The least k-th power non-residue

Transcription

1 The least k-th power non-resiue Enriue Treviño Department of Mathematics an Computer Science Lake Forest College Lake Forest, Illinois 60045, USA Abstract Let p be a prime number an let k 2 be a ivisor p. Norton prove that the least k-th power non-resiue mo p is at most 3.9p /4 log p unless k = 2 an p 3 mo 4, in which case the boun is 4.7p /4 log p. By improving the upper boun in the Burgess ineuality via a combinatorial iea, an by using some computing power, we improve the upper bouns to 0.9p /4 log p an.p /4 log p, respectively. Contents Introuction 2 Burgess Booker upper boun 3 3 Burgess Norton lower boun 9 4 Main theorem 7 5 Acknowlegements 27 6 Bibliography 27. Introuction Let p be a prime an let k 2 be a ivisor of p. Let gp, k be the least k-th power non-resiue mo p. The case k = 2, i.e., the uestion of how big the least uaratic non-resiue is, has been stuie extensively. Assuming the Generalize Riemann Hypothesis for Dirichlet L-functions, Ankeny [] Preprint submitte to Elsevier August 29, 204

2 showe that gp, 2 log p 2 an Bach [2] mae this explicit by proving uner GRH that gp, 2 2log p 2. The best unconitional results for gp, 2 are ue to Burgess [5], who, builing on work by Vinograov [9] an using Weil s boun for curves [20], showe that gp, k ε p 4 e +ε. For k 3 we have better estimates. Let ρ be Dickman s function, i.e., a continuous function that satisfies uρ u + ρu = 0 an ρu = for 0 u. Let α k be the uniue root of ρα =. Wang Yuan [22], builing k on work of Vinograov [9] an Buhštab [4], showe that, for real ε > 0, gp, k ε,k p 4α k +ε. It is worth noting that α 2 = e. Vinograov showe that α k e k k Buhštab prove, for k e 33, that α k > log k log log k + 2. All of the work escribe so far has been of aymptotic nature. In terms of getting explicit bouns, Karl Norton [0], builing on a techniue of Burgess [7], was able to show that gp, k 3.9p /4 log p unless k = 2 an p 3 mo 4 for which he showe gp, k 4.7p /4 log p. In this paper we will improve this result. Let h an w be any positive integers, let p mo k be a prime, an let χ be a character mo p of orer k, that is, k is the smallest positive integer such that χ k is the principal character. Define p h 2w S w p, h, χ, k := χm + l. m= Norton s proof uses an ineuality iscovere by Burgess [6], namely that l=0 S w p, h, χ, k < 4w w+ ph w + 2wp /2 h 2w. Norton mae some moifications to a clever argument of Burgess, to get an explicit lower boun for S w p, h, χ, k epening on gp, k. This allowe him to get the above state upper boun on gp, k. Inspire by a paper of Booker [3] that eals with the uaratic case in the Burgess ineuality, we improve the upper boun on as follows: 2 an

3 Theorem.. Let p be a prime. Let w, h an k be integers such that w 9h, h p, k 2 an k p. Let χ be a character mo p of orer k. Then p h 2w S w p, h, χ, k = χm + l < 2w! 2 w w! phw + 2w p /2 h 2w. m= l=0 This upper boun was state but not prove by Norton [] an prove for uaratic characters by Booker [3]. With a more careful combinatorial analysis we improve the term 4w w+ to 2w! 2 w w! w 2w 2 = ow w. e This improvement is the main result which allowe us to improve the upper boun on gp, k. This theorem has also allowe the author to get an explicit version of the Burgess ineuality in [7] an to improve the best known explicit boun on the largest string on which a Dirichlet character mo p is constant [8]. We state our main result in the following theorem: Theorem.2. Let p > 3 be an o prime. Let k 2 be an integer such that k p. Let gp, k be the least k-th power non-resiue mo p. Then gp, k < 0.9p /4 log p, unless k = 2 an p 3 mo 4, in which case gp, 2.p /4 log p. A similar boun was announce but not proven by Norton see [], namely that gp, k.p /4 log p + 4. In section 2 we will prove our upper boun on S w p, h, χ, k, i.e., Theorem.. In section 3 we will write own Norton s lower boun for S w p, h, χ, k with some moifications. In the last section of this paper we combine the upper boun from section 2 with the lower boun from section 3 to prove Theorem Burgess Booker upper boun Definition 2.. Let p > 2 be a prime an let l, l 2,..., l 2w be fixe integers. Then efine x F p x as follows: x = x + l x + l 2 x + l w x + l w+ p 2 x + l w+2 p 2 x + l 2w p 2. 3

4 Abusing notation, we will consier it as a rational function: x = x + l x + l 2 x + l w x + l w+ x + l w+2 x + l 2w. Note that if k p, the polynomial form for x is a k-th power if an only if the rational form for x is a k-th power. Definition 2.2. Let p be a prime. Let w, h an k be integers such that h p an k p. Let [h] = {0,, 2,..., h }. Let x be efine as in Definition 2.. Then efine b w p, h, k as follows: { b w p, h, k = l, l 2,..., l 2w [h] 2w x is a k-th power Fp x}. Lemma 2.. Let p be a prime. Let w, h an k be integers such that h p, k 2 an k p. Let b w p, h, k be efine as in Definition 2.2. Then b w p, h, k w k =0 w! 2 h w k 2!k! w k!. Proof. Let x be efine as in Definition 2.. One way of bouning how many 2w-tuples make x a k-th power in F p x is the following: given a tuple, we eliminate the terms from the numerator that appear also in the enominator. We o this until there are no more eliminations to be one. Let s say that the number of terms eliminate is t. Then t is an integer such that 0 t w. Now for x to be a k-th power the numerator an the enominator must each be a k-th power. Fix t. The number of ways of getting t eliminations is boune above by 2 w t!h t. 2 t The reason for this count is that we are picking t elements from the numerator to be matche up with t elements from the enominator. To pick the 2t factors that will be paire up we have w 2 t ways of oing it. But now we have t! ways of associating a one to one map between the t elements in the numerator an the t elements in the enominator. Once we have the t pairs, then there are at most h t ways of picking the values for each pair, giving us the state upper boun. 4

5 Now, the number of ways in which the remaining parts of the the numerator can be a k-th power is w t h w t! =! k, k, k,..., k!k! h. 3 We ivie by! because the multinomial associates an orer to the groups being picke. We multiply by h because each group of size k has h options. For the remaining parts of the enominator of x we woul have the same estimate with h replace by h since we alreay eliminate the common terms. Since we are intereste in an upper boun, to simplify calculations I will replace h with h. Combining 2 an 3 an summing over values of t w mo k we arrive at the following upper boun for b w p, h, k: 0 t w t w mo k w t!!k! 2 w t 2 t!h t+2 = 0 t w t w mo k w!!t!k! 2 t!h t+2. 4 Using that t = w k we can change variables an reach the esire ineuality. Definition 2.3. Let w, h an k be positive integers such that k 2. Then efine c w h, k as follows: c w h, k = w k =0 w! 2 h w k 2!k! w k!. Note that for any prime p with k p, Lemma 2. implies that b w p, h, k c w h, k. Lemma 2.2. Let w, h an k be positive integers such that k 2. Let c w h, k be efine as in Definition 2.3. If w 9h, then c w h, k is a ecreasing function in k. Proof. Since k is an integer greater than or eual to 2, it is enough to show that c w h, k c w h, k for all k 3. From Definition 2.3 we have c w h, k = w k =0 2 w! h w k 2!k! w k!. 5 5

6 Now, we arrange the right han sie of 5 to look more like c w h, k, getting: w k =0 w! 2 h w k 3!k! k 2 = w k =0 h w k! w! 2 h w k 3!k! w k! w k!. k 2 h w k! Now we use that w k! w k! w to get the ineuality c w h, k w k =0 w! 2 h w k 3!k! w k! The last step is true because w 9h an because k 3. The following corollary is an obvious conseuence: w cw h, k. k 2 h Corollary 2.. Let w, h an k be positive integers such that k 2. Let c w h, k be efine as in Definition 2.3. If w 9h, then c w h, k c w h, 2. Now we will prove a combinatorial ientity an a corollary that will be use later, but it is a cute result on its own. Lemma 2.3. Let w be a positive integer. Then w 2 2 w! = 2w! w 2! 2! 2 w w!. 6 =0 Proof. The proof will be one by counting the number of partitions of {, 2,..., 2w} into w pairs in two ways. It is worth noting that the way to count the left han sie of 6 was one in Lemma 2. when k = 2, however we ll give a ifferent exposition of the count below to perhaps make the combinatorics clearer. Let s count the number of partitions. There are 2w choices to pair the number. Then pick the next lowest number not picke. There are 2w 3 ways of choosing its partner. Then pick the next lowest number not 6

7 picke. There are 2w 5 ways of choosing its partner. If we continue with this process, we get 2w 2w 3 3 = 2w2w 2w 2 2 2w2w 42 = 2w! 2 w w!. Notice that this is the right han sie of the euation. Now, let s count the number of partitions ifferently. Consier the pairs as i, j with 0 < i < j 2w. Now let P be a partition of {, 2,... 2w} into w pairs. Define AP, BP an CP in the following way: AP = {i, j P 0 < i < j w} BP = {i, j P w < i < j 2w} an CP = {i, j P 0 < i w < j 2w} We can see by the construction that AP, BP an CP are pairwise isjoint. We can also notice that P = AP BP CP. Let AP =. Then the w 2 numbers w which are not in AP must be paire with numbers > w. Therefore CP = w 2 an BP =. Therefore a way of counting the number of partitions is by counting for each choice of with 0 w 2 the number of ways of getting AP, BP an CP. The number of ways of pairing up in this way is 2! 2! 2 2! w w 2! = 2! w 2 2 w! w 2! 2! Once we sum over all we get the left han sie of the euation, completing the proof. Corollary 2.2. Let w be a positive integer. Then w 2 =0 w w 2 w 2 = 2w. w Proof. Multiply both sies of euation 6 by 2w. The right han sie of the w! euation becomes 2w! 2 w = 2w! 2w 2 w w! w! w!w! =. w 7

8 The left han sie becomes w 2 w 2 2 w! = w! w 2! 2! =0 w 2 =0 w!2 w 2!!w 2! = w 2 =0 w w 2 w 2. Now we are reay to prove Theorem.. Proof of Theorem.. Let x be efine as in Definition 2.. Using that z 2 = z z an that χn = χn p 2 allows us to rewrite S w p, h, χ, k in terms of x as follows: p h 2w S w p, h, χ, k = χm + l = p χ x. m= l=0 l,...l 2w 0 l i h If x is not a k-th power F p x then using the Weil boun [4, Theorem 2C, page 43], we can boun the inner sum by r p, where r is the number of istinct roots of x which o not have multiplicity a multiple of k. In particular, we can boun the inner sum by 2w p. Using Lemma in [6], we have the better boun 2w 2 p +, but we shall not use it here. When x is a k-th power, then we use the trivial boun of p. Using this analysis, we can now boun S w p, h, χ, k by placing the boun 2w p when x is not a k-th power an p otherwise. Combining this with w 9h yiels S w p, h, χ, k 2w h 2w p + b w p, h, kp 2w h 2w p + c w h, 2p. 7 Now, let s calculate c w h, 2: c w h, 2 = w 2 =0 w!!2 x= 2 h w w 2! = 2w! 2 w w! hw, 8 the last euality coming from Lemma 2.3. Combining 7 an 8 we get the esire ineuality. Remark 2.. From the proof we coul erive a better upper boun when k > 2, which is p h 2w S w p, h, χ, k = χm + l < c w h, kp + 2w p /2 h 2w. 9 m= l=0 8

9 3. Burgess Norton lower boun Let s start with a couple of lemmas that will be reuire in our lower boun estimate. Lemma 3.. Let x be a real number. Then x x φ x φ 3 π 2 x2 x. 0 Proof. Let s estimate the sum. x x φ x = x µ φ = x x φ = x x x x = x x Now, writing x = x { x }, we get µ x = x2 2 = Now, since x x x 2 2 x 2 + { x} { x } 2 µ 2 = >x µ µ µ x µ x x x µ x x + 2 { x { x µ. } } = 6 π an since 0 { { x } x } 2, we have 4 φ φ 3 π 2 x2 x2 µ 2 x µ x >x x x suarefree.. Claim 3.. For real x, >x µ 2 x. 9

10 Proof of the Claim: Note that for any positive integer we have that is 2 +/2 t smaller than t. Thus 2 /2 >x µ 2 >x t t = 2 x 2 t t = 2 x. 2 To change x /2 into x, note that there is at least one missing in the interval [x, x + 4], since we only take suarefree s in the sum. Thus the absolute value of the sum is smaller than. This is smaller x x than once x, proving the claim for real x. To complete the x µ proof for x we use the fact that = 6, which implies that 2 π2 = µ = 6 2 π µ. One can now manually check the integer cases 2 >x x µ where x an note that 2 <, which implies the claim x+ >x for real x. Claim 3.2. For real x, the number of suarefree integers in [, x] is at most 2 3 x + 2. Proof of the Claim: The number of suarefree numbers up to x is at most x x x x x + 2. Claim 3.3. For real x, x µ x. Proof of the Claim: The proof here is a moifie version of a proof of Hilebran [9]. Let en = if n = an en = 0 otherwise. Let Sx = n x en. Then Sx =. However, we also have Sx = n x n µ = x µ = x µ { x µ. } x x x 0

11 Therefore, x x µ + x { x } µ 2 3 x + 3. To prove the last ineuality we use that the number of suarefree numbers up to x is at most 2 x + 2, which was proven in the previous claim. 3 Combining Claims 3., 3.2 an 3.3 with we have x x φ x φ 3 π 2 x2 2 x π 2 x2 x, where the last ineuality hols for x 2. For x 3, the right han sie of 0 is negative, while the left han sie is positive, therefore the ineuality is true for x 3. Now, for the integers 3 x 2 we can manually check that x x φ x φ 3 π 2 x + 2 x +. Since the right han sie of 0 is increasing for x 3, we have a proof for all real x 2. Lemma 3.2. Let x be a real number. Then 2x φ φ 9 log x π 2 x2 x 3 x x Proof. Doing the estimates the same way as in Lemma 3., we get 2x φ φ = 9 π 2 x2 3x2 µ 2 x x >x x µ 2 x µ { x + } { x { x µ. 3 2 } } x x x Claim 3.4. For real x, x x µ { x } 2 x { x { x µ } } x log x x

12 Proof of the Claim: We have x x µ { x } 2 x µ= x { x { x µ } } = { x } x µ x { x } 2. 4 Note, that all factors except µ are positive, which implies that we can boun 4 by { x } x { x } x 2 x. 5 µ= Note that log x x numbers: x suarefree However, an log x+. Now, let s boun the sum over suarefree x x x log x log 4 4 x suarefree x = + x µ= µ = x µ= x µ= + log 9 9 x µ= x 36 log log x log x + 3 5, x. 7 The last ineuality being true because of Claim 3.3. Aing 6 an 7, iviing by 2, an using 4 an 5 we get our claim. 2

13 Now, using the results of Claims 3., 3.3 an 3.4 in 3 yiels 2x x φ φ x 9 log x π 2 x2 x log x 0 π 2 x2 x + 3, 3 where the last ineuality is true for x 45. For x 3, the right han sie of 2 is negative, while the left han sie is positive, therefore the ineuality is true for x 3. Now, for the integers 3 x 45 we can manually check that 2x x φ φ 9 log x + π x x + 3 x + 3. Since the right han sie of 2 is increasing for x 3, we have a proof for all real x 45. To prove a lower boun on S w p, h, k we will nee an upper boun on gp, k. The next lemma is an elementary boun on gp, k proven for k = 2 in [2] Lemma. Lemma 3.3. Let gp, k be the least k-th power non-resiue mo p. Then gp, k < p + 2. Proof. The following proof is very similar to the one in [2] but the argument goes all the back to Western an Miller see [2] an Norton [0]. Let = gp, k an r =. Note that p < r < p +, therefore r is a k-th p power mo p. Since is a k-th power non-resiue mo p, then r is also a k-th power non-resiue mo p. By the minimality of, r. Therefore + p/ > r. Since p is an integer an p > 2, then an the lemma follows. p 2 + > 2 2, We now have the ingreients to prove the lower boun on S w p, h, k. 3

14 Theorem 3.. Let p 5 be a prime. Let χ be a character mo p of orer k. Assume that χa = for all a < H. Let h an w be positive integers such that 4 h H. Let X = H/h an let A = 3 π 2. Then S w p, h, χ, k = p h 2w χm + l 2h 2w AH 2 m= l=0 2AX Furthermore, if is a k-th power mo p, then log X S w p, h, χ, k 3h 2w AH AX 3AX AX 2 h Proof. We follow the proof in [0] with some minor moifications. The iea is to fin long intervals where χ is constant either or, making the inner sum as big as possible in a segment. For each pair of integers t, with efine I, t to be the close interval [ tp H I, t =. 0 t < X an gct, =, 8 Claim 3.5. The intervals I, t are isjoint., tp + H ]. Proof of the Claim: Let s assume that I, t an I 2, t 2 contain a common element s, so that s t ip i H i for i =, 2. Thus, t p t 2p 2 H + H 2. By Lemma 3.3, H gp, k < p + /2. Using that h 4 an p 2, we get t 2 t H p 2HX p = 2H2 hp < 2p + 2 p + /2 4p <. Now, since t 2 t 2 < an t, t 2,, 2 are integers, we have t 2 = t 2. But gc, t = an gc 2, t 2 =, therefore t = t 2 an = 2 proving the claim. Claim 3.6. Each I, t [ H, H + p. 4

15 Proof of the Claim: Since t 0 an p 2, we have tp H p H p H. Now, since t <, we have t, therefore tp + H p + H = p p H p p H X p Hh = p. H In the ineualities we use X an that X = H/h. To finish proving the claim we will use that h 4: tp + H p p Hh H p 4p H H < p H. The last ineuality is true because H < p+/2 an 4 h p an therefore H 2 + 4H < p + 5 p + 3 < 4p, which is true for p 5. Therefore, we have prove the claim. Using the perioicity of χ an Claims 3.5 an 3.6 we have the following: p h S w p, h, χ, k = χm + l m= l=0 h 2w χm + l =,t,t m I,t l=0 2w = h χm + l l=0 h χm + l tp H m< H+p m I,t l=0 2w 2w. 9 The sum is over all pairs, t satisfying 8. Note that χ = because 0 < X < H. The last euality in 9 comes from χm+l = χχm+ l = χm + l = χm + l tp it is only neee that χ 0, for 9 to be true. For, t satisfying 8, let J, t an K, t be efine as follows: [ tp H J, t =, tp h + an K, t = tp, tp + H ] h +. If m J, t, then for 0 l h we have 0 < tp m + l H, therefore χm + l tp = χ χtp m + l = χ. Similarly, if m K, t, then for 0 l h we have 0 < m+l tp H an hence χm + l tp = χ =. 5

16 Since each of J, t, K, t contains at least H h integers note that X = H an hence H h then we can place a lower boun on S h wp, h, χ, k as follows: hx h h 2w S w p, h, χ, k 2,t = 2h 2w+ X X φ X φ 2AX 2 h 2w+ 2AX. 20 The last ineuality being Lemma 3.. Once we make the substitution of X = H we get the esire ineuality. h If is a k-th power mo p, we can improve 20. Instea of using J, t an K, t, we simply consier the interval [ tp H L, t =, tp + H ] h +. If m L, t, then for 0 l h, we have H m + l tp H, an hence χm + l tp = unless m + l = tp. Since > t 0, then 0 m + l = t p < p. But p m + l implies that m + l = 0, an so t = 0. Because of the coprimality conition, t = 0 implies =. In this latter case, we omit those values of m for which there is an l with m + l = 0, an we get S w p, h, χ, k h 2w + h 2w h 2w + m H h+,t > 2H h + h 2w + H m h < X m L,t 0 t< gct,= 2H h h 2w. From this an X = H, it follows that if is a k-th power mo p, then h S w p, h, χ, k h 2w+ 2X φ φ + h X X log x 3AX 2 h 2w AX 3AX AX 2 h The last ineuality comes from Lemma 3.2. Once we make the substitution of X = H we get the esire ineuality. h 6

17 4. Main theorem Before we prove our main theorem, we nee a lemma: Lemma 4.. Let p be a prime. Let k > be an integer such that k p. If = gc k, p 2 an 2, then is a -th power mo p an furthermore gp, k gp,. Proof. Let r be a primitive root mop. Then r p 2 mo p. Since p, then is a -th power mo p. Now note that if a < gp, k, then a 2 is a k-th power mo p an hence a -th power mo p since k, therefore gp, gp, k. Note that 2 unless k = 2 an p 3 mo 4. The following theorem will eal with the large cases of our main theorem. The main theorem will be split into cases after proving this theorem. Theorem 4.. Let p be an o prime. Let k 2 be an integer such that k p an let p p 0. Then gp, k < βp 0 p /4 log p, unless k = 2 an p 3 mo 4, in which case gp, 2 αp 0 p /4 log p, where βp 0 an αp 0 are constants epening only on p 0 escribe in Table. π 2 We remark that from the proof, one can show that αp 0 e e = an βp 6 0 e = π 2A 6 e = as p0. 8A = Proof. Let χ be a character mo p of orer k. Assume that χa = for all a < H. Let h an w be positive integers such that 4 h H. Let X = H/h an let A = 3. Then by Theorem 3. π 2 p h 2w S w p, h, χ, k = χm + l 2h 2w AH 2. 2AX m= l=0 If w 9h, we have from Theorem. p h 2w S w p, h, χ, k = χm + l < 2w! 2 w w! phw + 2w p /2 h 2w. 2 m= l=0 7

18 p 0 βp 0 αp Table : Upper boun for the least k-th power non-resiue mo p for p p 0. Combining these two we get that 2AH 2 < 2w! 2AX 2 w w! ph w + 2w p /2 h = fw, h, 22 is true for all positive integers h an w satisfying 4 h H an w 9h. Note that if we want to have H as small as possible, then we want to minimize fw, h, because the left han sie is approximately 2AH 2, so H is approximately fw, h/2a, where A is a constant. To minimize fw, h one can use simple techniues from Calculus to figure out the best asymptotic choices of w an h. Below we have chosen h an w to match the optimal asymptotic choice an to simplify some of the ifficulties that come from ealing with the fact that h an w are integers. Let 2w! w w h = p 2 w 2w + 23 w! 2w an log p w =

19 Then fw, h = h 2w! p 2 w w! 2w! w < 2 w w! 2w < p h + 2w < h p 2w + + w w 2w + + w p 2w + + w w p 2 + 2w + p 2 2w + + e 2 2w! 2 w w! w 2w w w The last ineuality is true because p 2w < e 2. Note the following explicit ineualities on Stirling s formula [3] which will help us eal with the above expression: n n 2πn e 2n+ n n < n! < 2πn e 2n. e e Hence 2w! w 2 w w! < 2w e w 2 e 24w w 2w+ = 2w e 2 2w e 24w 2 2w 2 +w. 26 Combining 26 with 25 an using that w 2w < 2 fw, h < 2w + + p w < we get 2we 2 2w e 24w 2 2w + + w 2w 2 +w + 2we + p. Now, the right han sie is increasing in w, so we may just use an upper boun for w which woul be log p 4 +. Using this upper boun yiels e fw, h < = 4 log2 p + 5e + 2 e 4 + 5e + 2 log p + 8e + 3 log 2 p + 8e + 4 log 3 p log p + 8e e + 4 p log p p log 2 p = Kp p log 2 p, 27 where Kp epens on p an goes to e 4 as p. 9

20 Also note h < 2we + < e 2 log p + 2e + = e 2 + 2e + log p log p. Assume p p 0 an H αp 0 p /4 log p. We have αp 0 e 8A, hence X = H h Let Xp 0 be efine as αp 0p /4 log p e + 2e+ log p 2 log p e e 8A + 2e+ 2 log p p /4. Xp 0 = e e 8A p /4 + 2e+ 0, 2 log p 0 an let Kp 0 αp 0 =. 2A 2AXp 0 The left han sie of 22 can therefore be boune from below for p p 0 : 2AH 2 2A αp 0 2 p log 2 p 2AX 2AXp 0 Kp 0 p log 2 p Kp p log 2 p > fw, h, giving us a contraiction, proving that H < αp 0 p /4 log p, that is gp, k αp 0 p /4 log p. Now, if is a k-th power mo p we can o better, since by the secon part of Theorem 3. we have S w p, h, χ, k = p h χm + l m= l=0 2w 3h 2w AH 2 log X AX 3AX AX 2 h 20

21 Combining this with 2 we get log X 3AH AX 3AX + 2 3AX 2 h < fw, h. 28 Assume p p 0 an H βp 0 p /4 log p e 2A p/4 log p, then we can work just as before. Let Xp 0 = e e 2A p /4 + 2e+ 0, 2 log p 0 an let βp 0 3A Kp 0 log Xp AXp 0 3AXp AXp 0 2 h. The left han sie of 28 can therefore be boune from below for p p 0 : 3AH 2 3A βp 0 2 p log 2 p log X AX 3AX + 2 3AX 2 h log Xp AXp 0 Kp 0 p log 2 p Kp p log 2 p > fw, h, 3AXp AXp 0 2 h giving us a contraiction, proving that H < βp 0 p /4 log p, that is gp, k βp 0 p /4 log p. If gc k, p 2 = >, then Lemma 4. implies that is a -th power an gp, k gp, βp 0 p /4 log p. Note that we o nee > as the last ineuality is only true for 2. Since gc k, p 2 = if an only if k = 2 an p 3 mo 4, we conclue the statement of the theorem. The values of the table for αp 0 an βp 0 were compute by plugging in the respective values of p 0. 2

22 We have prove the main theorem for p To complete the proof we ll o it in four cases: when k = 2 an p mo 4 with p 0 25, when k = 2 an p mo 4 or k 3, where 0 25 < p < 0 60, when k 3 with p 0 25, an when k = 2 an p 3 mo 4 with p < To eal with the case where k = 2 an p mo 4 we first show that either p is a gp, 2 -pseuosuare or gp, 2 = 2. Let s recall what a pseuosuare is: Definition 4.. A positive integer n is calle a -pseuosuare if n mo 8 is not a suare an for all o primes r, we have n r =, where n r is the Legenre symbol. Lemma 4.2. For p a prime satisfying p mo 4 then either p is a gp, 2 -pseuosuare or gp, 2 = 2. Proof. If p 5 mo 8 then 2 is not a suare mo p, an hence gp, 2 = 2. Therefore, we may assume that p mo 8. Note that by the efinition of gp, 2, we have that = for all o primes r < gp, 2. Now, since p mo 8, by uaratic reciprocity we have p r = =. r p Therefore p is a gp, 2 -pseuosuare. r p Proposition 4.. Let p be a prime such that p mo 4 an p Then gp, 2 0.9p /4 log p. Proof. If p 5 mo 8, then gp, 2 = 2 an hence gp, 2 0.9p /4 log p as long as p 5, which is true. Therefore, we may assume p mo 8. We know from Lemma 4.2 that p is a gp, 2 -pseuosuare. In [5], it was shown that for 379, 379-pseuosuares are greater than Therefore if gp, 2 379, then p

23 w h p w h p w h p 6 76 [0 25, 0 27 ] 7 85 [0 27, 0 29 ] 7 99 [0 29, 0 3 ] 8 06 [0 3, 0 33 ] 8 2 [0 33, 0 35 ] 2 6 [0 35, 0 38 ] 22 3 [0 38, 0 4 ] [0 4, 0 44 ] [0 44, 0 47 ] 30 4 [0 47, 0 50 ] 3 59 [0 50, 0 54 ] [0 54, 0 58 ] [0 58, 0 60 ] Table 2: Values of h an w chosen to prove that gp, 2 0.9p /4 log p whenever p mo 4 an 0 25 p As an example on how to rea the table: when w = 6 an h = 76, then γp, w, h < 0.9 for all p [0 25, 0 27 ]. Since the solution to 0.9p /4 log p = 379 is below , then we nee only check up to for the cases where gp, A simple loop in the computer confirms that for all these cases we have gp, 2 0.9p /4 log p, completing the proof of the proposition. Proposition 4.2. Let p be prime such that 0 25 < p < If p mo 4 an k = 2 or if k 3, then gp, k 0.9p /4 log p. Proof. To eal with this gap, we ll choose particular w s an h s in fw, h see 22 instea of the values of h an w chosen in Theorem 4.. Let A = 3 as before π 2 Xp = e 2A h p/4. Let γp, w, h be efine in the following way: fw, h γp, w, h = 3A. p log 2 log Xp +3 3 p + 3AXp 3AXp 2 3AXp 2 h Then by similar arguments as in Theorem 4., we have that gp, k is less than γp, h, wp /4 log p. Hence, all we want is for γp, h, w to be less than or eual to 0.9. We ll attack this by picking particular h s an w s in ifferent intervals. To check whether γp, h, w 0.9, we nee only check the enpoints of the intervals since γp, h, w is concave up. Table 2 completes the proof. 23

24 Remark 4.. The metho can also yiel gp, p /4 log p when p mo 4. However, it woul reuire a much longer table to fill up the intervals all the way up to It is also worth noting that if we starte at 0 7 instea of 0 25 i.e., if we in t have the result on pseuosuares, then the ineuality we woul get woul be gp, p /4 log p, which is not much worse. Thus, the main ingreient in the improvement over Norton is not computational power, but improving the upper boun on the Burgess ineuality. Proposition 4.3. Let p 0 25 be a prime, an let k 3 be an integer. Then gp, k 0.9p /4 log p. Proof. Note that an upper boun for the least k-th power non-resiue is the least primitive root mo p, since a primitive root cannot be a k-th power. Running a loop where we check the least primitive root over primes up to 0 5 reveals that the only examples where the primitive root is greater than 0.9p /4 log p are p = 2, 3, 7 an 9. For p = 2, it oesn t make sense to efine k-th power non-resiue. For p = 3 it only makes sense when k = 2, but k 3. For p = 7 it makes sense for k = 2 an k = 3. Since k 3, we are left with the k = 3 case. For k = 3, the least cubic non-resiue is 2 < 0.97 /4 log 7. To check what happens with p = 9, I ran a program looping over the possible k s ivisors of 90 an foun that the least k-th power non-resiue is 2 for all k p with k 3. Therefore, for k 3 an p 0 5, gp, k 0.9p /4 log p. Therefore we are now in the case where 0 5 p Let s recall 9: p h 2w S w p, h, χ, k = χm + l < c w h, kph w + 2w p /2 h 2w. m= l=0 Since c w h, k is ecreasing on k an k 3, we can replace c w h, k by c w h, 3. Let f 2 w, h be efine as f 2 w, h := h p p 2w + c w h, 3 h w w 3 = h p 2w + =0 2 w! p.!3! h +w w 3! 24

25 w h p 3 2 [0 5, 0 7 ] 4 6 [0 7, 0 9 ] 6 2 [0 9, 0 2 ] 8 37 [0 2, 0 8 ] 2 47 [0 8, 0 25 ] Table 3: Values of h an w chosen to prove that gp, k 0.9p /4 log p whenever k 3 an 0 5 p As an example on how to rea the table: when w = 6 an h = 2, then γ 2 p, w, h < 0.9 for all p [0 9, 0 2 ]. Then by Theorem 3. combine with 9, we have that the ineuality 28 becomes log X 3AH AX 3AX + < f 2 3AX 2 2 w, h, h where A is the constant we ve been using, H gp, k an X = H. Now, let h e 2A Xp = h p/4. Let γ 2 p, w, h be efine in the following way: f 2 w, h γ 2 p, w, h = 3A. p log 2 log Xp +3 3 p + 3AXp 3AXp 2 3AXp 2 h Then by similar arguments as in Theorem 4., we have that gp, k is less than γ 2 p, h, wp /4 log p. Hence, all we want is for γ 2 p, h, w to be less than or eual to 0.9. We ll attack this by picking particular h s an w s in ifferent intervals. Table 3 completes the proof of the interval 0 5 p Proposition 4.4. Let p > 3 be a prime such that p 3 mo 4 an p < Then gp, 2.p /4 log p. Proof. Running a loop over primes p 3 mo 4 up to 0 7 reveals that there is only one counter example, p = 3. Hence for 3 < p 0 7, gp, 2.p /4 log p. 25

26 w h p w h p w h p 4 2 [0 7, ] 5 2 [0 7.6, 0 8 ] 5 24 [0 8, 0 9 ] 6 25 [0 9, 0 0 ] 7 27 [0 0, 0 ] 7 34 [0, 0 2 ] 8 35 [0 2, 0 3 ] 9 36 [0 3, 0 4 ] 8 44 [0 4, 0 5 ] 8 55 [0 5, 0 6 ] 9 56 [0 6, 0 7 ] 9 64 [0 7, 0 8 ] 0 64 [0 8, 0 9 ] 2 60 [0 9, 0 2 ] 3 67 [0 2, 0 23 ] 4 75 [0 23, 0 25 ] 6 77 [0 25, 0 27 ] 7 85 [0 27, 0 29 ] 8 93 [0 29, 0 3 ] 9 00 [0 3, 0 33 ] [0 33, 0 36 ] 2 2 [0 36, 0 39 ] [0 39, 0 42 ] [0 42, 0 45 ] [0 45, 0 48 ] [0 48, 0 5 ] [0 5, 0 54 ] [0 54, 0 58 ] [0 58, 0 60 ] Table 4: Values of h an w chosen to prove that gp, 2.p /4 log p whenever p 3 mo 4 an 0 7 p As an example on how to rea the table: when w = 0 an h = 64, then γ 3 p, w, h <. for all p [0 8, 0 9 ]. Therefore we are now in the case where 0 7 < p < To eal with this gap, we ll follow the same strategy as in Proposition 4., which is to choose particular w s an h s in fw, h an fill up gaps. As in the proof of Proposition 4., let A be the constant we ve been using an let e 8A Xp = h p/4. Let γ 3 p, w, h be efine in the following way: fw, h γ 3 p, w, h = 2A p log 2 p 2AXp Then by similar arguments as in Theorem 4., we have that gp, 2 is less than γ 3 p, h, wp /4 log p. Hence, all we want is for γ 3 p, h, w to be less than or eual to.. We ll attack this by picking particular h s an w s in ifferent intervals. To check whether γ 3 p, h, w., we nee only check the enpoints of the intervals, since γ 3 p, h, w is concave up. Table 4 completes the proof. Combining Propositions 4., 4.2, 4.3 an 4.4 yiels Theorem

27 5. Acknowlegements I woul like to thank my avisor Carl Pomerance for his guiance. He has been the riving force of my research. I woul also like to thank Paul Pollack for pointing out Hilebran s lecture notes an commenting on a raft of this paper. Finally, I like to thank the anonymous referee for his excellent suggestions that improve this paper. 6. Bibliography [] N. C. Ankeny. The least uaratic non resiue. Ann. of Math. 2, 55:65 72, 952. [2] E. Bach. Explicit bouns for primality testing an relate problems. Math. Comp., 559: , 990. [3] A. R. Booker. Quaratic class numbers an character sums. Math. Comp., 75255: electronic, [4] A. A. Buhštab [A. A. Buchstab]. On those numbers in an arithmetic progression all prime factors of which are small in orer of magnitue. Doklay Aka. Nauk SSSR N.S., 67:5 8, 949. [5] D. A. Burgess. The istribution of uaratic resiues an non-resiues. Mathematika, 4:06 2, 957. [6] D. A. Burgess. On character sums an primitive roots. Proc. Lonon Math. Soc. 3, 2:79 92, 962. [7] D. A. Burgess. A note on the istribution of resiues an non-resiues. J. Lonon Math. Soc., 38: , 963. [8] H. Davenport an P. Erös. The istribution of uaratic an higher resiues. Publ. Math. Debrecen, 2: , 952. [9] A. Hilebran. Introuction to Analytic Number Theory Lecture Notes [0] K. K. Norton. Numbers with small prime factors, an the least kth power non-resiue. Memoirs of the American Mathematical Society, No. 06. American Mathematical Society, Provience, R.I.,

28 [] K. K. Norton. Bouns for seuences of consecutive power resiues. I. In Analytic number theory Proc. Sympos. Pure Math., Vol. XXIV, St. Louis Univ., St. Louis, Mo., 972, pages Amer. Math. Soc., Provience, R.I., 973. [2] P. Pollack an E. Treviño. The primes that Eucli forgot. Amer. Math. Monthly, 2: , 204. [3] H. Robbins. A remark on Stirling s formula. Amer. Math. Monthly, 62:26 29, 955. [4] W. M. Schmit. Euations over finite fiels. An elementary approach. Lecture Notes in Mathematics, Vol Springer-Verlag, Berlin, 976. [5] J. P. Sorenson. Sieving for pseuosuares an pseuocubes in parallel using oubly-focuse enumeration an wheel atastructures. In Algorithmic number theory, volume 697 of Lecture Notes in Comput. Sci., pages [6] E. Treviño. Numerically explicit estimates for character sums. 20. Thesis Ph.D. Dartmouth College. [7] E. Treviño. The Burgess ineuality an the least k-th power non-resiue. Submitte. [8] E. Treviño. On the maximum number of consecutive integers on which a character is constant. Moscow Journal of Combinatorics an Number Theory 202, vol.2, iss., pp [9] I. M. Vinograov. Selecte works. Springer-Verlag, Berlin, 985. With a biography by K. K. Marzhanishvili, Translate from the Russian by Naiu Psv [P. S. V. Naiu], Translation eite by Yu. A. Bakhturin. [20] Anré Weil. Sur les courbes algébriues et les variétés ui s en éuisent Actualités Sci. In., no , Deuxième Partie, IV. [2] A. E. Western an J. C. P. Miller. Tables of inices an primitive roots. Royal Society Mathematical Tables, Vol. 9. Publishe for the Royal Society at the Cambrige University Press, Lonon, 968. [22] W. Yuan. Estimation an application of character sums. Shuxue Jinzhan, 7:78 83,