ON THE MAXIMUM NUMBER OF CONSECUTIVE INTEGERS ON WHICH A CHARACTER IS CONSTANT

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1 ON THE MAXIMUM NUMBER OF CONSECUTIVE INTEGERS ON WHICH A CHARACTER IS CONSTANT ENRIQUE TREVIÑO Abstract Let χ be a non-principal Dirichlet character to the prime moulus p In 1963, Burgess showe that the maximum number of consecutive integers H for which χ remains constant is O p 1/4 log p ) This is the best known asymptotic upper boun on this quantity Recently, McGown prove an explicit version of Burgess s theorem, namely that H < 706p 1/4 log p for p Preobrazhenskaya, ) in the Legenre symbol case, showe that H < π + o1) p 1/4 log p By improving an inequality of Burgess on charac- 6 log ter sums an using some ieas of Norton, we prove that H < 155p 1/4 log p whenever p , an H < 364p 1/4 log p for all p 1 introuction Let χ be a non-principal Dirichlet character to the prime moulus p In 1963, Burgess showe see [3]) that the maximum number of consecutive integers for which χ remains constant is Op 1/4 log p) This is the best known asymptotic upper boun on this quantity Recently, McGown see [9]) prove an explicit version of Burgess s theorem: Theorem A If χ is any non-principal Dirichlet character to the prime moulus p which is constant on N, N + H], then { πe } 6 H < + o1) p 1/4 log p, 3 where the o1) terms epens only on p Furthermore, 706p 1/4 log p, for p , H 7p 1/4 log p, for p Stronger bouns were announce but not proven by Norton in 1973 see [11]), namely that H 5p 1/4 log p for p > e an H < 41p 1/4 log p, in general Preobrazhenskaya in [1] showe that) when the Dirichlet character is the π Legenre symbol we have H < 6 log + o1) p 1/4 log p Mathematics Subject Classification Primary 11L40, 11Y60 Key wors an phrases Character Sums, Burgess inequality This paper is essentially Chapter 5 of the author s Ph D Dissertation [17] 1 The proof by Preobrazhenskaya can be extene to any Dirichlet character of prime moulus an is almost as strong as the inequality we have in this paper The constant provie by Preobrazhenskaya is approximately

2 ENRIQUE TREVIÑO The main ingreient in the proof of McGown is estimating p h 1 w S χ h, w) = χm + l), m=1 where p is a prime, χ is a non-principal character mo p, an h p is a positive integer Burgess, using Weil s work on the Riemann Hypothesis for function fiels see [18]), showe in [] that S χ h, w) < 4w) w+1 ph w + wp 1/ h w McGown improve this estimate see [9]) to S χ h, w) < 1 4 4w)w ph w + w 1)p 1/ h w In [16], with the further restriction that w 9h, the author improve the estimate to 1) S χ h, w) < w)! w w! phw + w 1)p 1/ h w In this paper, with help from the upper boun 1) on S χ h, w) an an improvement on McGown s lower boun for S χ h, w), we are able to prove Norton s claim an go a little further Theorem 1 If χ is any non-principal Dirichlet character to the prime moulus p which is constant on N, N + H], then { } π e H < 3 + o1) p 1/4 log p, where the o1) terms epens only on p Furthermore, 364p 1/4 log p, for all o p, H 155p 1/4 log p, for p l=0 Remark 1 The constant π e 3 = 1495 is 1 of McGown s asymptotic constant e = times the size One reason we stuy this problem is that it is a generalization of the problem of fining the least k-th power non-resiue mo p the case N = 0) an it allows one to boun the maximum number of consecutive integers that belong to a given coset C p /Cp k, where C p = Z/pZ) Furthermore, McGown see [8]) was able to use bouns on the least k-th power non-resiue an the secon least k-th power non-resiue to put a boun on the size of the iscriminant of a Norm-Eucliean Galois cubic fiel Regaring the least k-th power non-resiue, Norton prove that the least k-th power non-resiue is boune by 47p 1/4 log p in [10] The author improve this to 11p 1/4 log p in [16] Notation Most of our notation is stanar A possible exception is our notation for p n p the algebraic raical of n), written here as ran) some papers use the notation γn)) We write µn) for the Moebius function, a, b) for the greatest common ivisor of the integers a an b, an we write log x for the natural logarithm of x Some Lemmas To prove Theorem 1 we will nee a lower boun for S χ h, w) Before we can fin a lower boun for S χ h, w) we nee to prove four lemmas The first lemma is an estimate on the number of squarefree numbers up to a real number X The secon lemma is an upper boun on the tail of the sum of / The thir lemma is

3 ON THE MAXIMUM NUMBER OF CONSECUTIVE INTEGERS ON WHICH A CHARACTER IS CONSTANT3 a nice result concerning / The first three lemmas will be use to prove the main lemma, which is an estimate which will be crucial in giving a lower boun for S χ h, w) Lemma 1 For real X 1, the number of squarefree integers in [1, X] is at most 3 X + Proof The number of squarefree numbers up to X is at most X X X X X + Lemma For real X 1 an a a positive integer we have ) < 1 >X X, Proof Note that for any positive integer we have that 1 Thus >X >X t t = X 1 +1/ t is smaller than 1/ t t t = 1 X 1 To change X 1/ into X, note that there is at least one missing in the interval [X, X + 4], since we only take squarefree s in the sum Thus the absolute value 1 of the sum is smaller than 1 X 1 X+4) This is smaller than 1 X once X 11, proving the lemma for real X 11 To complete the proof for X 1 we nee to verify ) for X 11 To o this we use the fact that = 6 π 1 1 ) 1 p, which implies that >X = 6 π p a p a 1 1 ) 1 p = X p a a,)=1 1 1p ) X a,)=1 Let M = p = 310 an m = ra a), M) Hence, for X 11 we have p 11, a) =, m) We also have that p M m implies p a Since 1 1/p ) < 1, then 3) 1 1p ) 1 1p ) p a p M m X a,)=1 X,m)=1 Now, using that p a implies p m an that 1 1/p ) 1 > 1 yiels 6 4) π 1 1 ) 1 p 6 π 1 1 ) 1 p p a X p m X a,)=1 m,)=1

4 4 ENRIQUE TREVIÑO 1 One can now manually check that the right han sie of 3) is less than X+1 an that the right han sie of 4) is greater than 1 X+1 for integer X [1, 11] an the 3 possible values of m This shows that < 1 X+1 for all integers >X X [1, 11] Since this is true for all integers X [1, 11], then we have ) for any real X [1, 11] Tao in an expository article [15] prove the following lemma We give a ifferent proof inspire by the online lecture notes of Hilebran [6] Lemma 3 For X 1 a real number an a a positive integer: 5) 1 X Proof Let Now consier the sum S a X) = n X 1, if ra n) ra a), e a n) := 0, otherwise S a X) := n X e a n) First note that if S a X) = X, then the only term summe in 5) is = 1, showing that the sum is 1 Therefore we may assume that S a X) < X Now, X = Therefore 6) X X e a n) = n X = a X) + S X n { X } = X X ra) raa) 1 + X Note that the conitions ra) raa) an, a) = 1 overlap only when = 1 Therefore the right han sie of 6) is X + 1 Now, note that since S a X) < X, there is a prime j X such that j, a) = 1 Since µj) = 1, we can conclue that the right han sie of 6) is X This conclues the proof of 5) The following is the main lemma of the section: { X } Generalizations of this inequality can be foun in [5] an [14]

5 ON THE MAXIMUM NUMBER OF CONSECUTIVE INTEGERS ON WHICH A CHARACTER IS CONSTANT5 Lemma 4 For a an b coprime integers an X 1 a real number, we have ) X q 1 3 π X 13 1 X 1 4 q X 0 t<q gc at+b,q)=1 Proof Start by using inclusion-exclusion to get the sum equal to ) X q 1 q X 0 t<q gc at+b,q) Writing q = r an exchanging summation gives us ) X r 1 X r X 0 t<r at b mo Since gc a, b) = 1, the congruence at b mo has a solution if an only if gc, a) = 1 Note that in such a case, there are r values of t such that 0 t < r an at b mo Therefore the sum becomes 7) Writing X x gc = X X r) = X r X r X Therefore 7) becomes 8) X X gc 1 0 t<r at b mo X r 1 ) = X gc X r) r X { + X } we evaluate the insie sum of 7) as X X X + 1) = X X + { X } X X X + { X } 1 { X }) ) = >X X X +1 X 1 { X }) { } X 1 { }) X Now, 9) 1 = 6 π p a 1 1 p ) 1 6 π 10) Using Lemma 1 we get 1 { X X } 1 { }) X 1 8 X squarefree x 1 4 Combining 9), 10), Lemma an Lemma 3 with 8) yiels the proof

6 6 ENRIQUE TREVIÑO 3 Lower boun for S χ h, w) Now we are reay to fin a lower boun for S χ h, w) The proposition we shall prove improves Proposition 33 in [9] by a factor of 4 an it also has a smaller error term saving a log X) Another improvement is that the proposition has a less emaning conition for H, namely that H h ) /3 p 1/3 instea of H h 1) 1/3 p 1/3 Throughout, let A = 3 π Proposition 1 Let h an w be positive integers Let χ be a non-principal Dirichlet character to the prime moulus p which is constant on N, N + H] an such that ) /3 h 4h H p 1/3 Let X := H/h, then X 4 an ) 3 S χ h, w) π X h w+1 gx) = AH h w 1 gx), where 13 gx) = 1 1AX + 1 ) 4AX Proof The proof follows McGown s treatment of the metho of Burgess with some moifications inspire by the work of Norton By Dirichlet s Theorem in Diophantine approximation see Theorem 7 on p 101 of [4]), there exist coprime integers a an b satisfying 1 a H an 11) an p b H h 1 h + 1 H Let s efine the real interval: N + pt Iq, t) :=, N + H + pt ], q q for integers 0 t < q X an gc at + b, q) = 1 The reason Iq, t) is important, is that χ is constant insie the interval Inee, if m Iq, t), then χqm pt) = χn + i) for some i such that 0 < i H Therefore χm) = χq)χn + i) We will show that the Iq, t) are isjoint an that Iq, t) 0, p) First, let s show that the Iq, t) are isjoint If Iq 1, t 1 ) an Iq, t ) overlap then either N+pt1 q 1 N+pt q < N+H+pt1 q 1 or N+pt q N+pt1 q 1 < N+H+pt q In the first case, multiply all by q 1 q an then subtract Nq + pt 1 q This yiels Analogously, for the secon case, we get 1) Therefore, Nq 1 q ) p 0 Nq 1 q ) + pt q 1 t 1 q ) < Hq Hq 1 < Nq 1 q ) + pt q 1 + t 1 q ) 0 + t q 1 t 1 q < max {q 1, q }H p Therefore, combining 11) an 1), we get h XH p

7 ON THE MAXIMUM NUMBER OF CONSECUTIVE INTEGERS ON WHICH A CHARACTER IS CONSTANT7 b a q 1 q ) + t q 1 t 1 q N b = p + a N )) q 1 q ) + t q 1 t 1 q p N p q 1 q ) + t q 1 t 1 q b + a N ) q 1 q ) p Since a H h < XH p + h q 1 q ah XH p + Xh ah = X ah + hp ahp an H3 h p 4 by hypothesis, then H a + ph ahp 4H 3 h + ph ahp ph ahp = 1 a = ah + hp ahp Therefore b a q 1 q ) + t q 1 t 1 q < 1 a, implying that at 1 + b = at + b q 1 q However, since gc at 1 + b, q 1 ) = 1 an gc at + b, q ) = 1, then q 1 = q an therefore t 1 = t We have now prove that the Iq, t) are isjoint Since χp) = 0, we can assume without loss of generality that N + H < p Now let s prove that Iq, t) 0, p) If m Iq, t), then m > N+pt q 0 Also, m N+H+pt q < pt+1) q p Since the Iq, t) are isjoint an they are containe in 0, p), we have S χ h, w) = p 1 m=0 h 1 χm + l) l=0 h w q,t w h 1 χm + l) q,t m Iq,t) l=0 ) = h w+1 q X H q h w 0 t<q gc at+b,q)=1 ) X q 1 The last inequality is true since there are at least H q h subsets of h consecutive integers in Iq, t), an when there are h consecutive integers m, m+1, m+h 1, we have h 1 w χm + l) = h w l=0 To finish the proof of the Proposition we use Lemma 4 4 Proof of the main theorem Proof of Theorem 1 Let h an w be positive integers, an let A = 3 π Assume that 4h H h ) /3 p 1/3 Then by Proposition 1 an 1) we have for w 9h): Therefore, AH h w 1 gx) S χ h, w) < w)! w w! phw + w 1)p 1/ h w 13) AH gx) < fw, h),

8 8 ENRIQUE TREVIÑO where fw, h) = w)! w w! ph1 w + w 1)p 1/ h Using techniques from Calculus an an explicit version of Stirling s formula such as the one in [13] we can choose optimal w an h to make fw, h) as small as possible For the etails of this computation see [16] or [17] For large p, goo choices for h an w are e h = + e + 1 ) log p, log p an log p w = From there one can obtain e 14) fw, h) < 4 + 5e + 1 log p + 8e + 3 log p + 8e + 4 ) p log log 3 p = Kp) p log p p Assume that p p 0 an H Cp 0 ) p 1/4 log p We may assume Cp 0 ) π e 1, hence X = H h Let Xp 0 ) be efine as Cp 0)p 1/4 log p ) e + e+1 log p log p Xp 0 ) = e π e 1 e π e 1 + e+1 log p )p 1/4 + e+1 0 log p 0 Note that since H π e 1 p1/4 log p an h < long as p 1500 Now let Cp 0 ) = Kp 0 ) A gxp 0 )), )p 1/4 ) e + e+1 log p log p, then H 4h as with Kp) introuce in 14) The left han sie of 13) can therefore be boune from below for p p 0 : AH gx) A Cp 0 )) p log p gxp 0 )) Kp 0 ) p log p Kp) p log p > fw, h), giving us a contraiction, proving that H < Cp 0 )p 1/4 log p whenever H ) h /3 p 1/3 Note that if H > ) h /3 p 1/3, then χ is constant on a subset of H of carinality at most h ) /3 p 1/3 Therefore, H < Cp 0 )p 1/4 log p whenever h ) /3 p 1/3 Cp 0 )p 1/4 log p For p we have that Cp 0 )p 1/4 log p < ) h /3 p 1/3, which implies that for p 10 10, H < Cp 0 )p 1/4 log p It is not har to see that Cp 0 ) = π e 1 + o1), thus proving the first assertion in the Theorem Table 1 shows values of Cp 0 ) for ifferent values of p 0 We have been able to attack the problem with asymptotic choices for h an w, but we can fix the values of h an w an improve the bouns

9 ON THE MAXIMUM NUMBER OF CONSECUTIVE INTEGERS ON WHICH A CHARACTER IS CONSTANT9 p 0 Cp 0 ) Table 1 Upper boun H on the number of consecutive resiues with equal character value For p p 0, H < Cp 0 )p 1/4 log p From Table 1 we have establishe that H < 155p 1/4 log p when p Therefore to finish the proof of the theorem, we nee to eal with the interval < p < Let Xp) = π e 1 p 1/4, h an let γp, w, h) be efine in the following way: γp, w, h) = fw, h) A p log p gxp)) Then by similar arguments as before we have H < γp, h, w)p 1/4 log p whenever h an w are picke such that 4h γp, h, w)p 1/4 log p < ) h /3 p 1/3 an w 9h Hence, all we want is for γp, h, w) to be at most 155, for 4h 155p 1/4 log p < h /3 p 1/3 By picking w s an h s as in the Table, we complete the proof for p > noticing that with h an w fixe, γp, w, h) is concave up, allowing us to just check the enpoints of the intervals) Let s now prove that for all p we have H < 364p 1/4 log p It is true for p = since 364 1/4 log > 1 Now, for 19 p , it is true because of the following inequality of Brauer [1] establishe with elementary methos): H < p + < 364p 1/4 log p Assume p > We re going to show that in this case, in fact H < 3p 1/4 log p Note that we have a restriction on h since we want H < ) h /3 p 1/3 to be able to use our machinery If h = 94, then for p we have ) h /3 p 1/3 > 3p 1/4 log p Using w =, we have γp, w, h) < 3 whenever p [3 10 6, 10 8 ] Now picking w = 3 we get γp, w, h) < 3 whenever p [10 8, ] But, for p > , we can use the boun of H < 155p 1/4 log p, completing the proof Remark As mentione earlier, Norton announce but in t give etails) that he coul prove H < 41p 1/4 log p for all o p an H < 5p 1/4 log p for p > e In Theorem 1 we prove something slightly better than his first claim,

10 10 ENRIQUE TREVIÑO w h p w h p w h p 6 6 [5 10 9, ] 6 8 [10 10, ] 7 8 [ , ] 7 3 [10 11, 10 1 ] 7 37 [10 1, ] 8 41 [10 13, ] 8 44 [10 14, ] 9 45 [10 15, ] 9 51 [10 16, ] 9 59 [10 17, ] 10 6 [10 18, ] [10 19, 10 0 ] [10 0, 10 1 ] 1 7 [10 1, 10 3 ] [10 3, 10 5 ] 15 8 [10 5, 10 7 ] [10 7, 10 9 ] [10 9, ] [10 31, ] [10 33, ] [10 35, ] [10 37, ] [10 39, ] 150 [10 41, ] [10 43, ] [10 46, ] [10 49, 10 5 ] [10 5, ] [10 55, ] [10 58, 10 6 ] [10 6, ] Table As an example on how to rea the table: when w = 10 an h = 6, then γp, w, h) < 155 for all p [10 18, ] It is also worth noting that the inequalities 4h 155p 1/4 log p < h /3 p 1/3 an w 9h are also verifie for each choice of w an h but it is har to juge with his secon claim as our better boun kicks in later) To fill the gap, I will now show that H < 4p 1/4 log p for p > e 15 a slightly stronger claim than Norton s) Note that we nee only fill in the gap e 15 < p For h 67 we have 4p 1/4 log p < h ) /3 p 1/3 whenever p > e 15 Therefore we have H < γp, w, 67)p 1/4 log p for p > e 15 We note that γp,, 67) < 4 when p e 15, ) an γp, 3, 67) < 4 when p [10 75, ], completing the proof of our claim Remark 3 If we re looking for the maximum number of consecutive non-resiues for which χ remains constant, then we can o a little better than H < 364p 1/4 log p In fact we can prove H < 3p 1/4 log p for all o p Let s prove it It is true for p = 3 an for p = 5 since in both cases we have 3p 1/4 log p > p Now, for 7 p 10 6, it is true because of the following inequality of Huson [7] 3 : H < p 1/ + /3 p 1/3 + 1/3 p 1/6 + 1 < 3p 1/4 log p We can conclue by noting that for p > , H < 3p 1/4 log p Acknowlegements I woul like to thank Carl Pomerance for his useful comments regaring this paper an his encouragement I woul also like to thank Paul Pollack who irecte me to McGown s work Finally, I thank the anonymous referee for his helpful comments 3 This inequality was one using elementary methos that buil on the work of Brauer [1] The inequality uses a clever construction that is able to use information on gp, k) to boun the number of consecutive non-resiues for which χ remains constant However, it oes not appear to exten to inclue the case of the maximum number of consecutive resiues

11 ON THE MAXIMUM NUMBER OF CONSECUTIVE INTEGERS ON WHICH A CHARACTER IS CONSTANT 11 References 1 Alfre Brauer, Über ie Verteilung er Potenzreste, Math Z ), no 1, MR D A Burgess, On character sums an primitive roots, Proc Lonon Math Soc 3) 1 196), MR #A569) 3, A note on the istribution of resiues an non-resiues, J Lonon Math Soc ), MR #6135) 4 Paul Erős an János Surányi, Topics in the theory of numbers, Unergrauate Texts in Mathematics, Springer-Verlag, New York, 003, Translate from the secon Hungarian eition by Barry Guiuli MR j:11001) 5 Anrew Granville an K Sounararajan, Large character sums: pretentious characters an the Pólya-Vinograov theorem, J Amer Math Soc 0 007), no, electronic) MR k:1118) 6 Aolf Hilebran, Introuction to Analytic Number Theory Lecture Notes, Richar H Huson, On the istribution of k-th power nonresiues, Duke Math J ), MR #158) 8 Kevin J McGown, Norm-Eucliean cyclic fiels of prime egree, Int J Number Theory to appear) 9, On the constant in Burgess boun for the number of consecutive reiues or nonresiues, Funct Approx Comment Math to appear) 10 Karl K Norton, Numbers with small prime factors, an the least kth power non-resiue, Memoirs of the American Mathematical Society, No 106, American Mathematical Society, Provience, RI, 1971 MR #3948) 11, Bouns for sequences of consecutive power resiues I, Analytic number theory Proc Sympos Pure Math, Vol XXIV, St Louis Univ, St Louis, Mo, 197), Amer Math Soc, Provience, RI, 1973, pp 13 0 MR #1103) 1 T A Preobrazhenskaya, An estimate for the number of consecutive quarature resiues, Vestnik Moskov Univ Ser I Mat Mekh 009), no 1, 4 8, 71 MR Herbert Robbins, A remark on Stirling s formula, Amer Math Monthly ), 6 9 MR ,100e) 14 Mariusz Ska lba, On Euler-von Mangolt s equation, Colloq Math ), no 1, MR f:11149) 15 Terence Tao, A remark on partial sums involving the Möbius function, Bull Aust Math Soc ), no, MR Enrique Treviño, The least k-th power non-resiue, Submitte, , Numerically explicit estimates for character sums, 011, Thesis PhD) Dartmouth College 18 Anré Weil, On some exponential sums, Proc Nat Aca Sci U S A ), MR ,34e) Department of Mathematics an Statistics, Swarthmore College, Swarthmore, Pennsylvania aress: etrevin1@swarthmoreeu