BRANDON HANSON, ROBERT C. VAUGHAN, AND RUIXIANG ZHANG

Transcription

1 THE LEAST NUMBER WITH PRESCRIBED LEGENDRE SYMBOLS arxiv: v2 [math.nt] 29 Oct 2016 BRANDON HANSON, ROBERT C. VAUGHAN, AND RUIXIANG ZHANG Abstract. In this article we estimate the number of integers u to X which can be roerly reresented by a ositive-definite, binary, integral quadratic form of small discriminant. This estimate follows from understanding the vector of signs that arises from comuting the Legendre symbol of small integers n at multile rimes. 1. Introduction x A well-known and outstanding roblem in number theory is the estimation of the least quadratic non-residue modulo a rime. Recall, the least quadratic non-residue modulo is the integer { ( ) } n n = min n > 1 : = 1, ( ) where is the Legendre symbol of x modulo. Vinogradov conjectured that for any ε > 0, n should be at most ε rovided is sufficiently large relative to ε. This conjecture is still oen, however Linnik roved that any excetions to it are sarse. Secifically, he roved the conjecture holds for all but O(loglogN) of the rimes N, with the imlied constant deending only on ε. Now suose 1 and 2 are distinct, odd rimes and n is an integer (( not ) ( divisible by either. There are four ossibilities for the vector n n 1, 2 )). How large must N be so that all four vectors are realizedbyintegersboundedbyn? Ingeneral, onemightaskthefollowing. Problem 1. Given distinct odd rimes 1,..., k, how large must N be before one has seen each of the 2 k different vectors of signs in {1, 1} k realized by a vector of the from (( ) ( )) n n,..., 1 1 k

2 2 BRANDON HANSON, ROBERT C. VAUGHAN, AND RUIXIANG ZHANG with 1 n N? We believe that the above roblem is intrinsically interesting, but there is further motivation studying it. Recall that the quadratic form F(x,y) = Ax 2 +Bxy+Cy 2 has discriminant d = B 2 4AC, is said to reresent q if F(x,y) = q has a solution (x,y) Z 2 with x and y, and is said to roerly reresent q if furthermore x and y can be taken to be relatively rime. We say the form is definite if d < 0. One could then ask, Problem 2. What is the least ositive integer d such that q is reresented by a quadratic form of discriminant d? If we allow for indefinite forms, which is to say forms with ositive discriminant, then this roblem is less interesting. A number is a difference of squares if and only if it is not congruent to 2 mod 4. Thus, either q is not congruent to 2 mod 4 and q = x 2 y 2 has a solution, or else q/2 is not congruent to 2 mod 4 in which case q = 2x 2 2y 2 has a solution. As is outlined below, answering Problem 2 essentially amounts to the solution of Problem 1, because of the following theorem (see for instance [B, Proosition 4.1]). Theorem 1. The number q is roerly reresented by a binary quadratic form of discriminant d if and only if d is a square modulo 4q. When q is an odd rime and q 3 mod 4, Problem 2 boils down to that of finding the least quadratic non-residue modulo ( q. ) But when d handling Problem 2 for general q, rather than have = 1, we q ( ) require that = ( 1) 1 2 for each odd rime dividing q. Thus d we are interested in rescribing the Legendre symbol of d at several rimes, which returns us to Problem 1. One goal of this article is to extend Linnik s result on the least nonresidueandshowthatonecanusuallyrescribethesignofthelegendre symbol simultaneously at many rimes with a small integer, barring some local obstructions as described in the next section.

3 THE LEAST NUMBER WITH PRESCRIBED LEGENDRE SYMBOLS 3 2. Notation, reliminary observations, and statement of results To begin, let k 1 be an integer and 1,..., k N be distinct odd rimes. Let = ( 1,..., k ), q = 1 k and write ( ) (( ) ( )) n n n =,...,. 1 Finally, denote by G = {±1} k the (multilicative) grou of all k-tules with entries ±1. As described in the introduction, we are interested in the number n q = max e G min { n : n 1, k ( ) } n = e, n 1 mod 8. Thus n q is the least ( ositive ) integer such that we observe all ossible sign choices for with integers less than n q and congruent to n 1 mod 8. The congruence condition allows us to further insist that we deal only with squares modulo a ower of two, which is needed in alications. It is convenient to identify the grou G with F k 2 = (Z/2Z)k, the vector sace of dimension k over the field F 2, in the natural way. To be concrete, set U q (y) = {n : 1 n y, (n,q) = 1, n 1 mod 8}. Consider the ma θ q : U q (y) F k 2 given by ( ( ( )) 1 n θ q (n) = 1,..., 1 ( ( ))) n 1 mod Thei thentryofθ q (n)is1 mod 2ifnisaquadraticnon-residuemodulo i and0 mod 2 ifnis aquadratic residue modulo i. The maθ q isalso an additive function in the sense that θ q (mn) = θ q (m)+θ q (n), with the addition oeration belonging to F k 2. Moreover, for y sufficiently large the ma θ q is surjective by Chinese Remainder Theorem and the fact that the rimes i are distinct and odd. Suose we know that the integers in U q (y) san F k 2, in the sense that {θ q (n) : n U q (y)} contains a basis of F k 2. Then n q y k. Indeed, any vector e F k 2 is the sum of at most k basis vectors, each of which is of the form θ q (m) with m y. The roduct of these integers m gives k

4 4 BRANDON HANSON, ROBERT C. VAUGHAN, AND RUIXIANG ZHANG an integer n with θ q (n) = e. It is therefore sensible to consider the number g q = min { y : {θ q (n) : n U q (y)} sans F k 2}. We record the above observation as a lemma. Lemma 1. Let q = 1 k be an odd, square-free integer. Then n q g k q. We have so farreduced the roblem ofbounding n q to that of bounding g q. In the sirit of Linnik, we would like to estimate the number of q for which g q is large. However, we need to be mindful of the following obstruction. If g d > y for some divisor d of q then g q > y as well. Thus, if d is small and g d > y, then g q > y for at least [Q/d] numbers u to Q (the multiles of d), which is substantial. So, we need to restrict our attention to what we will call eligible q, namely those which are not divisible by some d for which g d is large. In fact, for technical reasons, we have need to consider the number g q,r defined as g q,r = min { y : {θ q (n) : n U qr (y)} sans F k 2}. Notice that in this latter definition, we want to generate the full grou of signs with numbers not just corime to q but also to r. Definition. Let y 1 be a arameter. We say q is y-eligible if for each divisor d of q with 1 < d < q, we have g d,q y. Otherwise, we say q is y-ineligible. If g q > y and q is y-eligible, we say q is y-excetional. So, anoddrimeisalways y-eligible, andis y-excetionalif n > y. It also becomes clear why we need to introduce the the notion of g q,r : it may be that g d y for each roer divisor d of q, but in order to get this full set of generators, we must use numbers which are corime to d but not corime to q. We now state our main results. Main Theorem. Let a 3 be fixed. Suose Q(Q,a) is the set of all integers q Q which are odd, square-free, and (logq) a -excetional. Then, for δ > 0, we have Q(Q,a) δ,a Q 2/a+δ.

5 THE LEAST NUMBER WITH PRESCRIBED LEGENDRE SYMBOLS 5 Recall that the square-free radical of q is r = q. Corollary 1. Let ε (0,1). There are ositive numbers a and c which deend only on ε and such that the following holds. For Q sufficiently large in terms of ε, there are at least cq ε (loglogq) 1 numbers q [Q,Q+Q ε ] such that g r (logr) a, where r is the square-free radical of q. As alications of the Main Theorem, we have the following corollaries. Corollary 2. Let ε > 0 and let Q be sufficiently large in terms of ε. There is an integer in the interval [Q,Q + Q ε ] which is roerly reresented by a definite, binary quadratic form of discriminant d with d Q ε. Corollary 3. Let ε > 0 and let Q be sufficiently large in terms of ε. There is an integer q in the interval [Q,Q+Q ε ] which can be written as with 0 u,v Q ε and u 0. q = 1 u x2 + v u y2 This final corollary can be comared with the roblem of bounding the gas between consecutive sums of two squares. Being that there are about x(logx) 1/2 integers u to x which are a sum of two squares, one might exect that the gas between such integers are at most (logx) c for some ositive constant c. However the best known bound, due to Bambah and Chowla [BC], is that there is a integers between x and x+o(x 1/4 ) which is a sum of two squares. Corollary 3 says that one has much smaller gas if we weaken squares to numbers which are in a sense almost-squares. 3. Facts from analytic number theory Here we recall some required background results from analytic number theory. Let S(x,y) = {n x : n = y and = 1 mod 8}

6 6 BRANDON HANSON, ROBERT C. VAUGHAN, AND RUIXIANG ZHANG and for a ositive integer q, let S q (x,y) = {n S(x,y) : (n,q) = 1}. We need to estimate these sets. To begin, we have the following which is a modification of Corollary 7.9 from [MV]. Theorem 2. Suose a is in the range 2 a < (logx) 1/2 /(2loglogx) and let δ > 0. Then for x sufficiently large and q x, x 1 1/a δ δ,a S q (x,(logx) a ) S(x,(logx) a ) δ,a x 1 1/a+δ. Proof. Let y (logx) 2 and let P = { 1,..., T } denote the set of rimes y with = 1 mod 8 and which do not divide q. Any roduct of rimes in P which does not exceed x belongs to S q (x,y). By the Prime Number Theorem in arithmetic rogressions (Corollary in [MV]), we have π(y;8,1) = y logy ( ) 1 4 +o(1). Since ω(n) is maximized when n is a rimorial number ω(q) logq loglogq logx loglogx. Because y (logx) 2, for x sufficiently large, T y. Now consider 5logy any roduct of rimes in P. Its logarithm is of the form T T v j log j logy v j. j=1 Thus a lower bound for S q (x,y) is the number of vectors { T [ logx N = (v 1,...,v T ) Z T : v j 0, v j logy j=1 [ Letting U = logx logy j=1 ] }. ], it is a simle combinatorial argument (see Lemma 7.7 of [MV]) that the number of such vectors is ( ) T +U N =. T By Stirling s Formula, n! n n+1/2 e n

7 so that THE LEAST NUMBER WITH PRESCRIBED LEGENDRE SYMBOLS 7 N (U +T)U+T+1/2 e U T U U+1/2 e U T T+1/2 e ( ) T U U +T 1. U U The right hand side above is increasing in T, thus ( ) U y 1 N 1+. 5U logy U If u = logx logy y then u 1 U u. Thus logy ( y N 5ulogy ) u 1 1 u. For the exonent 1 and the factor 1/ u we note that and thus N 1 y 1 5ulogy u y 1 y ( )logx ( y logy x = 5logx y ex logx ) logy log(5logx) Now we take y = (logx) a where 2 a < (logx) 1/2 /(2loglogx). Then we get ( N x 1 1 a ex aloglogx log5logx ) δ,a x 1 1/a δ. aloglogx Fortheuerbound,trivially S q (x,(logx) a ) islessthan S(x,(logx) a ), which is in turn at most the number of (logx) a -smooth numbers u to x. There are at most O δ,a (x 1 1/a+δ ) such numbers by Corollary 7.1 of [MV]. The main ingredient we need for the roof of our main theorems is a Large Sieve inequality. This one can be found in [IK, Theorem 7.13]. Theorem 3 (Large Sieve). Let Q 1 and let (a n ) n x be a sequence of comlex numbers. Then 2 q a n χ(n) (Q 2 +x) a n 2. ϕ(q) n x q Q χ mod q n x

8 8 BRANDON HANSON, ROBERT C. VAUGHAN, AND RUIXIANG ZHANG In the above sum over χ we mean that the summation occurs over all rimitive characters χ of modulus q. In order to make Lemma 1 useful, we need to have integers with a reasonable number of rime factors. The next lemma tells us such integers are ubiquitous. Lemma 2. Let ε (0,1) and K 1 be fixed. For all Q sufficiently large in terms of ε, there are at most O ( Q ε (loglogq) K) integers q [Q,Q+Q ε ] with at least (loglogq) K+1 rime factors. Proof. Let u [1,Q] be a number to be determined later. Write Then ω(q) ω u (q) = q >u ω u (q) = q u 1 1 logu 1. The number of integers q in question is at most (loglogq) K 1 (loglogq) K 1 Q q Q+Q ε ω(q) q log logq logu. Q q Q+Q ε ω u (q)+2q ε (loglogq) K 1 logq logu. To estimate the sum, Q ε ω u (q) +O(u) Qε loglogq+o(u). Q q Q+Q ε u Taking u = Q ε gives the bound ( ) Q ε O (loglogq) + Q ε K ε(loglogq) K+1 once Q is sufficiently large in terms of ε. ( = O Q ε (loglogq) K ) Finally, in order to guarantee that we can find y-eligible numbers in short intervals, we will need a basic consequence of Brun s Pure Sieve. This result can be read from Corollary 6.2 in [FI].

9 THE LEAST NUMBER WITH PRESCRIBED LEGENDRE SYMBOLS 9 Theorem 4. Let a 1 and ε > 0 be fixed. Then the number of integers in the interval [Q,Q+Q ε ] with no rime divisor less than z is asymtotic to Q ε V (z), where V(z) = <z (1 1 ), rovided z is in the range 1 z (logq) a. 4. Proofs of Main Theorems and Corollaries The following lemma rovides the key reduction to a roblem which is aroachable by the Large Sieve. Lemma 3. Let q be an odd, square-free, and y-excetional number. Then for any n S q (x,y), we have ( ) n = ( ) n = 1. q q Proof. For convenience, write k = ω(q), write q = 1 k, and denote by Θ q (y) = θ q (U q (y)) the image of U q (y) in F k 2. Since q is y-excetional we have g q > y, and so Θ q (y) sans a roer subsace of F k 2, say H. A number n S q(x,y) is a roduct of rimes in U q (y). Thus, writing n = n v, we have θ q (n) = v θ q () H since θ q () H for each occurring in the sum. Since H is a roer subsace, there is a non-zero vector in H. In other words, for some non-emty subset I {1,...,k}, each vector (x 1,...,x k ) H satisfies x i = 0. i I Infact, we must have I = {1,...,k}. To see this, suose thati were a roer subset with I = l < k. Let d I = i I i. Then the rojection π I : H F l 2 given by π I (x 1,...,x k ) = (x i ) i I is not surjective. Indeed, any vector in the image has co-ordinates whichsumto0 mod 2. Butthisrojectioncontainstheimageθ di (U q (n)).

10 10 BRANDON HANSON, ROBERT C. VAUGHAN, AND RUIXIANG ZHANG So the roer divisor d I satisfies g di,q > y, which violates the y- eligibility of q. Thus the each element of H satisfies k i=1 x i = 0 which means that ( ) n = 1 for any n S q (x,y). q We now rove our main theorem. The roof boils down to Linnik s result on the least number n χ for which a rimitive character χ satisfies χ(n χ ) 1. The result is stated but not roved, in [DK]. A roof is given in [Pol1] (Lemma 5.3), and we will will follow in much the same manner. Proof of Main Theorem. Fix δ > 0. Let C a,δ be a constant deending only on a and δ which is at our disosal, and let x > C a. We will work dyadically, and aly Theorem 3 with { 1 if n S(x 2,(logx) a ) a n = 0 otherwise. Suose q Q(Q,a) is in the range x < q 2x. Then S(x 2,(logx) a ) S(x 2,(logq) a ). So if n S(x 2,(logx) a ), then by Lemma 3, ( ) { n 1 if (n,q) = 1 = 0 otherwise. It follows that q n x 2 a n ( ) n S q (x 2,(logx) a ) q Suose x is large enough so that it satisfies 2 a > log(x 2 ) δ/16, which will be guaranteed by increasing C a,δ as necessary. By Theorem 2, S q (x 2,(logx) a ) = S q (x 2,2 a (logx 2 ) a ) S q (x 2,(logx 2 ) a δ/16 ) δ,a x 2 2/(a δ/16) δ/16 δ,a x 2 2/a δ/8

11 THE LEAST NUMBER WITH PRESCRIBED LEGENDRE SYMBOLS 11 byincreasingc a,δ ifnecessary. Nowwesumoverallq Q x = Q(Q,a) (x,2x] to get 2 x 4 4/a δ/4 q δ,a ϕ(q) a n χ(n) q Q x q Q x χ mod q n x 2 Rearranging, we see that δ,a x 2 S ( x 2,(logx 2 ) a) δ,a x 4 2/a+δ/4. Q x δ,a x 2/a+δ/2 Q 2/a+δ/2 for x Q. Summing over all x = 2 j in the range log 2 (C a,δ ) j log 2 Q we get Q(Q,a) δ,a Q 2 j δ,a Q 2/a+δ. j log Q Herewe have usedthat thereisacontributionofo a,δ (1)fromtheterms with 2 q log 2 (C a,δ ) and the fact that logq δ Q δ/2. Before roving the first corollary, we need some lemmas concerning eligibility. Lemma 4. If q is odd, square-free and y-ineligible, and if the least rime dividing q is, then q has a roer divisor d which is min{y, 1}-excetional. In articular, if q is odd, square-free and y-ineligible, and if all rime factors of q exceed y+1, then q has a roer divisor d which is y-excetional. Proof. Since q is y-ineligible it has a divisor d such that g d,q > y. Let d 0 = min{d : d q, 1 < d < q, g d,q > y}. Then d 0 is y-eligible. Indeed, if d 0 were not y-eligible, it would have a roer divisor d for which g d,d 0 > y. But then, since d 0 q, we have g d,q g d,d 0 > y contradicting the minimality of d 0. So d 0 is y-eligible but g d0,q > y. Either g d0 > y as well and so d 0 is y-excetional, or else g d0 y. In the latter case, the vectors θ d0 (n) with n U d0 (y) generate the full grou of signs for d 0, but those with n U q (y) do not. However, all of the elements of U d0 (y) \ U q (y) are divisible by some

12 12 BRANDON HANSON, ROBERT C. VAUGHAN, AND RUIXIANG ZHANG rime which is at least as big as, and so g d0 > 1. It may now be the case, since 1 < y, that d 0 is not 1-eligible. If so, let d 1 = min{d : d d 0, 1 < d < d 0, g d,d0 > 1}. As before, d 1 is 1-eligible and g d,d0 > 1. But in fact, since d 0 is only divisible by rimes greater than, U d0 ( 1) = U d1 ( 1), and g d1 > 1. Hence d 1 is 1-excetional. The second statement of the lemma follows immediately. Lemma 5. Ifq isan odd, square-freeinteger whichis (logq) a -ineligible, and if all rime factors of q are at least 2(logq) a, then q has a divisor d which is (logd) a -excetional. Proof. Since q is not (logq) a -eligible then it has a divisor d 1 which is (logq) a -excetional, bylemma4. Eitherd 1 isalso(logd 1 ) a -excetional (and we are done) or else it is not (logd 1 ) a -eligible. In the latter case it has a roer divisor d 2 which is either (logd 2 ) a -excetional, or else not (logd 2 ) a -eligible. Continuing in this fashion, we must arrive at a divisor dofq which is(logd) a -excetional. Indeed, eventually wewould either sto or arrive at a rime, and this rime is (log) a -excetional since all rimes are y-eligible for all y > 1. Proof of Corollary 1. Assume ε < 1/2 without any loss of generality and let a be a ositive number at our disosal, which will deend only on ε. Let B be the set of all numbers in [Q,Q + Q ε ] which have no rime factors smaller than 2(logQ) a. By Theorem 4, we have Now, by Mertens theorem, logv(z) = <z B Q ε V(2(logQ) a ). log(1 1 ) <z 1 1 = loglogz +O(1). So and hence V(2(logQ) a ) B 1 aloglogq Qε aloglogq.

13 THE LEAST NUMBER WITH PRESCRIBED LEGENDRE SYMBOLS 13 Let B B be the subset of all elements of q B which have a square-free radical r = q satisfying g r > (logr) a. For such q, r is either (logr) a -excetional or (logr) a -ineligible. Now, all rime factors of r exceed 2(logr) a. So, if r is (logr) a -ineligible, it follows from Lemma 5 that r is divisible by a number d which is (logd) a - excetional. In either case, q has a divisor d which is square-free and (logd) a -excetional. Moreover, this divisor d satisfies d (logq) a since it is a non-emty roduct of rimes exceeding (logq) a. Hence every integer in B is divisible by some d Q(2Q,a) which is at least (logq) a. It follows that the size of B is at most d (logq) a d Q(2Q,a) ( ) Q ε 2Q d +O(1) Q ε Q(u,a) du+o(q(2q,a)) (logq) u a 2 by artial summation. We aly our Main Theorem with δ = ε/4. The integral is at most O ε (1) Q ε while (logq) a u 2/a 2+ε/4 du ε,a Q ε (logq) a(1 ε/4) 2 ε,a Q(2Q,a) ε (2Q) 2/a+ε/4 ε Q ε/2 Q ε (logq) a/2, for a sufficiently large in terms of ε. Thus once a is sufficiently large in terms of ε and Q is sufficiently large in terms of a, we have B\B ε Q ε. The integers in B \ loglogq B all have a square-free radical r with g r (logr) a. Proof of Corollary 2. We can assume ε is small without any loss of generality. ByTheorem 1, it isenoughtofindsomenumber q [Q,Q+ Q ε ] and a discriminant d which is a square modulo 4q. If (d,4q) = 1, then in order for d to be a square modulo 4q, it suffices that (1) ( d ) 7 mod 8, d (2) = 1 for q and 1 mod 4, ( ) (3) = 1 for q and 3 mod 4. d By Corollary 1, there is a ositive number a such that the number of integers in q [Q,Q+Q ε ] with g q (logq) a is at least c ε Q ε (loglogq) 1

14 14 BRANDON HANSON, ROBERT C. VAUGHAN, AND RUIXIANG ZHANG for some constant c ε deending only on ε. By Lemma 2, one of these numbers will have ω(q) (loglogq) K for some K sufficiently large in terms of ε. Let 0 be the smallest ositive rime which is congruent to 7 modulo 8 and which does not divide q. Then, since q has k = ω(q) (loglogq) K rime factors, 0 is at most the k+1 thrime congruent to 7 mod 8 which is at most O(klogk) = O ε (Q ε/2 ) by the Prime Number Theorem. Since n q g ω(q) q (logq) a(loglogq)k = O ε (Q ε/2 ), we can find an integer d 0 ε Q ε/2 which is relatively rime to q, ( congruent to 1 modulo 8 so that d 0 0 = 7 mod 8, and such ) that is rescribed as needed for each dividing q. The number d = d 0 0 satisfies the desired roerties and the corollary is roved. Recall that a binary quadratic form q(x,y) = Ax 2 + Bxy + Cy 2 of discriminant d is called reduced if B A C. In this case d = 4AC B 2 3AC, so that all coefficients are bounded by d. We saytwoquadraticformsq 1 (x,y)andq 2 (x,y)areequivalent ifonecanbe obtained from the other by an invertible, integral change of variables. To be recise, we have q 2 (αx+βy,γx+δy) = q 1 (x,y) for some integers α,β,γ,δ with αδ βγ = 1. It is clear that equivalent forms reresent the same numbers. This, combined with the following theorem, [B, Theorem 2.3], shows that there is no loss of generality in working with reduced forms. Theorem 5. Every binary quadratic form of discriminant d is equivalent to a reduced form of the same discriminant. We are now ready to rove our final corollary. Proof of Corollary 3. By Corollary 2, we can reresent an integer q in the range Q q Q + Q ε by some ositive definite binary quadratic form Q of discriminant d with d at most Q ε. Without loss of generality we may assume this form is reduced, and thus write q = Ax 2 +Bxy+Cy 2 with B 2 4AC = d for some A,B,C bounded in absolute value by Q ε. It follows that A,C > 0, and by comleting d 0 0

15 THE LEAST NUMBER WITH PRESCRIBED LEGENDRE SYMBOLS 15 the square we see q = 1 ( (2Ax+By) 2 +dy 2). 4A Acknowledgment We thank the anonymous referee for ointing out errors in the original draft of this article, and for helful comments on the exosition. Part of this work was carried out while the first and third authors were attending the IPAM reunion conference for the rogram Algebraic Techniques for Combinatorial and Comutational Geometry. References [BC] R. P. Bambah and S. Chowla, On numbers which can be exressed as a sum of two squares, Proc. Nat. Inst. Sci. India 13, (1947) [B] D. A. Buell, Binary Quadratic Forms, Classical theory and modern comutations, Sringer-Verlag, New York, x+247. [DK] W. Duke and E. Kowalski, A roblem of Linnik for ellitic curves and mean-value estimates for automorhic reresentations, Invent. Math. 139 (2000) [FI] J. Friedlander and H. Iwaniec, Oera de Cribro, American Mathematical Society Colloquium Publications, 57. American Mathematical Society, Providence, RI, [IK] H. Iwaniec and E. Kowalski, Analytic Number Theory, American Mathematical Society Colloquium Publications, 53. American Mathematical Society, Providence, RI, [MV] H. L. Montgomery and R. C. Vaughan, Multilicative Number Theory 1. Classical Theory, Cambridge, [Pol1] P. Pollack, The average least quadratic nonresidue modulo m and other variations on a theme of Erdös, J. Number Theory 132 (2012),

16 16 BRANDON HANSON, ROBERT C. VAUGHAN, AND RUIXIANG ZHANG Pennsylvania State University, University Park, PA address: Pennsylvania State University, University Park, PA address: Princeton University, Princeton, NJ address: