MSC: Primary 11A15, Secondary 11A07, 11E25 Keywords: Reciprocity law; octic residue; congruence; quartic Jacobi symbol

Transcription

1 Act Arth , 1-5 Congruences for [/] mo ZHI-Hong Sun School of Mthemtcl Scences, Hun Norml Unverst, Hun, Jngsu 3001, PR Chn E-ml: Homege: htt://wwwhtceucn/xsjl/szh Abstrct Let Z be the set of ntegers, n let m, n be the gretest common vsor of ntegers m n n Let 1 mo be rme, Z, n c + x + wth c,, x, Z n c 1 mo Suose tht c, x + 1 or, x + c s ower of In the er, b usng the urtc recroct lw we etermne [/] mo n terms of c,, x n, where [ ] s the gretest nteger functon Hence we rtll solve some conjectures ose b the uthor n two revous ers MSC: Prmr 11A15, Seconr 11A07, 11E5 Kewors: Recroct lw; octc resue; congruence; urtc Jcob smbol 1 Introucton Let Z be the set of ntegers, 1 n Z[] + b, b Z} For n ostve o number m n Z let m be the urtc Jcob smbol We lso ssume 1 1 For our convenence we lso efne m m Then for n two o numbers m n n wth m > 0 or n > 0 we hve the followng generl urtc recroct lw: m m 1 n 1 n 1 n m For, b, c, Z wth c n, one cn efne the urtc Jcob smbol +b c+ s n [S] From [IR] we know tht +b c+ b c +b 1 c+, where x mens the comlex conjugte of x In Secton we lst mn roertes of the urtc Jcob smbol See lso [IR], [BEW] n [S] For rme k + 1 c + x + 3 wth k, c,, x, Z n c 1 mo, n [HW] n [H], b usng cclotomc numbers n Jcob sums Huson n Wllms rove tht ±1 mo f c ± mo 3, ± c mo f ± 1 mo 3 Let be rme of the form k + 1, Z, n Suose tht c + x + wth c,, x, Z n c 1 mo In [S] n [S5] the uthor ose mn conjectures on [/] mo n terms of c,, x n, where [ ] s the The uthor s suorte b the Ntonl Nturl Scences Founton of Chn No

2 gretest nteger functon For m, n Z let m, n be the gretest common vsor of m n n For m Z wth m α m 0 m 0 we s tht α m In the er, b evelong the clculton technue of urtc Jcob smbols we rtll solve mn conjectures from [S] n [S5], n estblsh new recroct lws for urtc n octc resues on conton tht c, x + 1 or, x + c α For the hstor of clsscl recroct lws, see [Lem] Suose r 0, t 0 n mo Assume c, x + 1 or 0, x + c 1 We then hve the followng tcl results 11 If 1 mo, s rme n + b wth, b Z, then x m c + b 1 mo c c b b m mo 1 If 7 mo s rme, then 1 [/] c m mo f 1 mo, 1 x 1 c m x mo f 5 mo c +1 m mo 13 If 1 mo, + b,, b Z, n, b 1, then 1 + x c m mo f n x, c m mo f n x, 1 b x+ c m 1 mo f n x, Bsc lemms 1 b x 1 c m 1 mo f n x c + b/x m b + Lemm 1 [S, Prooston 1] Let, b Z wth n b Then + b +b b / n b 1 1 b 1/ f b, 1 1 b 1 1 f b Lemm [S, Prooston ] Let, b Z wth n b Then 1 + b 1 b n 1 + b 1 b

3 Lemm 3 [S, Prooston 3] Let, b, c, Z wth c, b n If +b n c+ re reltvel rme elements of Z[], we hve the followng generl lw of urtc recroct: In rtculr, f b, then + b 1 b c 1 + +b 1 + b b + b Lemm [E], [S1, Lemm 1] Let, b, m Z wth m n m, +b 1 Then + b m + b m Lemm 5 [S3, Lemm 3] Let, b Z wth n b For n nteger x wth x, + b 1 we hve x + b x + b Lemm 6 Let, b Z wth b n, b 1 Then Proof B Lemms 1 n 3, b + b + b + b b 1 f b, + b 1 1 f b b + b + b 1 1 b + b + b 1 1 b 1 1 b b/ f b, 1 1 f b + b Thus the lemm s rove For gven o rme let Z enote the set of those rtonl numbers whose enomntor s not vsble b Followng [S1,S] we efne Q r k k Z, k + 3 r} for r 0, 1,, 3

4 Lemm 7 [S1, Theorem 3] Let be n o rme, r 0, 1,, 3}, k Z n k mo If 1 mo n t 1 mo wth t Z, then k Q r f n onl f k+t k t 1/ t r mo If 3 mo, then k Q r f n onl f k k+ +1/ r mo Lemm Let be n o rme, k Z n n k + 1 mo wth n Z n nn mo Then k+ nn+1 Proof For k 0 mo we hve k So the result s true Now ssume k 0 mo Then n 1 n+1 n 1 k 1 n so n 1 n+1 k+ B Lemm, k +1 1 n so k+ ±1 B [S1, Theorem ], k+ 1 k Q 0 nn+1 1 Hence k+ nn+1 Lemm 9 Suose c,, m, x Z, m, x c + mo m n m, xx + 1 Then xx + m m Proof Suose tht s rme vsor of m Then xx + If, then x c + 1 mo Thus, lng Lemm we obtn c + When, we hve c n so c Hence, m m x 1 + x xx + x xx + 1 xx + xx +, m m where n the roucts runs over ll rme vsors of m The roof s now comlete Lemm 10 Suose c,, x,, Z, c 1 mo,, c + x +, t 0, 0 1 mo n 0, xx + 1 If x, then c +x c+ If x, then c +x+ / 1 c x+ t t 1 c+ Proof Snce c + x + xx + + we see tht c + x +, 0 1 For even x we hve, c + x + 1 mo n so c + x + xx + c + x xx +

5 For o x we hve c + x + mo n so c + x + / t 0 c + x + 1 c +x+ / 1 t c + x + / / 0 1 c x+ t xx c x+ t xx Snce x c + mo 0, usng Lemms 9, 3 n we see tht xx t 1 t 1 Now combnng ll the bove we obtn the result Lemm 11 Let be rme of the form k + 1 n c + wth c, Z Suose Z, n x + wth x, Z Then x +, c x +, n, c + x + x +, c, x + + c x +, c Proof Snce x, we see tht x, 1 If x, then n so Ths contrcts the fct x, 1 Hence x Snce x, c + x + x, c + x, 1 n x + c c + x + xx +, we see tht x +, c x +, x + x +, n, c + x + xx +, c + x + x +, c + x + x +, c x + x +, c, c x +, c + x + x + x +, c x +, c, Thus the lemm s rove x +, c, c x +, c + x + x + x +, c c x +, c + x + Lemm 1 Let be rme of the form k + 1 n c + wth c, Z n c Suose Z,, x +, x, Z n x/ c+ 1 [ ]+n k Then 1 n [/] c k mo f 1 mo, 1 n c k x mo f 5 mo Proof It s cler tht c, 1, n so x 1 x/ 1 [ ]+n k 1 [ ]+n k mo c 5

6 Thus x 1 1 [ ]+n c k mo n so [ ] 1 [ ] [ ] 1 [ ] x [ ] 1 n c k mo f 1, 1 n c k x mo f 5 Ths roves the lemm 3 Determnton of [/] mo usng c+x+ or Theorem 31 Let be rme of the form k + 1, Z, n Suose tht c + x + wth c,, x, Z, c 1 mo, r 0, 0 1 mo, c, x + 1, x, 1 mo n c/x++ k If 1 mo, then 1 If 5 mo, then x c k mo f 1 mo, x c k+1 mo f 5 mo x c k+1 x x c k x mo f 1 mo, mo f 5 mo Proof Suose m x + n x s x 0 x 0 Snce x we hve 1 mo As c, x + 1, b Lemm 11 we hve, x + 1 n, c + x + 1 Note tht x, We lso hve x, 1 Usng Lemms 1-5, 10 n the fct tht n 1 for, n Z wth n n, n 1 we see tht c + x + k c + x + c + x + xx + + c + x + c + x + c + x + xx c + x + n xx + c + x + c + x + m+s+1 x0 x + / m x+ m+s+1 1 x 0 x+/m +1 x+ c + x + c + x + 1 x 0 x+/m +1 x+ x+ m+s+1 x 0 x 0 x + / m c x + / m 1 x 0 x+/m +1 x+ x+ m+s+1 1 x 0 1 x0 1 x 0 x+/m +1 x+ x+ m+s+1 1 x 0 1 s 1 x 0 x+/m +1 x+ x+ m+s+1 1 x 0 1 s x 6 x

7 Therefore, Observe tht n k 1 x 0 x+/m +1 x+ + x x+ m+s+1 s x/ x 1 f x, 1 f x x+ s m+s+1 x+ m+1 x s 1 m+1x+ x s m+1x+ 1 1 x+ f x, 1 m+1x+ f x From the bove we obtn x/ 31 x 0 x+/ 1 m +1 x+ + x + 1 m+1x+ + x+ k+1 f x, 1 x 0 x+/m +1 x+ + x m+1x+ k f x When x +, we hve 1 m+1x+ 1 x+ For 1 mo we hve, x n x + For 1 mo n 5 mo, we hve x,, x +, m 1 n 1 x 0 x+/+1 1 x 0+ x For 5 mo n 1 mo, we hve, x, x + n m For 5 mo, we see tht, x, x +, m 1 n 1 x 0 x+/+1 1 x x 0+ 0 x x + x 0 +1 Now, from the bove n 31 we euce tht x/ x k f 1 mo, x k+1 f 1 mo n 5 mo, x k+1 f 5 mo n 1 mo, x k f 5 mo Ths together wth Lemm 1 els the result Lemm 31 Let be rme of the form k + 1, Z, n Suose tht c + x + wth c,, x, Z, c 1 mo, r 0, t 0, mo, c, x + 1 n x Assume tht c/x++ k Then 1 x/ 1 [ ]+ x x x c +k f 1, 1 [ +1 ]+ + x x 1 x c +k 1 f 5 7

8 Proof As c, x + 1, b Lemm 11 we hve 0, x + 1 n 0, c + x + 1 Note tht x, We lso hve x, 1 It s esl seen tht c + x x + ± c ±x + c n so x + ± c ±x + c c + x + + Set ε x 1 Snce 1 we see tht x + ε mo n εx + c Usng Lemms 1-5, 10 n the bove we see tht n c + x + k ε x++εc + εx+ c x++εc + εx+ c 1 1 ε εx+ c x++εc + εx+ c x++εc + εx+ c c + x + xx + x++εc + εx+ c x++εc + εx+ c xx + x++εc + εx+ c x++εc + εx+ c x++εc/ 1 εx+ c 1 1 xx++1 εx+ c 1 c x+ t t 1 x++εc + εx+ c c + x + / x++εc + εx+ c xx + Obvousl 1b 1 b b, 1 x++εc 1 εx++c ε 1 εx+ c + 1 ε n so x++εc εx+ c 1 1 εx+ c + 1 ε εx+ c εx+ c 1 1+ε εx+ c εx+ c Also, 1 xx++1 1 x 1 + x n 1 c x+ 1 εx+ c Thus, x++εc/ 1 εx+ c 1 1 xx++1 εx+ c 1 c x+ t t 1 1+ε εx+ c εx+ c 1 εx+ c +1 1 εx+ c t t 1 1 ε + +t εx+ c εx+ c + t

9 It s esl seen tht x++εc + εx+ c xx + x + + εc + εx + c x + + εc + εx + c x x + + εc + ε c εc c ε ε x x + x x + ε 5+ε xx + x 1 + xx + x 5+ε 1 + x 1 1 xx + xx + x 1 x+x 1 5+ε 1 / xx+ 1 1 x 1 x Now combnng ll the bove we euce tht 3 It s cler tht k ε + 1 εx+ c ε + εx+ c +t εx+ c + t 1 x+x 1 5+ε + x 1 1 / xx+ 1 x/ 1 εx+ c 1 εx+ c εx++c 1 x+ c 1 +x 1 f 1 n so 0 mo, n 1 +x ε f 5 n so mo 1 x x / xx / x f 1 n so, Snce x + ε mo n 1 εx+ c + 1/ xx+ 1 + t 1 0 x+1 ε f 5 n so mo 1 1 we lso hve 1 1/ xx+ 1 εx+ c 1 1/ xx+ 1 + t 1 1/ xx+ 1 εx+ 1 1x 1/ x c 1 + t 1 1/ xx+ 1 + c x 1/ x + c 1 + t 1 + c 1 + t c 1x 1/ x f 1, ε 1 c 1 c 1x 1/ x ε + c 1 c 1x 1/ x +1 f 5 9

10 Note tht 1 c 1 1 c [ ] From the bove n 3 we euce the result Theorem 3 Let be rme of the form k + 1, Z, n Suose tht c + x + wth c,, x, Z, c 1 mo, r 0, t 0, mo, c, x + 1, x n c/x++ k If 1 mo, then 1 If 5 mo, then c k mo f 1 mo, x 1 + c k 1 mo f 3 mo, x 1 + c k 1 mo f 5 mo, c k mo f 7 mo x 1 c k 1 x 1 +5 c k 1 x 1 +3 c k x x 1 c k x mo f 1 mo, mo f 3 mo, mo f 5 mo, mo f 7 mo Proof Suose c 1 x 1/ x n m c 1 x 1/ x Then m n m+1 c xc + x As c xc + x we see tht m+1 We frst ssume 1 mo Snce n we hve m 3 n m 3 Thus, 1x 1/ x c 1 1x 1/ x c 1 m Ths s lso true when c 1 x 1/ x From the bove n Lemm 31 we euce tht x/ x 1 + k f ±1 mo, + k 1 f ±3 mo Now lng Lemm 1 we euce Suose 5 mo As 0 0 we get m Clerl mo Thus 3 mo f m, 5 mo f m 3, 1 mo f m > 3 10

11 For 1 mo we hve c 1 x 1 x n 1x 1/ x c 1 1x 1/ x c 1 m Ths s lso true when c 1 x 1/ x Thus, usng Lemm 31 we euce tht 5 x/ x 1 k 1 f 1 mo, k f 5 mo Now lng Lemm 1 we euce the result n the cse 5 mo n 1 mo For 3 mo we hve c 1 x 1/ x n so x c c x x/ x c 1 x 1 x c x +1 mo Thus, 1x 1/ x c +3 mo n so 1 x 1/ x c +3 Usng Lemm 31 we see tht k 1 f 3 mo, x 1 k f 7 mo Now lng Lemm 1 we obtn the result n the cse 5 mo n 3 mo Summrzng ll the bove we rove the theorem Remrk 31 We note tht the k n Theorems 31-3 eens onl on c x+ mo Corollr 31 Let 1, 9 mo 60 be rme n so c + x + 15 wth c,, x, Z Suose c 1 mo, r 0, t 0, mo n c, x + 1 If 1 mo, then 15 1 If 5 mo, then mo f c x+ 1 mo f c x+ ± 1 c mo f c x+ 1 x 1 1 x 1 1 x 1 x mo f c x+ x mo f c x+ cx mo f c 11 x+ 0, ±1 mo 15, ± mo 15, ±5, ±6 mo 15 0, ±1 mo 15, ± mo 15, ±5, ±6 mo 15

12 Proof Clerl x s o Thus, uttng 15 n Theorem 3 n notng tht see [S1, Exmle 1] n + n + n we euce the result 1 f n 0, ±1 mo 15, 1 f n ± mo 15, f n ±5, ±6 mo 15 For exmle, snce mo 15 we hve , 5, n mo 61 Theorem 33 Let be rme of the form k + 1, Z, n Suose tht c + x + wth c,, x, Z, c 1 mo, r 0, t 0, mo, 0, x + c 1 n / k If 1 mo, then 1 If 5 mo, then 5 1 +[ x ] c k mo f x, 1 1 x c k mo f x 1 [ x+ ] c k 1 x 1 +3 x+1 c k 1 x 1 3 x+1 c k x mo f x n so 1 mo, mo f x n 1 mo, mo f 3 mo Proof Suose x s x 0 x 0 n m x + c As x + c, 0 1, b Lemm 11 we hve 0, x + c 1 n 0, x + c + 1 Note tht x, We lso hve x, 1 If m < r, usng Lemms 1-5 n the fct 1 for Z wth, 1 we see tht 33 x + c m + m m m+1 + m m m+1 m + m c + x m+1 + m m 1 + m m + m

13 If m r, then clerl 3 x + c r r + r If m > r, usng Lemms 1-5 we see tht 35 x + c r r 1 1 r r+1 r r 1 + r r r r 1 1 x + c + xx + c r+1 r r 1 1 m r 1 xx + c r r + r+1 r+1 r + r x + c + / r B conserng the three cses m < r, m r n m > r n lng lemms n Secton, one m euce the result fter ong horrble long clcultons We onl rove the result n the cse m < r nclung x n x 1 The remnng two cses cn be rove smlrl For the etls n the cses m r n m > r, see the uthor s fourth verson of Qurtc, octc resues n bnr urtc forms n rxv: Now suose m < r Then c + x + m m xx + c + + x + c + m m xx + c + m m m+s+1 x 0 x + c/ m + m m + m m + m m 1 m 1/ m+1 m+s+1 1 x m 1 m+1 + m m x 0 1 m 1/ m+1 m+s+1 1 m +x 0 x + m+1 x 0 m Clerl m n m x + x 0 x 0 1 x0 1 1 x 0 1 x0 x 1 13 s m + m m x 0 1 s x

14 Thus, c + x m + m 1 m +x 0 m+1 + x m 1/ m+1 m+s+1 s m x As x + c + / m 1 + / m+1 mo n x c + mo 0, usng Lemm 9 we see tht x + c + / m t 0 x + c + / m 1 t x + c + m+1 1 t xx + c m t + c m t + c m t m t 0 m+1 1 t 1 m t t 1 m+1 1 m+1 t t 1 From the bove n 33 we euce tht 36 x + c m+1 1 m +x 0 m+1 + x 0 1 m 1 m 1/ m+1 m+s+1 s 1 m+1 t t x/ If m 0, then x + c, x, n so 1 mo Thus, from 36 we euce tht x + c x +1 x/ x + c Snce k, c 1+ x 1 x 1 x n c 1 + x x +c x x +1 1 [ x+ ]+[ ], x + c from the bove n the fct 0 1 mo we erve tht x/ 1 x + x +1 1 [ 1 [ ]+ x x+ [ x+ ]+[ ] k ]+ k 1 [ ]+ +[ x ] k f 1 mo, 1 [ ]+[ x+ ] k 1 f 5 mo 1

15 Now lng Lemm 1 we obtn the result n the cse m 0 If m 1 < r, then x 1 mo, s 0, n 1 mo x 1 mo we hve n so Thus, 1 c x + c x + c x + c c + + mo Snce n so c mo Therefore 1 mo n 1 c c c Hence, from 36 we euce tht x + c x x/ x/ Snce k, we get x/ c k Now lng Lemm 1 we obtn c k mo s sserte Now we ssume m < r Then x 3 mo, s 0, n so 1 mo Snce x + c + cx + c we see tht m+1 r m 1 0 m 1 x + c x + c + c m m As m n r m + 1 m +, we must hve m+1, m + 1, t m+1 n m 1 + c mo Thus m 3, m c m 1 + m c 1 + m 1 + mo n so mo Therefore, b 36 we hve m x + c m+1 m+1 c 1 + m 1 1 t x/ + m+1 1 c 1 +m 3 x/ Note tht 1 c 1 1 c 1 1 1, k n 1 f m > 3, 1 f t >, 1 m 3 1 f m f t We then get x/ c k Recll tht Alng Lemm 1 we obtn 1 1 c k 1 +1 the cse m < r Summrzng ll the bove we rove the theorem 15 + c k mo Ths roves the result n

16 New recroct lws for urtc n octc resues Theorem 1 Let n be rmes such tht 1 mo n 3 mo Suose c + x +, c,, x, Z, c 1 mo, r 0, t 0, mo n c x +1 m mo Assume c, x + 1 or 0, x + c 1 Then 1 [/] + +1 x 1 c m mo f 1 mo, 1 3 x 1 c m x mo f 5 mo Proof Snce 1 mo n 3 mo we see tht x n x s o We frst ssume c, x + 1 B Lemm 11 we hve, x + c + x + 1 It s esl seen tht c/x+ c/x++ c x+ c+x+ c x mo Thus, for k 0, 1,, 3, usng Lemm 7 we get c + x + Snce c x +1 k c c x + Q x+ k c x+ + c +1 k mo x +1 k mo c x +1 m mo, from the bove we euce tht c + x + m k mo 1 +5 m+1 f 3 mo, 1 +1 m f 7 mo Now, lng Theorem 3 we erve the result Now we ssume 0, 1 B Lemm 11,,, + 1 It s esl seen tht + c x mo Thus, for k 0, 1,, 3, usng Lemm 7 we get x + c k x + c Q k + x + c x + c c +1 x k mo +1 +k mo k mo c +1 x k mo Snce c x +1 m mo, from the bove we euce tht +1 m 1 +1 m Now lng Theorem 33 we euce the result The roof s now comlete 16

17 Corollr 1 Let 1 mo n 3 mo be rmes such tht c + x + wth c,, x, Z n c Suose c 1 mo, r 0, t 0 n mo Assume c, + x 1 or 0, x + c 1 If 1 mo, then 1 If 5 mo, then ± 1 x 1 + mo f x ±c mo, x 1 + c mo f x ± mo 5 ± x mo 1 3 cx mo f x ±c mo, f x ± mo Proof If x ±c mo, then n so c x +1 ±1 +1 ±1 mo If x ± mo, then c n so c x mo Now lng Theorem 1 we euce the result We note tht Corollr 1 rtll settles [S5, Conjecture 3] For exmle, let be rme such tht 13 mo n hence c + x + 3 wth c,, x, Z Suose c 1 mo, r 0, t 0 n mo If c, x + 1 or 0, x + c 1, then 3 5 ± x cx Ths rtll solves [S, Conjecture 91] mo f x ±c mo 3, mo f x ± mo 3 Theorem Let n be rmes such tht 1 mo, 7 mo, c + x +, c,, x, Z, c 1 mo, r 0, t 0 n mo Assume c, x + 1 or 0, x + c 1 Suose c c+ +1 m mo Then 1 [/] c m mo f 1 mo, 1 x 1 c m x mo f 5 mo Proof Observe tht c c c c x + +1 c x +1 mo The result follows from Theorem 1 We note tht f, then the m n Theorem eens onl on c 17 mo

18 Corollr Let 1 mo n 7 mo be rmes such tht c + x + wth c,, x, Z n cc Suose c 1 mo, r 0, t 0 n mo Assume c, x+ 1 or 0, 1 If 1 mo, then mo f c, 1 mo f, ± c mo f 16 7 n c ± mo, mo f n c ± mo If 5 mo, then x 1 1 x 1 x x ± x x 1 x mo f c, mo f, cx mo mo f 16 7 n c ± mo, f n c ± mo Proof Clerl 1 mo f c, c 1 mo f, mo f c mo, mo f c mo Thus the result follows from Theorem Theorem 3 Let n be stnct rmes of the form k + 1, c + x +, + b,, b, c,, x, Z, c 1 mo, r 0, t 0 n mo Assume c, x + 1 or 0, x + c 1 Suose c+b x 1 b m mo If 1 mo, then 1 If 5 mo, then 5 x+ 1 +[ ] c m mo f x, 1 1 x c m mo f x 1 [ x ] c m+1 x mo f x, 1 +3 x 1 c m+1 x mo f x 1

19 Proof Clerl x We frst ssume c, x + 1 B Lemm 11,, x + c + x + 1 It s esl seen tht c+bx+ c bx+ c+b x b mo Thus, for k 0, 1,, 3, usng Lemm 7 we get c + x + k c x + Q k c + bx + c bx + c + b b x c + b 1 x 1 1 c x+ + b 1 c x+ b b k mo b k mo b k 1 mo Snce c+b x Now the result follows from Theorems 31 n 3 mmetel 1 b m mo, from the bove we get c+x+ b k mo 1 m+ Suose 0, x + c 1 B Lemm 11,, x + c + x + c 1 It s esl seen tht b +b c+b x mo Thus, for k 0, 1,, 3, usng Lemm 7 we get x + c k x + c Q k bx + c + bx + c c + b 1 x 1 + b b 1 b b 1 +k mo b k mo k mo c + b x 1 b k mo Snce c+b x 1 b m mo, b the bove we get 1 m Thus, lng Theorem 33 n the fct x x c mo for even x we erve the result The roof s now comlete Corollr 3 Let 1 mo n 5 mo be rmes such tht c + x + wth c,, x, Z n c Suose c 1 mo, r 0, t 0 n mo Assume c, x + 1 or 0, x + c 1 If 1 mo, then 1 ± 1 + x+ mo f x n x ±c mo, ± 1 + x 1 + mo f x n x ±c mo, ± x+ c ± x 1 + c mo f x n x ± mo, mo f x n x ± mo 19

20 If 5 mo, then 5 ±δx cx mo 1 5 δx x where δx 1 or 1 ccorng s x or not f x ±c mo, mo f x ± mo, Proof If x ±c mo, then n so c+b x 1 c x 1 ±1 1 ±1 mo If x ± mo, then c n so c+b ± b 1 ± 1 5 b euce the result x 1 b x 1 mo Now uttng the bove wth Theorem 3 we Theorem Let n be stnct rmes such tht 1 mo, 1 mo, c + x +, + b,, b, c,, x, Z, c 1 mo, r 0, t 0 n mo Assume c, x+ 1 or 0, 1 Suose c+b c b 1 b m mo If 1 mo, then If 5 mo, then x m mo c 1 x c m+1 x 1 x+1 c m+1 x mo f x, mo f x Proof Observe tht b mo, x mo n so c + b 1 c b 1 c + b c b 1 1 c + b 1 c + b x 1 mo The result follows from Theorem 3 We note tht f, then the m n Theorem eens onl on c mo Corollr Let 1 mo n 1 mo be stnct rmes such tht c + x + wth c,, x, Z n cc Suose c 1 mo, r 0, t 0 n mo Assume c, x+ 1 or 0, 1 If 1 mo, then x mo f c, 1 + x mo f, x mo f 16 1 n c ± mo, ± x c mo f 16 9 n c ± mo 0

21 If 5 mo n εx 1 x or 1 x+1 ccorng s x or x, then εx cx εx cx mo f c, mo f, εx cx εx x mo f 16 1 n c ± mo, mo f 16 9 n c ± mo Proof Suose tht + b wth, b Z Then clerl c + b 1 c b 1 1 mo f c, 1 mo f, mo f 16 1 n c ± mo, ± b mo f 16 9 n c ± mo Thus the result follows from Theorem Corollr 5 Let 1 mo be rme such tht 17 n c + x + 17 wth c,, x, Z Suose c 1 mo, r 0, t 0 n mo Assume c, x + 1 or 0, x + c 1 If 1 mo, then then x mo f 17 c, 1 + x mo f c ± mo 17, ± 1 + x c mo f c ±5, ±10 mo 17 If 5 mo n εx 1 x or 1 x+1 ccorng s x or x, 17 5 εx cx εx cx ±εx x mo f 17 c, mo f c ± mo 17, mo f c ±5, ±10 mo 17 Proof Snce n 17 1 ±5+ ± ±10+ ±10 mo 17, from Theorem n Corollr we euce the result Theorem 5 Let 1 mo be rme, c + x + +b +b,, b, c,, x, Z, 0,,, b 1, c 1 mo, r 0, t 0 n mo Assume c, x + 1 or 0, x + c 1 Suose c+b/x b+ m 1

22 If 1 mo, then + b x c m mo f n x, 1 + c m mo f n x, 1 b x c m 1 mo f n x, 1 b x 1 c m 1 mo f n x If 5 mo, then + b 5 1 x c m 1 x 1 x+1 c m 1 x 1 x + b+1 c m x 1 b 1 c m x mo f n x, mo f n x, mo f n x, mo f n x Proof Set +b Then clerl n We frst ssume c, x+ 1 B Lemm 11,, x +, c + x + 1 Snce c x+ c+x+ c x mo, we see tht c/x + + c + x + c + x + c + x + b + b c + x + c x + c + x + c x + b + b + b + b + c x+ c+x+ 1 c 1 x c /x 1 b + b + b + b + b c + b/x 1 1 b+1 b + b + b + b m 1 b+1 b m 1 b+1 + b 1 m 1 b m 1 b+1 +[ ] m 1 Ths together wth Theorems 31 n 3 els the result uner the conton c, x+ 1 Now we ssume 0, 1 B Lemm 11,,, + 1

23 Snce + c x mo, usng Lemm 6 we see tht /x + c x + c x + c x + c b + b x + c + x + c x + c + x + c b + b + b + b c 1 x c /x 1 b + b + b + b + 1 c + b/x 1 1 b + b + b + m 1 b 1 m 1 b+1 1 m f, 1 1 m m f Combnng ths wth Theorem 33 we euce the result uner the conton 0, x + c 1 The roof s now comlete Remrk 1 Let be rme of the form k + 1, Z,,, n c + x + wth c,, x, Z, c 1 mo, r 0 n 0 1 mo, we conjecture tht one cn lws choose the sgn of x such tht c, x + 1 or 0, x + c 1 Thus the conton c, x + 1 or 0, x + c 1 n Theorems 1-5 n Corollres 1-5 cn be cncele See lso relte conjectures n [S] n [S5] References [BEW] BC Bernt, RJ Evns n KS Wllms, Guss n Jcob Sums, Wle, New York, 199 [E] RJ Evns, Resuct of rmes, Rock Mountn J Mth , [H] RH Huson, Dohntne etermntons of 3 1/ n 5 1/, Pcfc J Mth , 9-55 [HW] RH Huson n KS Wllms, Some new resuct crter, Pcfc J Mth , [IR] K Ireln n M Rosen, A Clsscl Introucton to Moern Number Theor, n e, Srnger, New York, 1990 [Lem] F Lemmermeer, Recroct Lws: From Euler to Esensten, Srnger, Berln, 000 [S1] ZH Sun, Sulements to the theor of urtc resues, Act Arth , [S] ZH Sun, Qurtc resues n bnr urtc forms, J Number Theor , 10-5 [S3] ZH Sun, On the urtc chrcter of urtc unts, J Number Theor 1 00, [S] ZH Sun, Qurtc, octc resues n Lucs seuences, J Number Theor , [S5] ZH Sun, Congruences for A + A + mb 1/ n b + + b 1/ mo, Act Arth ,