THE LEAST PRIME QUADRATIC NONRESIDUE IN A PRESCRIBED RESIDUE CLASS MOD 4

Transcription

1 THE LEAST PRIME QUADRATIC NONRESIDUE IN A PRESCRIBED RESIDUE CLASS MOD 4 PAUL POLLACK Abstract For all rimes 5, there is a rime quadratic nonresidue q < with q 3 (mod 4 For all rimes 3, there is a rime quadratic nonresidue q < with q (mod 4 Introduction Let be a rime It is easy to see that there is always a rime in the interval [, ] that is a quadratic nonresidue modulo Indeed, since a roduct of squares is a square, the least quadratic nonresidue modulo is necessarily rime Since there is some quadratic nonresidue less than (in fact, there are of them, the claim follows With a bit more work, one can show that whenever > 3 there is a rime quadratic nonresidue less than (This statement, accomanied by a clever elementary roof, aears as Lemma 58 in [9] In this note, we show that both of the corime residue classes modulo 4 contain quadratic nonresidues smaller than once 3 Theorem For every rime 5, there is a rime quadratic nonresidue q 3 (mod 4 with q < Theorem For every rime 3, there is a rime quadratic nonresidue q (mod 4 with q < In both of these theorems, the lower bound on is easily seen to be shar These theorems are comlementary to those of Gica [6], who roved the analogous results for rime quadratic residues (by methods different from those used here Secifically, if 9, then there is a rime quadratic residue q < with q 3 (mod 4, while if 4, then there is a rime quadratic residue q < with q (mod 4 Again, the bounds are shar Theorems and are rimarily of interest as examles of numerically exlicit results; from an asymtotic ersective, sharer results are already available For r = or 3, let n r ( denote the least rime quadratic nonresidue modulo lying in the rogression r modulo 4 One can deduce from the machinery of [] that n 3 ( 4 +o(, as The same methods give n ( +o(, but in fact Friedlander showed already in 973 that n ( e +o( [4] In view of the results at infinity just described, one might guess that the roofs of Theorems and should be entirely routine This is not so A major obstacle is that the estimate on n 3 ( quoted above is ineffective; the method of [] does not allow one to determine, even in rincile, an exlicit 0 for which n 3 ( < once > 0 As regards n (, the methods of [] and [4] are effective But the roof of Theorem still resents challenges For examle, the Burgess-tye character sum estimates in Z[i] roved in [4] have not (as far as I know ever been made numerically exlicit It would not be difficult to use the method of [], in conjunction with an exlicit version of the 00 Mathematics Subject Classification A5 (rimary, E6, L40 (secondary Key words and hrases quadratic nonresidue, genus theory, binary quadratic forms, Pólya Vinogradov inequality

2 PAUL POLLACK classical Burgess bound (eg, that roved in [3], to find some value of 0 such that n ( < for all > 0 However, 0 would be large enough that it would be far from routine to check Theorem in the remaining range [3, 0 ] Our aroach to these roblems is as follows: We avoid character sum analysis altogether in the roof of Theorem Instead, we aeal to classical results on binary quadratic forms (all of which were known already to Gauss The roof of Theorem goes through character sum estimates, but the Burgess bounds are eschewed in favor of the Pólya Vinogradov inequality, for which we have exlicit versions with remarkably agreeable leading terms Notation The letters, q, and l are reserved for rime variables unless otherwise noted It is convenient to adot the following modified O-notation: f = O (g indicates that f g for all values of the arguments under consideration Proof of Theorem We first disose of those cases when 3 (mod 4 Since 5 is assumed, we in fact have 7, so that 4 3 Since 4 3 (mod 4, there is a rime q 4 with q 3 (mod 4 By quadratic recirocity, ( ( ( q 4 = = =, q q and so q is our desired rime To rove Theorem when (mod 4, we aeal to the reduction theory and genus theory of binary quadratic forms; useful references for this material are the books of Cox [] and Flath [3] For rimes (mod 4, the genus characters associated to the discriminant 4 are the Legendre symbol ( and χ, the nontrivial Dirichlet character modulo 4 By Gauss s theorem on the existence of genera (cf Theorem 36(ii on [3, 58], there is a rimitive, ositive definite binary quadratic form F of discriminant 4 such that for any integer a corime to 4 reresented by F, ( a ( = and χ(a = (For the discriminant 4, there are exactly two genera, and the condition ( characterizes the forms F in the nonrincial genus Alying Gauss Lagrange reduction, we can assume F (x, y = Ax + Bxy + Cy, where B A C Since B 4AC = 4, the integer B is even Moreover, A > : Indeed, if A =, then B = 0, and C =, so that F is the rincial form X + Y But F is not in the rincial genus! Thus, A Now Since C = B 4 A + A 4 A + A 4 = 4AC B 4A A = 3A, we have that A 4/3 The exression 4 t + t is decreasing as a function of the real variable t for t [, 4/3], and we conclude that ( C 4 t + = t t= + As A C, the same uer bound holds for A (This uer bound also follows from A 4/3 In articular, neither A nor C is a multile of Since B is even and Ax + Bxy + Cy is rimitive, at least one of A or C is odd Pick a integer from {A, C}, say a Then a is rime to 4, and a is reresented by

3 THE LEAST PRIME QUADRATIC NONRESIDUE 3 F Thus, ( holds Since χ(a =, we have a 3 (mod 4, and so there is a rime q a with q 3 (mod 4 Then 4 = B 4ac B (mod q, so that ( ( ( ( ( 4 4 q = = = = q q q q Hence, ( q = Since q a + <, the rime q has the roerties claimed in Theorem Remarks (i In the (easier case 3 (mod 4, sharer results have been obtained by Nagell [0]: If 3 (mod 8 and > 3, one can find a q 3 (mod 4 with ( q = and q < + When 7 (mod 8, one can find such a q < (ii In the case (mod 4, our roof shows that there is a nonresidue q 3 (mod 4 with q + One can do a bit better for large while maintaining effectivity! as follows Let h be the number of classes of rimitive, ositive definite forms of discriminant 4, so that each of the two corresonding genera is comrised of h inequivalent forms By a straightforward counting argument, we can show that each genus contains a reduced form Ax + Bxy + Cy with A h/ log h: If Ax + Bxy + Cy is a reduced form of discriminant 4, then B 4 (mod A Given A, the number of B satisfying this congruence with B A is O(d(A, where d( is the number-of-divisors function Moreover, since B 4AC = 4, the integers B and A determine C Thus, the number of ossible forms with A T is A T d(a T log T Now taking T as the maximum value of A for all reduced forms in the genus, we deduce that T log T h, so that T h/ log h Choose a form G(x, y = Ax + Bxy + Cy in the nontrivial genus with A h/ log h Since B A 4/3, we have AC = (B + 4/4 4/3, and so A C 4 3A log h h Following our roof of Theorem reveals that there is a rime quadratic nonresidue q 3 (mod 4 with q max{a, C} log h/h The lower bound h ɛ (log ɛ of Goldfeld Gross Zagier (see [8, Chater 3] thus yields q ɛ /(log ɛ, where the imlied constant is effectively comutable for each ɛ > 0 3 Proof of Theorem We continue to use χ to denote the nontrivial Dirichlet character modulo 4 Let r(n := de=n χ(d, so that r(n is multilicative in n and r(n = 4 #{(x, y Z : x + y = n} (see [7, Theorem 78, 34] We begin with an estimate for the artial sums of r(n over odd n Lemma 3 For all real x, n x r(n = x π 4 + O ( x + 4

4 4 PAUL POLLACK Proof Inserting the definition of r(n and interchanging the order of summation, we find that n x, r(n = de x, d, χ(d Alying Dirichlet s hyerbola method, χ(d = χ(d + χ(d χ(d de x d, e x/d = + 3, e x d x/e d,e x d, say Since χ is suorted o integers, and since the artial sums of χ( are all 0 or, ( + x ; similarly, (3 e x = 3 χ(d + x e x We turn now to Noting that the number of odd natural numbers not exceeding t, for real t, is t/ + O (/, we have that = χ(d e x/d = = x Since d χ(d/d = π/4, we also have that = π 4 χ(d d d> x χ(d d ( x χ(d d + O χ(d d + O ( ( + x 4 = π 4 + O ( x, using in the last ste that the nonzero terms χ(d/d are alternating in sign and decreasing in absolute value Substituting this estimate above reveals that (4 = x π ( + 3 x 4 + O 4 Combining (, (3, and (4 yields the lemma Remark As ointed out by the referee, the O term could be imroved to 5 4 x by taking account of the fact that and 3 are nonnegative Motivated by Lemma 3, we let RSUM + (x = x π x + 4 4, and RSUM (x = x π x 4 4 To roceed, we also require an estimate for the artial sums of r(n when weighted by the Legendre symbol ( n We derive a suitable bound from the following exlicit version of the Pólya Vinogradov inequality due to Frolenkov and Soundararajan (see [5, Theorem ] Proosition 4 Let χ be a rimitive Dirichlet character modulo m, where m 00 For all M Z and N Z +, M+N χ(n PV(m, n=m+

5 THE LEAST PRIME QUADRATIC NONRESIDUE 5 where PV(m := π m log m + m In fact, when χ( =, one can relace by the smaller constant π π We will not need this, however It is useful to observe that the estimate of Proosition 4 continues to hold with the sum on n restricted to odd values When m is even, this claim is trivial, since in that case χ is suorted o n If m is odd, write for an inverse of modulo m, and note that χ(j + = χ(χ(j + ; thus, a sum over odd n has the same absolute value as an unrestricted sum on j taken over a different interval Since the bound of Proosition 4 alies to sums over arbitrary intervals [M +, M + N], our assertion follows Lemma 5 Let be a rime, and let M Then ( n r(n WRSUM + (, M, n M where WRSUM + (, M := + M de M d, (PV( + PV(4 + PV( PV(4 Proof Inserting the definition of r(n and interchanging the order of summation, we find that ( n ( (d ( n M, r(n = de M, d, χ(d e By the hyerbola method, (( ( d e χ(d = +, 3 where now = d M = e M 3 = d,e M d, ( d χ(d ( e d M/e ( d χ(d e M/d ( d ( e, χ(d, ( e Now ( is a rimitive character of conductor while ( χ( is a rimitive character of conductor 4 Alying Proosition 4 to the inner sums in and yields + M PV(, and + M PV(4 Writing 3 as a roduct of two sums, we see that 3 PV( PV(4 The lemma follows from combining these estimates for,, and 3 We now rove Theorem Proof A straightforward comutation with PARI/GP shows that for every rime with 3 < 3 0, there is a rime q (mod 4 with q < and ( q = (The

6 6 PAUL POLLACK comutation took roughly 6 hours running under gc on a Core i5-600u machine So it is enough to rove the theorem under the assumtion that 3 0 We take Using our lower bound on, we find that M := 60 (log RSUM (M 78 WRSUM + (, M (We used Mathematica to comute the minimum of RSUM (M/WRSUM + (, M for 3 0 So by Lemmas 3 and 5, ( n r(n r(n 78 n M Hence, n M ( r(n n M ( n n M Considering the ossibilities for ( n, we deduce that (5 r(n 34 r(n 78 n M, ( n = n M r(n n M, n r(n The sum being subtracted on the right-hand side of (5 is quite small Indeed, since M < for the values of under consideration, n M, n r(n = r( m M m odd (We used here that r( = 0 or In our range of, r(m RSUM + (M RSUM + (M 0000 RSUM (M (Indeed, this holds already for 0 4 Referring back to (5, and noting that 34/78 /0000 > 04357, we find that (6 r(n r(n n M, ( n = n M To finish things off, we suose for a contradiction that ( q = for all rimes q < with q (mod 4 Under this assumtion, if r(n > 0 for a number n M with ( n =, then n must be divisible by some rime q (mod 4 with q > Moreover, writing n = qk, we have r(n = r(k Therefore, r(n r(n (7 n M, ( n = We claim that for every real t 6, <q M n M, q (mod 4 q n = <q M k M/q q (mod 4 k odd r(k t k t k odd r(k

7 THE LEAST PRIME QUADRATIC NONRESIDUE 7 It clearly suffices to verify this for integers t For t 300, it follows from Lemma 3, since RSUM + (t/t / in that range; for 6 t < 300, it is verified directly with a short comuter rogram Using this estimate to bound the inner sum in (7, n M, ( n = r(n M/5<q M q (mod 4 + M/6<q M/5 q (mod M <q M/6 q (mod 4 To bound the sums on q, we use that for ɛ := /400 and all t 0 0, log q = t( + O (ɛ; q t q (mod 4 q this is a result of Ramaré and Rumely [] (see the entry for k = 4, x 0 = 0 0 in their Table Thus, M/5<q M q (mod 4 Similarly, Moreover, M/6<q M/5 q (mod 4 <q M/6 q (mod 4 this is at most M/6 q = = t log t log (M/5 M/5<q M q (mod 4 log q M( + ɛ (M/5( ɛ = log(m/5 log (M/6 M/6<q M/5 q (mod 4 log q (M/5( + ɛ (M/6( ɛ log(m/6 ( t log t d q t q (mod 4 q t q (mod 4 M/6 log q M/6 ( + ɛ log(m/6 ( ɛ log + ( + ɛ which in turn equals ( + 3 ɛ M/5 log(m/5 M/60 = ( + ɛ log(m/6 q t q (mod 4 M/6 ( log q d t log t ; ( t log t + t log(t dt, ( + ɛ log(m/6 ( ɛ log + ( ( + ɛ log log(m/6 log log( + log log(m/6 Putting these estimates together, we see that (8 r(n S(, n M, ( n =

8 8 PAUL POLLACK where S( := ( + 3 ɛ M/5 M/0 + ( + ɛ log(m/5 log(m/6 + ( + ɛ M/ log(m/6 ( ɛm/ log + ( 4 M( + ɛ log log(m/6 log log( + log log(m/6 Comaring (8 and (6, we see that (9 S( 0785 RSUM (M Again using Mathematica, we see this inequality fails for 3 0 Remark for the sketical All of the inequalities asserted above can be verified with a finite number of high-recision calculations In articular, there is no need to trust the outut of Mathematica s Minimize or NMinimize functions The details are not articularly insiring, so we discuss how this goes only for the key inequality (9 (meaning that we exlain how to rigorously refute (9 Simlifying, S( M = ( + 3 ɛ /5 log(m/5 +(+ɛ /0 / +ɛ log(m/6 log + ( (+ɛ log + log(m/6 4 log In our range of, all four summands are easily checked to be decreasing as functions of the real variable, so that S(/M is decreasing On the other hand, RSUM (M M = π M 5 4M, which is ositive and increasing in our range of Thus, S( RSUM (M = S( M M RSUM (M is a roduct of ositive, decreasing functions, so is decreasing At = 3 0, It follows that (9 fails for all 3 0 S( RSUM (M < 07 4 Concluding thoughts: Do we need character sums? It is natural to wonder if the intricate analysis above is necessary to rove Theorem, esecially as certain secial cases of that theorem do admit algebraic treatments For examle, let 7 (mod 8 From work of Dickson [], we can write = x + y + z for some x, y, z Z (In fact, Dickson shows that x + y + z reresents all integers not of the form 4 k (6m + 4 Since a sum of two squares cannot be 7 mod 8, it must be that z is odd, so that x + y = z 5 (mod 8 Any rimes that aear to a ower in a sum of two squares are mod 4, and so are or 5 mod 8 Since x + y 5 (mod 8, it follows that there is some rime q 5 (mod 8 that divides x + y Then ( q = ( q ( z = = q ( q ( ( z z = = 0 or q q

9 THE LEAST PRIME QUADRATIC NONRESIDUE 9 But q z <, so ( q 0, and hence ( q = So q is our desired rime (Comare with the roof of [6, Theorem ] This argument can be extended somewhat, but the author has not so far succeeded in establishing all of Theorem this way Acknowledgments Research of the author is suorted by NSF award DMS-4068 He thanks David Jensen for suggesting the roblem and Enrique Treviño for useful conversations He also thanks the referee for a careful reading of the manuscrit References [] D A Cox, Primes of the form x + ny, second ed, Pure and Alied Mathematics (Hoboken, John Wiley & Sons, Inc, Hoboken, NJ, 03 [] L E Dickson, Integers reresented by ositive ternary quadratic forms, Bull Amer Math Soc 33 (97, [3] D E Flath, Introduction to number theory, A Wiley-Interscience Publication, John Wiley & Sons, Inc, New York, 989 [4] J B Friedlander, On characters and olynomials, Acta Arith 5 (973/74, 3 37 [5] D A Frolenkov and K Soundararajan, A generalization of the Pólya-Vinogradov inequality, Ramanujan J 3 (03, 7 79 [6] A Gica, Quadratic residues of certain tyes, Rocky Mountain J Math 36 (006, [7] G H Hardy and E M Wright, An introduction to the theory of numbers, sixth ed, Oxford University Press, Oxford, 008 [8] H Iwaniec and E Kowalski, Analytic number theory, American Mathematical Society Colloquium Publications, vol 53, American Mathematical Society, Providence, RI, 004 [9] M Köcher, Elementare Abschätzungen für rime quadratische Reste und Nichtreste, PhD thesis, Universität Tübingen, 00, available online at htts://ublikationenuni-tuebingende/ xmlui/handle/0900/4844 [0] T Nagell, Sur les restes et les non-restes quadratiques suivant un module remier, Ark Mat (950, [] P Pollack, Prime slitting in abelian number fields and linear combinations of Dirichlet characters, Int J Number Theory 0 (04, [] O Ramaré and R Rumely, Primes in arithmetic rogressions, Math Com 65 (996, no 3, [3] E Treviño, The Burgess inequality and the least kth ower non-residue, Int J Number Theory (05, Deartment of Mathematics, University of Georgia, Athens, GA address: ollack@ugaedu