 mod if x + 15y 1, mod 15, 0 mod if 11, 1 mod 15 and 1 0 7 { x mod if x + 15y 1, mod 15, 0 mod if 11, 1 mod 15 Poof Fom Theoem 1, Lemma 7 and Theoem 1 with m 7 and t 5 we deduce the esult Lemma 8 Let be a ime such that ±1 mod 5 Then 1 0n + 00 + 108 5n + 51 5 5 n0 { x x if 1, mod 15 and so x + 75y, 0 if 11, 1 mod 15 Poof Fom [I, 1] we now that the ellitic cuve defined by the equation y x + 160 + 08 5x + 10 107 5 has comlex multilication by the ode of disciminant 75 By and [I, Theoem 1] we have 1 n + 160 + 08 5n + 10 107 5 n0 { 5 1 5 x x if 1, mod 15 and so x + 75y, 0 if 11, 1 mod 15 Obseve that 160 + 08 5 00 + 108 5 7 + 5 and 10 107 5 50 + 10 5 7 + 5 Using Lemma 1 we see that 1 n + 160 + 08 5n + 10 107 5 n0 7 + 5/ 1 n + 00 + 108 5n 50 + 10 5 n0 0
Since 7+ 5 5 1 5 5 + 5 and 1 n + 00 + 108 5n 50 + 10 5 n0 1 5n + 00 + 108 5 5n 50 + 10 5 n0 5 1 0n + 00 + 108 5n + 51 5 5, n0 fom the above we deduce the esult Theoem 7 Let be a ime such that 1, mod 5 Then 9 { x 5 ] x mod if 1, mod 15 and so x + 75y, 0 0 mod if 11, 1 mod 15 and 1 0 { x 860 mod if 1, mod 15 and so x + 75y, 0 mod if 11, 1 mod 15 Poof Fom Theoem 1, Lemma 8 and Theoem 1 with m 860 and t 9 5 we deduce the esult Rema Let > 5 be a ime In [S] the autho made a conjectue equivalent to 1 0 x mod if 1, mod 15 and so x + 75y, 860 x mod if 7, 1 mod 15 and so x + 5y, 0 mod if mod Let b {17, 1, 89} and fb 1, 8, 00 accoding as b 17, 1, 89 In [Su1, Conjectues A0, A and A], ZW Sun conjectued that fo any odd ime, b, 5 1 0 fb x mod if b 1 and so x + by, x mod if b 1 and so x + by, 0 mod if b Now we atially solve 5 Theoem 8 Let be an odd ime such that 17 1 Then ] 17 { x x mod if 1 mod and so x + 51y, 0 mod if mod 1
and 1 0 1 { x mod if 1 mod and so x + 51y, 0 mod if mod Poof Fom [I, 1] we now that the ellitic cuve defined by the equation y x 60+1 17x 10 56 17 has comlex multilication by the ode of disciminant 51 Thus, by and [I, Theoem 1] we have 1 n 60 + 1 17n 10 56 17 n0 { x x if 1 mod and so x + 51y, 0 if mod It then follows fom 1 and Theoem 1 that 17 ] P [ ] 1 n0 17 1 n + 17 5n + 17 + + 7 17 n0 n 5 + 17 n + 105+8 17 1 n 60 + 1 17n 10 56 17 n0 { x x mod if 1 mod and so x + 51y, 0 mod if mod Taing m 1 and t 17 in Theoem 1 and then alying the above we deduce the emaining esult Using [I, 1-15] and the method in the oof of Theoem 8 one can similaly ove Theoems 9 and 10 Theoem 9 Let be an odd ime such that 1 1 Then 5 1 { x ] x mod if 1 mod and so x + 1y, 0 mod if mod and 1 0 8 { x mod if 1 mod and so x + 1y, 0 mod if mod
Theoem 10 Let > 5 be a ime such that 89 1 Then 5 89 { x ] x mod if 1 mod and so x + 67y, 500 0 mod if mod and 1 0 { x 00 mod if 1 mod and so x + 67y, 0 mod if mod To conclude, we ose the following conjectues Conjectue 1 Fo any ime > 5 we have 1 0 1 0 1 0 9 + 1 860 + 15 15 + 158 mod 15, mod, 1 1 mod Conjectue Let be a ime with ±1 mod 5 Then 1 n + 15 + 5n + 151 10 5 n0 { x x if 1, mod 15 and so x + 15y, 0 if 11, 1 mod 15 Conjectue Let > be a ime Then { 5 1 1 x x ] mod if 1 mod and so x + y, 9 0 mod if mod Refeences [BE] [BM] [C] [G] B C Bendt and R J Evans, Sums of Gauss, Eisenstein, Jacobi, Jacobsthal, and Bewe, Illinois J Math 1979, 7-7 J Billhat and P Moton, Class numbes of quadatic fields, Hasse invaiants of ellitic cuves, and the suesingula olynomial, J Numbe Theoy 106 00, 79-111 DA Cox, Pimes of the Fom x + ny : Femat, Class Field Theoy, and Comlex Multilication, Wiley, New Yo, 1989 HW Gould, Combinatoial Identities, A Standadized Set of Tables Listing 500 Binomial Coefficient Summations, Mogantown, W Va, 197
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