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arXiv:1011.4492v1 [math.NT] 19 Nov 2010

On the second smallest prime non-residue

Kevin J. McGown1

Department of Mathematics, University of California, San Diego,

9500 Gilman Drive, La Jolla, CA 92093

Abstract

Let χ be a non-principal Dirichlet character modulo a prime p. Let q1< q2

denote the two smallest prime non-residues of χ. We give explicit upper

bounds on q2that improve upon all known results. We also provide a good

upper estimate on the product q1q2which has an upcoming application to

the study of norm-Euclidean Galois fields.

Keywords:

2010 MSC: Primary 11A15, 11N25; Secondary 11A05

Dirichlet character, non-residues, power residues

1. Introduction and Summary

Let χ be a non-principal Dirichlet character modulo a prime p. We call

a positive integer n a non-residue of χ if χ(n) / ∈ {0,1}, and denote by q1<

q2< ··· < qnthe n smallest prime non-residues of χ. The question of putting

an upper bound on q1is a classical problem which goes all the way back to

the study of the least quadratic non-residue.

The literature on this problem is extensive and we will not review it here

except to say that the work of Burgess in the 1960’s significantly advanced

existing knowledge on this matter. Burgess’ famous character sum estimate

(see [1]) implies that qn= O(p1/4+ε) for all n.2For the case of q1, one can

apply the “Vinogradov trick” (see [3, 4, 5]) to Burgess’ result, which gives

the stronger bound of q1= O(p

1

4√e+ε) (see [1]).

Email address: mcgownk@math.oregonstate.edu (Kevin J. McGown)

1Current address: Department of Mathematics, Oregon State University, 368 Kidder

Hall, Corvallis, OR 97331

2The O constant here depends upon ε and n; see [2] for more detail.

Preprint submitted to ElsevierNovember 22, 2010

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p0

107

108

109

1010

1011

1012

1013

C

11.0421

8.2760

7.2906

6.8121

6.5496

6.3964

6.3033

p0

1014

1015

1016

1017

1018

1019

1020

C

6.2452

6.2077

6.1829

6.1659

6.1536

6.1445

6.1374

Table 1: Values of C for various choices of p0

Making these results explicit with constants of a reasonable magnitude

turns out to be difficult, and often times it is results of this nature that one

requires in application. In this paper, we will restrict ourselves to the study

of q1and q2, and we will only be interested in bounds which are completely

explicit and independent of the order of χ.3

The best known explicit bound on q1was given by Norton (see [6]) by

applying Burgess’ method (see [1, 7]) with some modifications.

Theorem 1 (Norton). Suppose that χ is a non-principal Dirichlet charac-

ter modulo a prime p, and that q1is the smallest (prime) non-residue of χ.

Then q1< 4.7p1/4logp, and moreover, the constant can be improved to 3.9

when the order of χ and (p − 1)/2 have a common factor.

We prove the following theorem, which can be viewed as a generalization of

Norton’s result but with a slightly larger constant.

Theorem 2. Fix a real constant p0≥ 107. There exists an explicit constant

C (see Table 1) such that if χ is a non-principal Dirichlet character modulo

a prime p ≥ p0and u is a prime with u ≥ e2logp, then there exists n ∈ Z+

with (n,u) = 1, χ(n) ?= 1, and

n < C p1/4logp.

Provided that q1is not too small, the above theorem immediately gives an

explicit bound on q2.

3In Corollary 3 we do assume that χ has odd order, but we emphasize that none of our

constants depend upon the order of χ.

2

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Corollary 1. Fix a real constant p0≥ 107. Let χ be a non-principal Dirich-

let character modulo a prime p ≥ p0. Suppose that q1 < q2 are the two

smallest prime non-residues of χ. If q1> e2logp, then

q2< C p1/4logp,

where the constant C is the same constant as in the statement of Theorem 2

(see Table 1).

Using a lemma of Hudson and an explicit result of the author on con-

secutive non-residues, we can remove the restriction on q1for a small price.

Corollary 2. Let χ be a non-principal Dirichlet character modulo a prime

p ≥ 1019. Suppose that q1< q2are the two smallest prime non-residues of χ.

Then

q2< 53p1/4(logp)2.

The value q2has not been as extensively studied as q1, and it appears that

prior to now, the best explicit bound was essentially q2 ≤ cp2/5for some

absolute constant c (see [8, 9, 10, 11]). Corollary 2 constitutes an explicit

bound on q2 which even improves slightly on the best known O-bound of

p1/4+ε.

For the application the author has in mind to norm-Euclidean Galois

fields (see [12]), the following corollary is more useful.

Corollary 3. Let χ be a non-principal Dirichlet character modulo a prime

p ≥ 1018having odd order. Suppose that q1< q2are the two smallest prime

non-residues of χ. Then

q1q2< 24p1/2(logp)2.

2. Outline of the Proof

We will establish our results using a generalization of Burgess’ method.

The approach will be similar to a previous paper of the author (see [13]),

but it will be sufficiently different as these results do not follow from the

aforementioned ones or vice versa. The main idea behind Burgess’ method

is to combine upper and lower bounds for the following sum:

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Definition 1. If h,r ∈ Z+and χ is a Dirichlet character modulo p, then we

define

p−1

?

We will use the following lemma, proven in [13], which is a slight improve-

ment on Lemma 2 of [1].

S(χ,h,r) :=

x=0

?????

h

?

m=1

χ(x + m)

?????

2r

.

Lemma 1. Suppose χ is any non-principal Dirichlet character to the prime

modulus p. If r,h ∈ Z+, then

S(χ,h,r) <1

4(4r)rphr+ (2r − 1)p1/2h2r.

Apart from the use of Lemma 1, the proofs of Theorem 2 and Corollary 1

are completely self-contained; in particular, they do not rely on Theorem 1.

However, the derivation of Corollary 2 will use Theorem 1.2 of [13], and the

derivation of Corollary 3 will use Theorem 1 and an explicit version of the

P´ olya–Vinogradov inequality given in [14].

The meat of the proof of our results is to give a lower bound on S(χ,h,r),

under some extra conditions on the involved parameters. In §3 we prove the

following:

Proposition 1. Let h,r,u ∈ Z+with u prime and h ≤ u. Suppose that

χ is a Dirichlet character modulo a prime p ≥ 5 such that χ(n) = 1 for

all n ∈ [1,H] satisfying (n,u) = 1. Assume 2h < H ≤ (2hp)1/2and set

X := H/(2h) > 1. Then

S(χ,h,r) ≥

6

π2(1 − u−1)h(h − 2)2rX2f(X,u).

For each fixed u we have f(X,u) → 1 as X → ∞; the function f(X,u) is

explicitly defined in Lemma 5.

Combining Lemma 1 and Proposition 1 with a careful choice of the pa-

rameters h and r gives our main result from which Theorem 2 follows:

Theorem 3. Suppose that χ is a non-principal Dirichlet character modulo

a prime p ≥ 107, and that u is a prime with u ≥ e2logp. Suppose χ(n) = 1

for all n ∈ [1,H] with (n,u) = 1. If

H ≤ (2e2logp − 2)1/2p1/2,

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then

H ≤ Kg(p)p1/4logp,

where

K =πe

√2≈ 6.0385

and

g(p) =

?

?

?

?

?

?

1 +

4

3logp

?

?

2e2 ,89

?

1 −

1

e2logp

?

f

Kp1/4

? .

The function g(p) is positive and decreasing for p ≥ 107, with g(p) → 1 as

p → ∞. The function f(X,u) is defined in Lemma 5.

The proofs of Theorems 2 and 3 are carried out in §4. Finally in §5 we

derive Corollaries 1, 2, and 3.

3. Proof of Proposition 1

The idea is to locate a large number of disjoint intervals on which χ is

“almost constant.” For the remainder of this section p will denote a prime

with p ≥ 5, and h,H will denote positive integers. The following are the

intervals that will be of interest to us:

Definition 2. For integers with 0 ≤ t < q, we define the intervals

?pt

J(q,t) =

qq

I(q,t) =

q,H + pt

−H + pt

q

?

,

I(q,t)⋆=

?

?pt

q,H + pt

?

q

− h

?

,

?

,−pt

,

J(q,t)⋆=

−H + pt

q

,−pt

q− h

?

.

We note that the intervals I(q,t)⋆, J(q,t)⋆might be empty. In fact,

they are non-empty exactly when h < H/q, which will always be the case

whenever we employ them.

Lemma 2. Let X > 1 be a real number and suppose XH < p. Then the

intervals I(q,t) where 0 ≤ t < q ≤ X with (t,q) = 1 are disjoint, and

similarly for J(q,t).

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Proof. If I(q1,t1) and I(q2,t2) intersect, then we have:

pt1/q1≤ (H + pt2)/q2

pt2/q2≤ (H + pt1)/q1

It follows that

|t1q2− t2q1| ≤XH

p

< 1;

whence t1q2 = t2q1 which implies t1 = t2, q1 = q2. (When t1 = t2 = 0,

the condition (q1,t1) = (q2= t2) = 1 forces q1= q2= 1, so the argument

goes through in this case as well.) The proof for the intervals J(q,t) is the

same. ?

Lemma 3. Let h,u ∈ Z+with u prime and h ≤ u. Suppose that χ is a

Dirichlet character modulo p such that χ(n) = 1 for all n ∈ [1,H] with

(n,u) = 1. If z ∈ I(q,t)⋆∪ J(q,t)⋆and (q,u) = 1, then

?????

H/q. First suppose z ∈ I(q,t)⋆. We will show that the values χ(z + n) for

n = 0,...,h − 1 are all equal except for possibly one value of n. This will

immediately give the result upon application of the triangle inequality.

For n = 0,...,h − 1, we have z + n ∈ I(q,t) and hence q(z + n) − pt ∈

(0,H]. Provided u does not divide q(z + n) − pt, we have

χ(z + n) = χ(q)χ(q(z + n)) = χ(q)χ(q(z + n) − pt) = χ(q).

But if u divides q(z + n) − pt for two distinct values of n, say n1and n2, we

find that u divides q(n1− n2). Since (u,q) = 1, we conclude that u divides

n1− n2and hence |n1− n2| ≥ u. This leads to h ≤ u ≤ |n1− n2| ≤ h − 1, a

contradiction. The proof for z ∈ J(q,t)⋆is similar. ?

Lemma 4. Suppose that X > 1 is a real number and u ∈ Z+is prime. Then

?

(n,u)=1

h−1

?

m=0

χ(z + m)

?????≥ h − 2.

Proof. We note that by hypothesis I(q,t)⋆∪ J(q,t)⋆?= ∅ and hence h <

n≤X

n =(1 − u−1)

2

X2+ θX,uX ,

where the sum is taken over positive integers and θX,udenotes a real number,

depending on X and u, that belongs to the interval (−1,1).

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Proof. For any Y > 0 we have

?

n≤Y

n =⌊Y ⌋(⌊Y ⌋ + 1)

2

.

Upon an application of the obvious inequality Y − 1 < ⌊Y ⌋ ≤ Y , we obtain

the identity

n =Y2

2

?

n≤Y

+Y

2θY,

where θY ∈ (−1,1]. Now we write

?

(n,u)=1

n≤X

n =

?

X2

2(1 − u−1) +X

n≤X

n − u

?

n≤X/u

n

=

2(θX− θX/u),

and observe that

−2 < θX− θX/u< 2.

The result follows. ?

Lemma 5. Suppose X > 1 and u ∈ Z+is prime. Then

?

(q,u)=1

1≤q≤X

φ(q) ≥

3

π2(1 − u−1)X2f(X,u),

where

f(X,u) = 1 −π2

3

?

1

2X2+

1

2X+

1

1 − u−1·1 + logX

X

?

.

Proof. First we observe:

?

(q,u)=1

1≤q≤X

φ(q) =

?

(q,u)=1

1≤q≤X

?

m|q

q

mµ(m)

=

?

(m,u)=1

1≤m≤X

µ(m)

?

(r,u)=1

1≤r≤X/m

r

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Applying Lemma 4 to the above gives:

?

(q,u)=1

1≤q≤X

φ(q) =

X2

2

?1 − u−1?

?

(m,u)=1

1≤m≤X

µ(m)

m2

+ X

?

(m,u)=1

1≤m≤X

µ(m)

m

θX/m,u

Now we use the bounds:

?

(m,u)=1

1≤m≤X

µ(m)

m2

≥

6

π2−

1

X2−1

X,

?

(m,u)=1

1≤m≤X

µ(m)

m

θX/m,u ≤

?

1≤m≤X

1

m≤ 1 + logX

The result follows from an application of the triangle inequality and some

rearrangement. ?

Proof of Proposition 1. We begin by noting that H/q ≥ H/X = 2h.

Using Lemma 2 and Lemma 3 we have:

S(χ,h,r) =

p−1

?

x=0

?????

h−1

?

m=0

χ(x + m)

?????

2r

≥

?

0≤t<q≤X

(q,u)=(q,t)=1

?

z∈I⋆

q,t∪J⋆

q,t

?????

h−1

?

m=0

χ(z + m)

?????

2r

≥

?

(q,tu)=1

0≤t<q≤X

2

?H

q− h

?

(h − 2)2r

≥

?

(q,tu)=1

0≤t<q≤X

2h(h − 2)2r

= 2h(h − 2)2r

?

(q,u)=1

1≤q≤X

φ(q)

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Now the result follows from Lemma 5. ?

4. Proofs of the Theorems

Before launching the proof of Theorem 3, we establish the following simple

convexity result:

Lemma 6. Suppose h,r ≥ 1. We have the following implications:

1

2h

h ≥ 6r + 5 =⇒

?

h

4r

h − 2

?r

?r

<7

≤

1

h + 1

?

4r

h + 1

?r

h ≥ 16r + 2 =⇒

?

2r − 1

h

h − 26

h ≥ 2r − 1 =⇒

≤

2r

h + 1

Proof. By the convexity of the logarithm, we have logt ≥ (2log2)(t−1) for

all t ∈ [1/2,1]. Applying this, together with the hypothesis that 6(r + 1) ≤

h + 1, we get

?h − 2

This yields

1

2≤

h + 1

log

h + 1

?

≥ −6log2

h + 1≥ −log2

r + 1.

?h − 2

?r+1

,

and first implication follows. For the proof of the second implication, we

observe (again by convexity) that logt ≤ t − 1 for all t and hence

?

this leads to

?

The third implication is trivial. ?

rlog

h

h − 2

?

≤

2r

h − 2≤1

8;

h

h − 2

?r

≤ exp

?1

8

?

<7

6.

Proof of Theorem 3. First, we may assume H ≥ Kp1/4logp, or else

there is nothing to prove. We set h = ⌊Alogp⌋, r = ⌊B logp⌋ with A = e2,

B = 1/4 and verify that r,h satisfy all three conditions in Lemma 6. The

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constants A and B were chosen to minimize the quantity AB subject to the

constraint A ≥ 4B exp(1/(2B)).

One verifies that Kp1/4> 28e2for p ≥ 107and hence H > 28h. We set

X := H/(2h) and observe that we have the a priori lower bound

X =H

2h≥Kp1/4logp

2e2logp

=Kp1/4

2e2

,

and, in particular, X > 14 from the previous sentence. Since p ≥ 105and

e2log(105) ≈ 85.1, we know u ≥ 89 and hence f(X,u) ≥ f(X,89). For

notational convenience, we will write f(X) := f(X,89).

Combining Lemma 1 and Proposition 1, we obtain

6

π2

?1 − u−1?h(h − 2)2r

Rearranging the above and applying Lemma 6 gives

?H

2h

?2

f(X) ≤

1

4(4r)rphr+ (2r − 1)p1/2h2r.

6

π2

?1 − u−1?H2f(X)

≤ 4h2p1/2

?

1

4h

?

4r

h − 2

?

?r?

h

h − 2

?r

?r

p1/2+2r − 1

h

?

h

h − 2

?2r?

≤ 4h2p1/2

?

1

h + 1

4r

h + 1

p1/2+

3r

h + 1

?

. (1)

Plugging in our choices of r,h and using the fact that

A ≥ 4B exp

?1

2B

?

=⇒

?4B

A

?r

≤ p−1/2

we obtain

6

π2

?1 − u−1?H2f(X) ≤ 4A2(logp)2p1/2

?

?

1

Alogp

1

Alogp+3B

?

?4B

A

?r

?

p1/2+3B

A

?

≤ 4A2p1/2(logp)2

A

= 12ABp1/2(logp)2

1 +

1

3B logp

?

. (2)

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Plugging in our choices of A and B yields:4

6

π2

?1 − u−1?H2f(X) ≤ 3e2p1/2(logp)2

As f(X) is increasing and positive for X ≥ 14, the result now follows upon

solving (3) for H. ?

?

1 +

4

3logp

?

(3)

Proof of Theorem 2. Suppose p ≥ 107. Let n0denote the smallest n ∈ Z+

such that (n,u) = 1 and χ(n) ?= 1. Set H := n0− 1 so that χ(n) = 1 for all

n ∈ [1,H] with (n,u) = 1.

First we show that H ≤ (2e2logp − 2)1/2p1/2. By way of contradiction,

suppose H > (2e2logp − 2)1/2p1/2. In this case we set H0= ⌊(2e2logp −

2)1/2p1/2⌋, and note that we still have χ(n) = 1 for all n ∈ [1,H0] with

(n,u) = 1 for this smaller value H0. We invoke Theorem 3 to conclude that

H0< Kg(p)p1/4logp where Kg(p) ≤ Kg(107) < 12. Using again the fact

that p ≥ 107, we have

H0< 12p1/4logp < (2e2logp − 2)1/2p1/2− 1 < H0,

which is a contradiction. This proves that H ≤ (2e2logp − 2)1/2p1/2.

Having shown that H satisfies the required condition, we apply Theorem 3

to find H ≤ Kg(p0)p1/4logp when p ≥ p0≥ 107. Therefore

n0≤ Kg(p0)p1/4logp + 1,

for p ≥ p0≥ 107. Computation of the table of constants is routine; for each

value of p0, we compute (being careful to round up) the quantity

Kg(p0) +

1

p1/4

0

logp0

. ?

5. Proofs of the Corollaries

Proof of Corollary 1. Apply Theorem 2 with u = q1and observe that the

smallest n ∈ Z+with (n,q1) = 1 and χ(n) ?= 1 is equal to q2. ?

4At this point our choices of A and B are properly motivated – the condition

A ≥ 4B exp(1/(2B)) was to ensure that the quantity in the square brackets of (1) re-

mains bounded as p → ∞, and we wanted to minimize AB so that the constant appearing

in (2) was as small as possible.

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The following is a lemma due to Hudson (see [11]) that will allow us to

prove Corollary 2. The proof is brief and so we include it for the sake of

completeness.

Lemma 7 (Hudson). Let χ be a non-principal Dirichlet character modulo

a prime p ≥ 5. Suppose that q1< q2are the two smallest prime non-residues

of χ, and that q1 ?= 2 or q2 ?= 3. Let S denote the maximal number of

consecutive integers for which χ takes the same value. Then q2≤ Sq1+ 1.

Proof. Let t ∈ Z+be maximal such that 1 + tq1 < q2. (This is always

possible unless q1= 2 and q2= 3.) Then the t + 1 integers

1, 1 + q1, ..., 1 + tq1

(4)

are residues with respect to χ. Let x be denote the unique inverse of q1

modulo p in the interval (0,p). Multiplying (4) by x allows us to see that

the t + 1 consecutive integers

x, x + 1, ..., x + t

take on the same character value; hence t + 1 ≤ S. By the maximality of t,

we conclude that q2≤ (t + 1)q1+ 1 ≤ Sq1+ 1. ?

We note that the above Lemma can be improved if χ(−1) = 1 (see [11])

but we will not require this. The other result we we use in the proof of

Corollary 2 is the following, which is a special case of Theorem 1.2 of [13].

Theorem 4. If χ is any non-principal Dirichlet character to the prime mod-

ulus p ≥ 1019which is constant on (N,N + H], then H < 7.1p1/4logp.

Proof of Corollary 2. If q1 > e2logp, then we apply Corollary 1 and

we are done. Hence we may assume that q1≤ e2logp. If q2= 3, then we

are clearly done, so we may also assume q2?= 3. In this case, we combine

Lemma 7 and Theorem 4 to conclude that q2≤ (7.1p1/4logp)(e2logp)+1 <

53p1/4(logp)2. ?

In order to prove Corollary 3, we will use the following result which gives

a weak bound on q2, but requires no extra hypotheses on q1.

Lemma 8. Let χ be a non-principal Dirichlet character modulo m ≥ 1015.

Suppose that q1< q2are the two smallest prime non-residues of χ. Then

q2< 2m1/2logm.

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Page 13

Proof. Using the explicit version of the P´ olya–Vinogradov inequality proven

in [14], we find

?

(n,q1)=1

n<x

χ(n)

=

??????

?????

?

n<x

χ(n) − χ(q1)

?

n<x/q1

χ(n)

??????

≤

?

?

n<x

χ(n)

?????+

??????

?

n<x/q1

χ(n)

??????

≤ 2

1

3log3m1/2logm + 6.5m1/2

?

.

If χ(n) = 1 for all n ≤ x with (n,q1) = 1, then

?

(n,q1)=1

n<x

χ(n) ≥ (1 − q−1

1)x − 1.

Thus for 1 < x < q2, we have

(1 − q−1

1)x − 1 ≤ 2

?

1

3log3m1/2logm + 6.5m1/2

?

.

Using the fact that q1≥ 2 and letting x approach q2from the left, we obtain

?

and the result follows. ?

q2≤ 4

1

3log3m1/2logm + 6.5m1/2

?

+ 2,

Proof of Corollary 3. If q1< e2logp, we use Lemma 8 to obtain q2< 2p1/2logp

and hence q1q2< 2e2p1/2(logp)2< 15p1/2(logp)2. If q1≥ e2logp, then we

apply Theorem 1 (using the fact that χ has odd order) and Corollary 1 to

find q1q2≤ C′p1/2(logp)2with C′= (3.9)(6.1536) < 24. ?

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13

Page 14

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