2 E. TESKE AND H.C. WILLIAMS with h(?) = and () = : It is imortant to realize that at this time, the fast methods for evaluating class num

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1 A PROBLEM CONCERNING A CHARACTER SUM E. TESKE AND H.C. WILLIAMS Dedicated to the memory of Daniel Shanks ( ) Abstract. Let be a rime congruent to?1 modulo 4, n the Legendre symbol and S(k) = P?1 n=1 nk n. The roblem of nding a rime such that S(3) > 0 was one of the motivating forces behind the develoment of several of Shanks' ideas for comuting in algebraic number elds, although neither he nor D. H. and Emma Lehmer were ever successful in nding such a. In this aer we exhibit some techniues which were successful in roducing, for each k such that 3 k 2000, a value for such that S(k) > Introduction Shortly after the death of Daniel Shanks, the second author received a collection of corresondence between Shanks and D. H. and Emma Lehmer. This material covers the eriod between when Shanks was very active in develoing ideas that would be of great signicance to the develoment of comutational algebraic number theory. Furthermore, it is evident from this corresondence that a rather simle looking roblem served as a focus for his and the Lehmers' investigations during this time. In order to discuss their roblem we rst reuire some notation. We let d denote a fundamental discriminant of an imaginary uadratic eld K = Q( d ) and let h(d) denote the class number of K. In a brief letter, dated Ar.2, 1968, from the Lehmers to Shanks, the roblem of trying to roduce a small value for the ratio () = h(?)= ; where is a rime congruent to 3 modulo 8, is mentioned. We are trying to get this ratio down to.041. According to a theorem of Chowla there are innitely many such rimes, but we have not seen one yet. Any candidates? During the following year the Lehmers and Shanks made a concerted eort to nd small values for (). By Aug. 23 they had found = with h(?) = 2925 and () = :05940, breaking the \0:06 barrier"; and by Set. 26 they had found the best candidate that they ever discovered, namely = Date: January 7, Mathematics Subject Classication. 11Y40, 11Y99. Research suorted by NSERC of Canada grant A

2 2 E. TESKE AND H.C. WILLIAMS with h(?) = and () = : It is imortant to realize that at this time, the fast methods for evaluating class numbers that are used today did not exist. Indeed, Shanks was motivated by this roblem to develo fast methods because the Lehmers were roducing large values of as ossible candidates. Throughout this corresondence it is ossible to see Shanks develo and rene the ideas which were to culminate in a very imortant aer (Shanks [13]), where he introduced the baby ste-giant ste method for evaluating h(d) and his method of factorization of d, based essentially on the determination of ambiguous ideal classes in the class grou of K. He even recognized that his techniue for evaluating h(d) was likely to be of comlexity O(jdj 1=4+" ) for any " > 0, but it was Lenstra [8] who showed later that it was of comlexity O(jdj 1=5+" ) under the Extended Riemann Hyothesis (ERH). This reresented a considerable imrovement of the revious method of counting classes, a techniue of comlexity O(jdj 1=2+" ). Insired by his success with imaginary uadratic elds, Shanks [14] went on to discover what he called the \infrastructure" of the class grou of a real uadratic eld and how it could be alied to solve the roblem of determining its regulator and class number. In resonse to a uestion by Shanks concerning the location of the \Theorem of Chowla", the Lehmers mentioned that it in fact aeared in a aer by Ayoub, Chowla and Walum [1]. In this aer the authors discussed the character sum (1.1) S(k) = where is a rime ( 3(mod 4)) and?1 n k n n=1 n ; is the Legendre symbol. They ointed out that, while S(1) =?h(?), S(2) =? 2 h(?) and S(k) < 0 whenever k? 2, they could rove that S(3) > 0 innitely often. It is not immediately clear why this should mean that () < :041 innitely often and no roof of this was ever rovided by the Lehmers; however, the basic idea behind their thinking is suggested in the aer by the Lehmers and Shanks [7] which originated as a result of their collaboration. We illustrate this below. As usual, we dene the Dirichlet L-function by L(s; ) = 1 n?s (n): n=1? d Also, if (n) is the Kronecker symbol n, then the analytic class number formula for K = Q( d ) asserts that (1.2) 2h(d)=(w jdj ) = L(1; ); where w is the number of roots of unity in K (w = 2 if jdj > 4). When d =? 1(mod 4), then (n) =? d n (1.3) = n S(3) = 3 2 hence S(3) > 0 if and only if (1.4). In [1] it is shown that for this character?l(1; ) L(3; ) ; L(1; ) < 3 22 L(3; ):

3 A PROBLEM CONCERNING A CHARACTER SUM 3 Recalling the Euler roduct reresentation Y of L(s; ), s (1.5) L(s; ) = s? () ; where the roduct is taken over Y all the rimes, we see that 3 Y 3 L(3; ) = 3? () 3? () 41 Y ? () >41 Y >41 3 3? 1 : Now let a be 4 times the roduct of all the rimes less than or eual to 41 and b be a xed integer such that the Kronecker symbol =?1 for all the rime divisors of a. We have (a; b) = 1 and for any rime = ax + b we get () =?1 for 41 and Y 3? 1 L(3; ) < (3) = :84644: 41 Since from (1.2) and (1.4) we get () < 3L(3; )=2 3 when S(3) > 0, we see that () < :041 for such rimes. Unfortunately, it is not roved in [1] that S(3) > 0 for innitely many rimes selected from the arithmetic rogression fax + bg. However, it is ossible, by referring to a later theorem of Joshi [6], to rove the Lehmers' assertion that () < :041 innitely often without even reuiring that S(3) > 0 innitely often. The Lehmers and Shanks never did nd a value of for which either () < :041 or S(3) > 0. They did, however nd several values of for which S(4) > 0 [7], and in a letter dated June 5, 1969, they noted that S(5); S(6) > 0 for = 163. In fact the authors of [1] state at the end of their aer that results similar to the existence of an innitude of rimes such that S(3) > 0 hold for other small values of k. Later Fine [4] roved the following result. Theorem 1.1. For each real k > 2 there are innitely many rimes 3(mod 4) for which S(k) > 0 and innitely many for which S(k) < 0. b Unfortunately, Fine's method is not easily adated to the roblem of nding values for such that S(k) > 0. The urose of this aer is to show how to nd such values of for small integer values of k. Our initial objective was to discover values of such that S(k) > 0 for 3 < k 50, but we were somewhat surrised to learn that we could extend our method to do this for all 3 < k We also exhibit a value of for which S(3) > 0 and () < :041 under the ERH. 2. Our initial strategy As was done in [1], we can exand x k in a Fourier exansion with eriod 1 to obtain 1 S(k) = k m b m (k) ; m=1

4 4 E. TESKE AND H.C. WILLIAMS where b m (k) 2 Now, on integrating by arts b m (k) 2 Also, b m (1) = b m (2) =?1=(m). Hence, and S(k) =?k where (m) = m =?k?? = m. = = Z 1 x k sin 2mx dx: 0 Z 1 2m (2m) k+1 y k sin y dy 0 = 1 k(k? 1) b m (k? 2)?? 2m (2m) 2 : 2 b m (k) = b i=0 i=0 b In order to get S(k) > 0, we need or (2.1) where we dene i=0 A(k) = i=0 ()!? k i 2()!? k i+1 (2m) +1 ()!? k i (2) ()!? k i 1 1 m m m= (2) L( + 1; ) A ; (2) L( + 1; ) < 0 L(1; ) < A(k); ()!? k i+1 (2) L( + 1; ): This is a simle generalization of (1.4). Notice that, although we have converted a sum of? 1 summands to one of about k=2 summands, a diculty emerges in that the coecients of L( + 1; ) become very large and therefore dicult to work with. This freuently necessitates the use of routines which can calculate to very high levels of recision. Our rst strategy was to extend the idea the Lehmers emloyed to nd the numbers mentioned in x1; that is, we try to nd such that () =?1 for as many small rimes as ossible. Suose that () =?1 for all rimes Q; then from (1.5) where F s (Q) = Y Q L(s; ) = F s (Q)T s (Q; ); s s + 1 ; T s(q; ) = Y >Q s s? () : 1 A

5 A PROBLEM CONCERNING A CHARACTER SUM 5 We now need to estimate T s (Q; ). To this end we note that hence, (2.2)? log T s (Q; ) = >Q log(1? ()= s ) = >Q jlog T s (Q; )j 1 >Q 1 i is : 1?() i i is ; P We next examine the sum >Q?s (s > 1). If we let (x) reresent the usual rime counting function, then by artial summation 1 = s (m)[m?s? (m + 1)?s ]? (Q)=Q s >Q mq Z m+1 = s (x)x?s?1 dx? (Q)=Q s mq m Z 1 = s (x)x?s?1 dx? (Q)=Q s : Q By a result of Rosser and Schoenfeld [12], we have x= log x < (x) < x log x x log x Hence, >Q 1 s < Substituting this into (2.2) we get j log T s (Q; )j < s log Q 2 log Q Z 1 x?s dx? Q < Q?s+1 = log Q (s 3; Q 90): Q log Q 1 Q?si Since e x < 1 + 2x and e?x > 1? 2x for 0 < x < 1, we get or (2.3) i! jt s (Q; )? 1j < 3Q?s+1 = log Q (x > 17): Q Q s log Q < 3Q?s+1 ; (s 3; Q 90): 2 log Q j T s (Q; ) > j? 3Q?s+1 = log Q for any j 2 Z as long as s 3 and Q 90. From (2.3) it follows that where A(k) = B(k; Q) =? k ()! (2) F +1 (Q) i+1 T +1 (Q; ) > B(k; Q); ()!? k (2) i+1 F +1 (Q)? 3 log Q b ()!? k (2Q) F +1(Q);

6 6 E. TESKE AND H.C. WILLIAMS thus, if L(1; ) < B(k; Q), then S(k) > 0. In order to nd values of 3(mod 4) such that () =?1 for all Q we made use of the number sieve MSSU (see Lukes et al. [10] or [11]). We let r be a rime and dene N r by?nr 1. N r 3(mod 8) 2. =?1 for all odd rimes r 3. N r is the least ositive rime integer satisfying 1. and 2. By the heuristic reasoning of [11], we would exect log N r to be roughly (r log 2= log r) 1+o(1). In somewhat over a month of MSSU time, Jacobson [5],.128, comuted Table 1 below. This table is an extension of arts of Tables III and IIIa in [7]. r N r h(?n r ) L(1; ) , , ; : : : ; , , , , , 101, , 109, , ; : : : ; ; : : : ; , 197, , ; : : : ; , 263, Table 1. N r { Least Prime Solutions Note that for 41 r 269 we have (2.4) log N r > (r log 2)= log r:

7 A PROBLEM CONCERNING A CHARACTER SUM 7 We next comuted a table of values for B(k; Q) for Q = 270. Because of the growth rate of the terms of B(k; Q), we comuted it to 800 digits of recision. We found that if k 10, the values of B(k; 270) decreased monotonically for k < 400 and B(400; 270) = : Furthermore, B(4; 270) = : < B(k; 270) (4 < k 10) and B(142; 270) = : ; hence, we see that for = N 257, we have S(k) > 0 for all k such that 4 k A second aroach As the comutation of the N r values is very exensive and the concomitant rate of decrease of L(1; ) is very slow, we develoed a second strategy for nding values of such that S(k) > 0. The idea here was to allow for a greater degree of freedom than that aorded by insisting that () =?1 for all rimes Q. To this end we dene F s (Q; ) by Y s F s (Q; ) = s? () and B(k; Q; ) = Q ()!? k i+1 F+1 (Q; ) (2)? 3 log Q ()!? k F+1 (Q; ) (2Q) : By using the same reasoning as that emloyed in x2, we see that S(k) > 0 if L(1; ) < B(k; Q; ) or (3.1) If we dene and C(k; Q; ) = T 1 (Q; ) < B(k; Q; )=F 1 (Q; ): G s (Q; ) = F s (Q; )=F 1 (Q; ) = Y then by (3.1) we see that S(k) > 0 if (3.2) Q ()!? k i+1 (2) G +1 (Q; )? 3 log Q s?1 (? ()) s? () T 1 (Q; ) = L(1; )=F 1 (Q; ) < C(k; Q; ): ()!? k G+1 (Q; ) (2Q) ; Now a result of Elliott (see [13]) asserts that if Q > 2 and F (Q; z; ) is the density of all ositive d (here () =?d ) such that T 1 (Q; ) 1=(1 + z) or T 1 (Q; ) 1 + z (0 < z < 2) then there exist constants A, B such that F (Q; z; ) < 2A exf?bq log 2 (1 + z)g: This strongly suggests that if and Q are chosen with sucient care, we should get C(k; Q; ) 1 + z and the chance that T 1 (Q; ) < 1 + z is very good, esecially

8 8 E. TESKE AND H.C. WILLIAMS if z is small. For examle, if is selected such that () = +1 for = 2; 3; 5 and () =?1 for 7 < Q = 220, then C(k; Q; ) > 1:011 for 26 < k 800. (This was determined by comuting C(k; Q; ) to 2000 digits of recision.) But, by using the MSSU we found that = satises the conditions above and h(?) = ; hence, since F 1 (220; ) = 1: , we get T 1 (220; ) = L(1; )=F 1 (220; ) = 1: < 1:011: Naturally, this leads to the uestion of how best to secify the values of () for the small rimes. After conducting a number of numerical exeriments we discovered that the values of C(k; Q; ) tended to be largest over the longest interval for k when () = 1 for only the rst few rimes and () =?1 for the remainder of the rimes Q. For examle, if () = 1 for = 2; 3; 5; 7; Q = 90 and () =?1 for all other rimes Q, then C(k; Q; ) > 1:05 for 57 k 325 and if () = 1 for = 2; 3; 5 only, then C(k; Q; ) > 1:05 for 27 k 319 and if () = 1 for = 2; 3; 5; 7; 11, then C(k; Q; ) > 1:05 for 122 k 273. If () = 1 for = 79; 83; 89 and () =?1 for the remaining < 220, then C(k; Q; ) > 1:05 only when 4 k 67. We also found it useful to make Q in (3.2) much larger than the limit to which we can sieve with the MSSU. This is because if Q denotes the uer bound on the rime moduli used by MSSU, then F 1 (Q ; ) will likely not dier very much from F 1 (Q; ) when Q is much larger than Q, and this likely means that C(k; Q ; ) will not dier very much from C(k; Q; ). For examle, if we ut Q = 230 and Q = 1000 and secify that () = 1 for = 2; 3; 5; 7 and () =?1 for all the remaining Q, then (3.3) = satises our conditions on () for Q. Here while F 1 (Q ; ) = 2: F 1 (Q; ) = 2: : For this value of we get h(?) = ; hence, T 1 (Q; ) = L(1; )=F 1 (Q; ) = 1: On tabulating C(k; Q; ) for the values roduced by and Q = 1000, we found that if 29 k 35, then C(k; Q; ) > 1:0128 and C(k; Q; ) > 1:085 for 35 k Thus, since 1:085 > 1:011021, we see that for given by (3.3) we get S(k) > 0 for 29 k That the value 1:085 is uite a lot larger than 1:011 suggests that if we had tabulated C(k; Q; ) even further, we would likely have roduced an even larger value for k such that S(k) > 0; however, at this oint the comutation of the C(k; Q; ) values was very exensive because we were using 6000 digits of recision.

9 A PROBLEM CONCERNING A CHARACTER SUM 9 4. The roblem of S(3) We now know values of for which S(k) > 0 for all 4 k 2000, but we have not yet found a value of for which S(3) > 0. In [7] it is shown that if (4.1) L(1; ) < (6)=(4(2)(3)) = :12863; then S(3) > 0, and it follows that () < 0: Jacobson [5], , used the MSSU to roduce numbers N?1(mod 4) such that?n =?1 for = 2; 3; : : : ; 199 and comuted F 1 (Q; ) for Q = 1000 to search for likely values of N with small L(1; ). For those that were rime, he comuted h(?n) and an accurate value of (N). The best result he obtained was for the 20 digit (4.2) = : Here h(?) = , L(1; ) = : , and () = : While this value of () may seem rather close to :041, it is still a long distance away. An attemt by the authors to get an imroved value by searching u to the limit 2: , but secifying that?n =?1 for = 2; 3; : : : ; 149 only did not roduce a better result. This is actually not very surrising because for given by (4.2) we have () =?1 for all rimes 211, but (223) = 1. However, () =?1 for This is still not as good as N 257 in x2, but then (251) = 1, (257) =?1, (263) = 1, (269) = (271) = (277) =?1, (281) = 1, (283) = (293) = (307) =?1, (311) = 1; furthermore, () =?1 for Thus, given its size, is a most remarkable number because there are so few values of 401 for which () = 1; this hels to exlain the good value of (). In a letter of March 5, 1969, the Lehmers described a method which they used to nd a value of N such that L(1; ) is small. They resecied N to be such that N 1(mod ) and N 3(mod ); that is, they found A by the Chinese remainder theorem such that A 1(mod ) and A 3(mod ) and ut N = A + B where B = They then emloyed their sieving device, the DLS-127, to nd values for such that A + B 3(mod 8) and?(a+b) =?1 for all rimes (3 127). By this rocess they roduced the 20 digit number N = for which () =?1 for all 149 and L(1; ) = :17009, but N is unfortunately comosite. In an attemt Q to nd a better result than Jacobson's, we used the 353 Lehmers' idea with B = 251 3: and A such that?a =?1 for all j B. We then used the MSSU to sieve for values of such that for u to 241. Our best result was the 62 digit rime?(a+b) = =?1 for which F 1 (Q; ) = : with Q = This very likely means that the value of L(1; ) for this is less than that for Jacobson's number, but it is not suciently small to guarantee that S(3) > 0 or () < :041. Indeed, by August of 1968 Shanks and the Lehmers had reached the conclusion that the DLS-127 would never be able to nd a value for such that S(3) > 0. Shanks went so far as to

10 10 E. TESKE AND H.C. WILLIAMS estimate that such a value for might have to satisfy () =?1 for all rimes However, the value of F 1 (1283) = : which is already less than the value of L(1; ) needed by (4.1). If we use (2.4) to estimate a lower bound on N 1283 we get log N 1283 > 124, suggesting that N 1283 is likely to be a number of at least 54 digits, a number far too large for any current sieve device to nd. There is, however, thanks to a recent result of Bach [2], another way to nd a candidate for. Because F 1 (1279) is uite close to (6)=(4(2)(3)), we simly?n found values for N such that =?1 for all We did this by secifying that for all rime values of 1279, N 3(mod 8); N 1(mod ) when?1(mod 4); N r()(mod ) when 1(mod 4): Here r() denotes a randomly selected uadratic nonresidue of. Notice that if N satises these conditions we have?n =?1 for all The diculty with this rocess is that the values we get for N are very large, 535 or more digits. However, testing the numbers for rimality is very easy because N? 1 is divisible by all the rimes?1(mod 4) ( 1279). Thus it is easy to nd a comletely factored art of N?1 which exceeds N, and the method of Pocklington mentioned in Brillhart, Lehmer and Selfridge [3] (Theorem 4) can easily be used to establish the rimality of N. We roduced 10 rime values for in this way and selected that one such that F 1 (200000) was least. This value is the 535 digit = (4.3) n n n n n n n n : It remains to show that L(1; ) for this satises (4.1). We used the method of [2] to estimate L(1; ). We dene C(Q) = 2Q?1 i=q i log i; (j=0,1, : : :,Q-1). Bach showed that under the ERH j log L(1; )? Q?1 i=0 a j = (Q + j) log(q + i)=c(q) a i log F 1 (Q + i? 1)j A(Q; d);

11 A PROBLEM CONCERNING A CHARACTER SUM 11 where A(Q; d) = (A log jdj + B)=( log Q) and A, B are exlicit constants tabulated in [2], Table 3. For Q = and d =?, we comuted S(Q; ) = Q?1 i=0 a i log F 1 (Q + i? 1) by carrying 40 digits of recision, and found that We also found that Hence, since we get S(Q; ) =?2: : A(Q; ) = : : e S(Q;) e?a(q;) L(1; ) e S(Q;) e A(Q;) ; 0: L(1; ) 0: : Thus, for given by (4.3) we get S(3) > 0 and () < :041 under the ERH. The reader may feel that this value for is a less worthy candidate than the others mentioned here because our estimate for L(1; ) is conditional on the ERH. However, it must in all honesty be mentioned that all of the other results on () or L(1; ) here and in [7] are imlicitly deendent either on the truth of the ERH or some heuristic estimation of L(1; ) by the Euler roduct F 1 (Q; ). The fact is that we currently have no algorithm for evaluating the class number h of a uadratic eld of discriminant d (real or imaginary) which is rovably better than O(jdj 1=2+" ) in comlexity. This comlexity measure is far too large to allow for the rigorous comutation of h for the size of jdj that we have to work with here. Acknowledgment. The authors wish to thank the LiDIA Grou [9] and the SIMATH Research Grou [15] in Darmstadt and Saarbrucken, resectively, for roviding software and comuting time. References 1. R. Ayoub, S. Chowla, and H. Walum, On sums involving uadratic characters, J. London Math. Soc. 42 (1967), 152{ E. Bach, Imroved aroximations for Euler roducts, Number Theory: CMS Conference Proceedings, vol. 15, AMS, Providence, R.I., 1995,. 13{ J. Brillhart, D.H. Lehmer, and J. Selfridge, New rimality criteria and factorizations of 2 m 1, Math. Com. 29 (1975), 620{ N.J. Fine, On a uestion of Ayoub, Chowla and Walum concerning character sums, Illinois J. Math. 14 (1970), 88{ M.J. Jacobson, Jr., Comutational techniues in uadratic elds, M.Sc. Thesis, University of Manitoba, P.T. Joshi, The size of L(1; ) for real nonrincial residue characters with rime modulus, J. Number Theory 2 (1970), 58{ D.H. Lehmer, E. Lehmer, and D. Shanks, Integer seuences having rescribed uadratic character, Math. Com. 24 (1970), 433{ H.W. Lenstra, Jr., On the calculation of regulators and class numbers of uadratic elds, London Math. Soc. Lecture Note Series 56 (1982), 123{150.

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