Differences between powers of a primitive root

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DIFFERENCES BETWEEN POWERS OF A PRIMITIVE ROOT

MARIAN VÂJÂITU and ALEXANDRU ZAHARESCU

Received 14 May 2001

We study the set of differences{gx_−gy_{(modp): 1≤x, y≤N}wherepis a large prime} number,gis a primitive root(modp), andp2/3_{< N < p.}

2000 Mathematics Subject Classification: 11A07.

1. Introduction. Letp be a large prime number andg a primitive root(modp). The distribution of powersgn_{(modp), 1≤n≤N, for a given integerN < phas been} investigated in [1,2,4]. In this paper, we use techniques from [4] to study the set of differences

A:=gx_−gy_{(modp): 1≤x, y≤N. (1.1)}

A natural question, attributed to Andrew Odlyzko, asks for which values ofNcan we be sure that any residueh(modp)belongs to A? He conjectured that one can take

Nto be as small asp1/2+_{, for any fixed >0 andplarge enough in terms of . If} true, this would be essentially best possible sinceAhas at mostN2_{elements. For any} residuea(modp), denote

ν(N,a)=#1≤x, y≤N:gx_−gy_{≡a(modp). (1.2)}

If a≡0(modp)we have the diagonal solutions x=y, thusν(N,0)=N. For a≡

0(modp)it is proved in [4, Theorem 2] that

ν(N,a)=N_p2+Oplog2p. (1.3)

It follows that we can takeN=c0p3/4logpin Odlyzko’s problem, for some absolute constantc0. The exponent 3/4 is a natural barrier in this problem, as well as in other similar ones. An example of another such problem is the following: given a large prime numberp, for which values ofNcan we be sure that any residueh≡0(modp)

belongs to the set{xy(modp): 1≤x, y≤N}? Again we expect thatNcan be taken to be as small asp1/2+_{. As with the other problem, it is known that we can take} N=c1p3/4logp for some absolute constantc1, and this is proved by using Weil’s bounds for Kloosterman sums [5]. If one assumes the well-known H∗ conjecture of Hooley which gives square root cancellation in short exponential sums of the form

fractional parts of the form({n2_α})

n∈N, which would be completely solved precisely

if one could deal with the case whenN=p2/3−_{(see [3] and the references therein).} Returning to the setA, its structure is also relevant to the pair correlation problem for the set{gn_{(modp),1≤n≤N}. Here one wants an asymptotic formula for}

#

1≤x=y≤N:gx_−gy_{≡h(modp), h∈}p NJ

, (1.4)

for any fixed intervalJ⊂R. The pair correlation problem is similar to Odlyzko’s prob-lem, but it is more tractable due to the extra average overh. This problem is solved in [4] forN > p5/7+_{, the result being that the pair correlation is Poissonian asp→ ∞} (here we needN/p→0). It is also proved in [4] that under the assumption of the gen-eralized Riemann hypothesis (for DirichletL-functions) the exponent can be reduced from 5/7+to 2/3+. We mention that by assuming square root type cancellation in certain short character sums with polynomials1≤n≤Nχ(P(n)), the exponent 3/4 in Odlyzko’s problem can be reduced to 2/3+as well. Taking into account the difficulty of the conjectures which would reduce the exponent to 2/3+in all these problems, it might be of interest to have some more modest, but unconditional results, valid in the rangeN > p2/3+_.

Our first objective, in this paper, is to provide a good upper bound for the second moment

M2(N):= a(modp)

ν(N,a)−

N2

p

2

. (1.5)

From (1.3), it follows thatM2(N) p2log4p. The following theorem gives a sharper upper bound forM2(N).

Theorem1.1. For any prime numberp, any primitive rootgmodp, and any posi-tive integerN < p,

M2(N) pNlogp. (1.6)

Since each residueh(modp)which does not belong toAcontributes anN4_/p2_in

M2(N), we obtain the following corollary.

Corollary 1.2. For any prime numberp, any primitive root gmodp, and any positive integerN < p,

#h(modp):h∈A p3logp

N3 . (1.7)

Thus, forN > p2/3+_{, it follows that almost all the residuesa(modp)belong toA.} Although by its nature the inequality (1.6) does not give any indication on where the possible residuesh∈Amight be located, there is a way of obtaining results as in Corollary 1.2, withhrestricted to a smaller set.

Theorem1.3. For any prime numberp, any primitive rootgmodp, and any posi-tive integerN < p,

#1≤h <p:hprime, h(modp)∈A

_p3_logp

N3 1/2

Corollary1.4. For any >0, any prime numberp, and any primitive rootgmodp, almost all the prime numbersh <√p(in the sense that the exceptional set has p1/2− elements) can be represented in the form

h≡gx_−gy_{(modp) (1.9)}

with1≤x,y≤p2/3+_.

Note that a weaker form of Corollary 1.4, with the range 1≤x, y ≤p2/3+ re-placed by the larger range 1≤x,y≤p5/6+_{, follows directly by takingN=p}5/6+_in Corollary 1.2. The point inCorollary 1.4is that it gives a result wherehis restricted to belong to a small set, at no cost of increasing the range 1≤x,y≤p2/3+_.

2. Proof of Theorem1.1. Let p be a prime number, g a primitive root modp, andNa positive integer smaller thanp. We know thata≡0(modp)contributes an

(N−N2_/p)2_{< N}2_inM

2(N). Fora≡0(modp)define a functionhaonZ/(p−1)Z× Z/(p−1)Zby

ha(x,y)=   

1, ifgx_−gy_≡a(modp),

0, else. (2.1)

Thusν(N,a)=1≤x,y≤Nha(x,y). Expandingha in a Fourier series onZ/(p−1)Z× Z/(p−1)Zwe get

ν(N,a)=

r ,s(modp−1) ˆ

ha(r ,s) 1≤x,y≤N

er x+sy p−1

, (2.2)

where the Fourier coefficients are given by

ˆ_{ha(r ,s)=} 1

(p−1)2

x,y(modp−1)

ha(x,y)e

−r x_p+sy −1

. (2.3)

The main contribution in (2.2) comes from the terms withr≡s≡0(modp−1), and this equals ˆha(0,0)N2. It is easy to see that ˆha(0,0)=1/p+O(1/p2). Thus

ν(N,a)=N_p2

1+O

₁

p

+R(a), (2.4)

where

R(a)=

(r ,s)≡(0,0)

ˆ_{ha(r ,s)FN(r )FN(s), (2.5)}

FN(r )= 1≤x≤N

e r x p−1

, FN(s)= 1≤y≤N

e sy p−1

From (2.4) and the definition ofM2(N), it follows that in order to proveTheorem 1.1 it will be enough to show that

p−1

a=1

R(a)2 _pN

logp. (2.7)

From [4, Lemma 7] it follows that

ˆ_{ha(r ,s)=}χs(−1)τ

χr_τχs_τχ−(r+s)

p(p−1)2 χr+s(a), (2.8)

whereτ(χr_{), τ(χ}s_{), τ(χ}−(r+s)_{) are Gauss sums associated with the corresponding} multiplicative charactersχr_,χs_,χ−(r+s)_{defined modp, andχis the unique character} modpwhich corresponds to our primitive rootgby

χgm_=e m p−1

, (2.9)

for any integerm. From (2.5) and (2.8) we derive

R(a)=

m(modp−1)

bmχm(a), (2.10)

where

bm= τ

χ−m p(p−1)2

(r ,s)≡(0,0)(modp−1) r+s=m(modp−1)

FN(r )FN(s)χs(−1)τχrτχs. (2.11)

Since

τχn₌   

_p,

ifn≡0(modp−1),

1, ifn≡0(modp−1), (2.12)

it follows that

_{bm p}−3/2

r+s=m(modp−1)

_{FN(r )FN(s). (2.13)}

HereFN(r )and FN(s)are geometric progressions and can be estimated accurately. We allowr,s, andmto run over the set{−(p−1)/2+1,−(p−1)/2+2,...,(p−1)/2}. Then

FN(r ) min

N, p |r|

, (2.14)

and similarly for|FN(s)|. From (2.13) and (2.14) it follows that

bm p−3/2

r+s≡m(modp−1) |r|,|s|≤(p−1)/2

min

N, p |r|

min

N, p |s|

By Cauchy’s inequality we derive

bm p−3/2  

|r|≤(p−1)/2 min

N2_, p2

|r|2 

 1/2



|s|≤(p−1)/2 min

N2_, p2

|s|2 

 1/2

=p−3/2

|r|≤(p−1)/2 min

N2_,p2

r2

p−1/2_N.

(2.16)

Ignoring the two termsr=0,s=mandr=m,s=0 which contribute in (2.15) at most 2p−3/2_N2_≤2p−1/2_{N, the rest of the sum in (2.15) is less than or equal to}

r+s≡m(modp−1) 0<|r|,|s|≤(p−1)/2

p2

|r||s|=S1+S2, (2.17)

where we denote byS1the sum of the terms with|r| ≤ |s|and byS2the sum of the terms with|r|>|s|. Note that inS1we have|s| ≥ |m|/2 and so

S1

0<|r|≤(p−1)/2

p2

|m||r|

p2_logp

|m| (2.18)

and similarly forS2. From (2.16), (2.17), and (2.18) we conclude that

bm √1_pmin

N,plogp |m|

. (2.19)

We now return to (2.10) and compute

p−1

a=1

R(a)2

= p−1

a=1m1(modp−1) m2(modp−1)

bm1b¯m2χ

m1−m2(a)

=

m1,m2(modp−1) bm1¯bm2

p−1

a=1

χm1−m2(a).

(2.20)

The orthogonality of characters(modp)shows that the last inner sum is zero unless

m1=m2when it equalsp−1, hence

p−1

a=1

R(a)2

=(p−1)

m(modp−1)

bm2. (2.21)

Using (2.19) in (2.21) we obtain

p−1

a=1

R(a)2

|m|≤(p−1)/2 min

N2_,p2log 2_p

|m|2

pNlogp. (2.22)

3. Proof of Theorem1.3. Letp,g, andN be as in the statement of the theorem. We will combine the second moment estimate fromTheorem 1.1with two new ideas. The first idea is to restrict the range ofx,yto 1≤x,y≤N1=[N/2]in the definition ofAin order to increase the number of residues which do not belong to the set. To be precise, we consider the set

A1=

gx_−gy_{(modp): 1≤x, y≤N} 1

, (3.1)

and note that, for any residueh(modp)which does not belong toAand any integer 0≤n≤N1, the residuehg−n _{will not belong to A1. Indeed, if there were integers} x,y∈ {1,2,...,N1}such thatgx−gy≡hg−n(modp), thengx+n−gy+n≡h(modp) which is not the case since 1≤x+n,y+n≤N, andhdoes not belong toA. Therefore, if Ᏼis a set of residues (modp)which do not belong toA, no element of the set ᏹ= {hg−n_{(modp):h∈Ᏼ,0≤n≤N1}will belong toA1. The second idea is captured} in the following lemma.

Lemma 3.1. Let p be a prime number, g a primitive root modp, Ᏼ a set of prime numbers smaller than √p, N1 an integer larger than |Ᏼ|, and denote ᏹ=

{hg−n_{(modp):h∈Ᏼ,0≤n≤N1}. Then}

|ᏹ| ≥|Ᏼ|

|Ᏼ|+1

2 . (3.2)

Proof. The setᏹbecomes larger if one increasesN1thus it is enough to deal with the caseN1= |Ᏼ|. Consider the sets

Ᏼn=hg−n(modp):h∈Ᏼ. (3.3)

Each of these sets has exactly|Ᏼ|elements and we have

ᏹ=

0≤n≤N1

Ᏼn. (3.4)

We claim that for any 1≤n1≠n2≤N1, the intersectionᏴn1∩Ᏼn2 has at most one

element. Indeed, assume that for some distinctn1,n2∈ {1,2,...,N1}, the setᏴn1∩

Ᏼn2 has at least two elements, call them a and b. There are then prime numbers p1,p2,p3,p4∈Ᏼsuch that

a≡p1g−n1≡p2g−n2(modp),

b≡p3g−n1≡p4g−n2(modp).

(3.5)

Note that sincen1≡n2(modp−1)we haveg−n1≡g−n2(modp)hence the numbers

p1 and p2 are distinct. Also, p1 and p3 are distinct because a and b are distinct. We have

ab≡p1p4g−n1−n2≡p2p3g−n1−n2(modp), (3.6)

thus

Now the point is that p1p4and p2p3 are positive integers less thanp, and so the above congruence implies the equalityp1p4=p2p3. Since these four factors are prime numbers,p1coincides with eitherp2orp3, which is not the case. This proves the claim. We now count inᏹall the elements ofᏴ0, all the elements ofᏴ1with possibly one exception if this was already counted inᏴ0, fromᏴ2we count all the elements with at most two exceptions, and so on. Thus

|ᏹ| ≥ |Ᏼ|+|Ᏼ|−1+···+1=|Ᏼ|

|Ᏼ|+1

2 , (3.8)

which proves the lemma.

We now applyLemma 3.1to the setᏴof prime numbers<√p which do not be-long toA, and withN1=[N/2]. It follows that the corresponding setᏹhas at least

|Ᏼ|2_{/2 elements. As we know, none of them belongs toA1. Thus each such element} contributes anN4

1/p2inM2(N1), and combining this withTheorem 1.1we find that

|Ᏼ|2 2

N4 1

p2 ≤M2 _N

1 pN1logp. (3.9)

This implies

|Ᏼ|

_p3_logp

N3 1/2

, (3.10)

which completes the proof ofTheorem 1.3.

References

[1] C. I. Cobeli, S. M. Gonek, and A. Zaharescu,On the distribution of small powers of a primitive root, J. Number Theory88(2001), no. 1, 49–58.

[2] H. L. Montgomery,Distribution of small powers of a primitive root, Advances in Number Theory (Kingston, ON, 1991), Oxford Sci. Publ., Oxford University Press, New York, 1993, pp. 137–149.

[3] Z. Rudnick, P. Sarnak, and A. Zaharescu,The distribution of spacings between the fractional parts ofn2_{α, Invent. Math.145(2001), no. 1, 37–57.}

[4] Z. Rudnick and A. Zaharescu,The distribution of spacings between small powers of a prim-itive root, Israel J. Math.120(2000), 271–287.

[5] A. Weil,On some exponential sums, Proc. Nat. Acad. Sci. U.S.A.34(1948), 204–207.

Marian Vâjâitu: Institute of Mathematics of the Romanian Academy, P.O. Box_1-764, 70700Bucharest, Romania

E-mail address:mvajaitu@stoilow.imar.ro

Alexandru Zaharescu: Institute of Mathematics of the Romanian Academy, P.O. Box 1-764,70700Bucharest, Romania

Current address: Department of Mathematics, University of Illinois at Urbana-Champaign, Altgeld Hall,₁₄₀₉W. Green Street, Urbana, IL₆₁₈₀₁, USA