When the number of divisors is a quadratic residue

WHEN THE NUMBER OF DIVISORS IS A QUADRATIC RESIDUE

1

WHEN THE NUMBER OF DIVISORS IS A QUADRATIC RESIDUE ` OLIVIER BORDELLES   Abstract. Let q > 2 be a prime number and define λq := τq where τ (n) is the number of   divisors of n and q· is the Legendre symbol. When τ (n) is a quadratic residue modulo q, then (λq ⋆ 1) (n) could be close to the number of divisors of n. This is the aim of this work to compare the mean value of the function λq ⋆ 1 to the well average order of τ . The proof reveals   known

arXiv:1701.02286v1 [math.NT] 9 Jan 2017

that the results depend heavily on the value of q2 . A bound for short sums in the case q = 5 is also given, using profound results from the theory of integer points close to certain smooth curves.

1. Introduction and main result Ω

If λ = (−1) is the Liouville function, then L(s, λ) = This implies the convolution identity X

n6x

ζ(2s) ζ(s)

(σ > 1) .

k j (λ ⋆ 1) (n) = x1/2 .


τ

Define λ3 := 3 where where τ (n) is the number of divisors of n and symbol modulo 3. Then from Proposition 3 below L(s, λ3 ) =

ζ(3s) ζ(s)

· 3




is the Legendre

(σ > 1)

implying the convolution identity X

n6x

k j (λ3 ⋆ 1) (n) = x1/3 .

Now let q > 2 be a prime number and define λq := modulo q. Our main aim is to investigate the sum X (λq ⋆ 1) (n).

  τ q

where

  · q

is the Legendre symbol

n6x

When τ (n) is a quadratic residue modulo q, one may wonder if (λq ⋆ 1) (n) has a high probability to be equal to the number of divisors of n. It then could be interesting to study its average order and to compare it to that of τ , i.e. X
(1) τ (n) = x (log x + 2γ − 1) + O xθ+ε n6x

where 14 6 θ 6 131 416 , the left-hand side being established by Hardy [5], the right-hand side being the best estimate to date due to Huxley [6]. The main result of this paper can be stated as follows. Theorem 1. Let q > 3 be a prime number. ⊲

If q ≡ ±1 (mod 8)

X

n6x

    Pq′ ζ′ (1) + Oq,ε xmax(1/cq ,θ)+ε (λq ⋆ 1) (n) = xζ(q)Pq (1) log x + 2γ − 1 + q (q) + ζ Pq

2010 Mathematics Subject Classification. Primary 11N37; Secondary 11A25, 11M41. Key words and phrases. Number of divisors, Legendre symbol, mean values, Riemann hypothesis.

` OLIVIER BORDELLES

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where θ is defined in (1), cq is given in (2) and      q−1  X Y m 1  m+1 1 + − Pq (1) = m q q p m=cq p   o  Pq−1 n  m m+1 ′ − m X Pq m=cq q q pm . n   o (1) = − log p  P m+1 1 Pq 1 + q−1 − m p m m=cq

⊲

If q ≡ ±11 (mod 24) X (λq ⋆ 1) (n) = x1/2 ζ

q 2

n6x

where

1 2

Rq ⊲




:=

Y p




Rq

q

1 2




q

p

  + Oq,ε x1/3+ε

!    q−1  X m+1 m 1 . 1+ + q q pm/2 m=3

If q ≡ ±5 (mod 24), there exists c > 0 such that X 1/4 −1/5 1/4 3/5 . (λq ⋆ 1) (n) ≪q x1/2 e−c(log x ) (log log x ) n6x

Furthermore, if the Riemann hypothesis is true, then for x sufficiently large X √ 1/2 √ 5/2+ε . (λq ⋆ 1) (n) ≪q,ε x1/4 e(log x) (log log x) n6x

Example 2. X

n6x

X

n6x

X

n6x

X

n6x

  . (λ7 ⋆ 1) (n) = 0.454 x (log x + 2γ + 0.784) + Oε x1/2+ε .

  . (λ23 ⋆ 1) (n) = 0.899 x (log x + 2γ − 0.678) + Oε x131/416+ε .   . (λ13 ⋆ 1) (n) = 1.969 x1/2 + Oε x1/3+ε . (λ5 ⋆ 1) (n) ≪ x1/2 e−c(log x

1/4 3/5

)

(log log x1/4 )

−1/5

.

2. Notation
In what follows, x > e is a large real number, ε ∈ 0, 14 is a small real number which does not need   to be the same at each occurrence, s := σ + it ∈ C, q always denotes an odd prime number, · is the Legendre symbol modulo q and define q   τ λq := q P where τ (n) := d|n 1. Also, 1 is the constant arithmetic function equal to 1. 4

For any arithmetic functions F and G, L(s, F ) is the Dirichlet series of F , the Dirichlet convolution product F ⋆ G is defined by X F (d)G(n/d) (F ⋆ G)(n) := d|n

and F

−1

is the Dirichlet convolution inverse of F . If r ∈ Z>2 , then ( 1, if n = mr ; ar (n) := 0, otherwise.

For some c > 0, set δc (x) := e−c(log x)

3/5

(log log x)−1/5

1/2

and ω(x) := e(log x)

(log log x)5/2+ε

.

Finally, let M (x) and L(x) be respectively the Mertens function and the summatory function of the Liouville function, i.e. X X M (x) := µ(n) and L(x) := λ(n). n6x

n6x

WHEN THE NUMBER OF DIVISORS IS A QUADRATIC RESIDUE

3. The Dirichlet series of λq Proposition 3. Let q > 3 be a prime number. For any s ∈ C such that σ > 1 ⊲

If q ≡ ±1 (mod 8)

L (s, λq ) = ζ(qs)ζ(s)

Y p

where

    q−1  X m 1 m + 1  1 + − q q pms m=c 

( 2, cq := > 4,

(2) ⊲

q

if q ≡ ±7 (mod 24) ; if q ≡ ±1 (mod 24).

If q ≡ ±3 (mod 8)      q−1  X m 1  ζ(qs)ζ(2s) Y  m+1 + L (s, λq ) = 1+ ζ(s) q q pms p m=dq

where

dq := Proof. Set χq :=

  · q

(

2, 3,

if q ≡ ±5 (mod 24) or q = 3 ; if q ≡ ±11 (mod 24).

for convenience. From [8, Lemma 2.1], we have

L (s, λq ) =

=

=

= If q ≡ ±1 (mod 8), then

  2 q

! ! ∞ ∞ Y X X χq (α) χq (α + 1) s = 1+p 1+ psα psα p p α=2 α=1 ( !)   −1 X q−1  Y 1 m 1 s −s 1 − qs 1+p −p p q pms p m=1 ) (  −1 X q−1  Y  m 1 1 1 − qs p q p(m−1)s p m=1 !  q−1  X Y m 1 ζ(qs) 1+ . q p(m−1)s p m=2

Y

= 1 and

L (s, λq ) = ζ(qs)ζ(s)

Y p

where  q−1    1 1 X m 1− s p m=2 q p(m−1)s

  q−1  X 1 m 1 − q pms p(m−1)s m=2   q−2  q−1  X X m+1 m 1 1 − ms ms q p q p m=1 m=2         q−1 X m+1 m 1 1 q 2 1 − + − s ms (q−1)s q p q q p q p m=2     q−1 X m+1 m 1 1 − + s. ms q q p p m=2

=

=

=

= Similarly, if q ≡ ±3 (mod 8), then

!  q−1    1 1 1 X m 1− s + 1− s p p m=2 q p(m−1)s

  2 q

= −1 and

!  q−1    ζ(qs)ζ(2s) Y 1 1 1 X m L (s, λq ) = 1+ s + 1+ s ζ(s) p p m=2 q p(m−1)s p

3

` OLIVIER BORDELLES

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where  q−1    1 1 X m 1+ s p m=2 q p(m−1)s

  q−1  X 1 m 1 + q pms p(m−1)s m=2   q−2  q−1  X X m+1 m 1 1 + ms ms q p q p m=1 m=2         q−1 X 2 1 m+1 q m 1 1 + − + s ms (q−1)s q p q q p q p m=2     q−1 X m+1 m 1 1 + − s. ms q q p p m=2

=

=

=

=

        We achieve the proof noting that, if q ≡ ±1 (mod 24), then q3 − 2q = 4q − 3q = 0 and,         similarly, if q ≡ ±11 (mod 24), then 3q + q2 = 0 whereas 4q + q3 = 2.  4. Proof of Theorem 1 4.1. The case q ≡ ±1 (mod 8). For σ > 1, we set      q−1  ∞ X Y X gq (n) m 1  m+1 1 + − := ζ(qs)P (s) := . Gq (s) = ζ(qs) q ms q q p ns p m=c n=1 q

First observe that cq < q in the case q ≡ ±1 (mod 24). Indeed, among the q − 4 integers   m ∈ 1 m {4, . . . , q−1}, it is known from [3, p.76] that there are 2 (q−3)−3 of them such that q = m+1 . q     m m+1 1 , and the Consequently there are 2 (q + 1) integers m ∈ {4, . . . , q − 1} verifying q 6= q inequality follows. Thus this Dirichlet series is absolutely convergent in the half-plane σ > (2), so that X |gq (n)| ≪q,ε x1/cq +ε .

1 cq

where cq is given in

n6x

By partial summation, we infer X gq (n) n

n6x

X gq (n) x log n n

n6x

  = ζ(q)Pq (1) + O x−1+1/cq +ε   = ζ(q)Pq (1) log x + qPq (1)ζ ′ (q) + Pq′ (1)ζ(q) + O x−1+1/cq +ε .

From Proposition 3, λq ⋆ 1 = gq ⋆ τ . Consequently X X X τ (k) (λq ⋆ 1) (n) = gq (d) n6x

d6x

=

X

k6x/d

gq (d)

d6x



   x x θ+ε x x log + (2γ − 1) + O d d d d

 = x ζ(q)Pq (1) log x + qPq (1)ζ ′ (q) + Pq′ (1)ζ(q) + (2γ − 1)ζ(q)Pq (1)   + O xmax(1/cq ,θ)+ε

where θ is defined in (1) and where we used −ε

x

X |gq (d)| ≪ dθ

d6x

( x1/cq −θ , if c−1 q > θ; 1, otherwise.

WHEN THE NUMBER OF DIVISORS IS A QUADRATIC RESIDUE

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4.2. The case q ≡ ±11 (mod 24). For σ > 1, we set !    q−1  ∞ X X Y hq (n) m 1 m+1 + := ζ(qs)R (s) := . 1+ Hq (s) = ζ(qs) q ms q q p ns p n=1 m=3

Since q > 5, this Dirichlet series is absolutely convergent in the half-plane σ > 13 , so that X |hq (n)| ≪q,ε x1/3+ε . n6x

From Proposition 3, λq ⋆ 1 = hq ⋆ a2 , hence X

(λq ⋆ 1) (n) =

n6x

X

d6x

= =

r  x hq (d) d

  X hq (d) √ + O x1/3+ε d d6x  
x1/2 Hq 21 + O x1/3+ε .

x1/2

4.3. The case q ≡ ±5 (mod 24). In this case, it is necessary to rewrite L(s, λq ) in the following shape. Lemma 4. Assume q ≡ ±5 (mod 24). For any σ > 1, L(s, λq ) =  ζ(5s),         Kq (s) := ζ(4s)Lq (s),       L (s)   q , ζ(4s)

where Lq (s) := ζ(qs)

Kq (s) with ζ(s)ζ(2s)

if q = 5 if q ≡ ±19, ±29 (mod 120) if q ≡ ±43, ±53 (mod 120)

Y

!
   q−1  2 p2s + ps + 1 m+1 m 1 p2s + 1 X + + 2s 1+ p7s − p5s p − 1 m=6 q q pms

:=

ζ(qs)

p

and Lq (s)

Y

1−

2p2s − 1

(p2s − 1)3 (p2s + 1) !    q−1  X m+1 m 1 p8s + . + 3 q q pms (p2s − 1) (p2s + 1) m=6 p

The Dirichlet series Lq is absolutely convergent in the half-plane σ > 51 , and the Dirichlet series Lq is absolutely convergent in the half-plane σ > 16 . Proof. From Proposition 3, we immediately get ζ(5s) . ζ(s)ζ(2s)         Now suppose q > 5 and q ≡ ±5 (mod 24). In this case, 3q + 2q = −2 and 4q + q3 = 0 so that we may write by Proposition 3 !    q−1  X m+1 ζ(qs)ζ(2s) Y m 1 2 L (s, λq ) = + 1 − 2s + ζ(s) p q q pms p m=4 (3)

L(s, λ5 ) =

Kq (s) ζ(s)ζ(2s)

= where Kq (s) := ζ(qs)

Y p

!    q−1  X m+1 m 1 + + . 1− q q pms (p2s − 1)2 (p2s − 1)2 m=4 1

p4s

` OLIVIER BORDELLES

6

Assume q ≡ ±19, ±29 (mod 120). Then         4 6 5 5 + = + = 2. q q q q Kq (s) can therefore be written as Kq (s)

!    q−1  X m+1 m 1 1+ = ζ(qs) + + q q pms ps (p2s − 1)2 (p2s − 1)2 m=6 p !
   q−1  Y 2 p2s + ps + 1 m+1 m 1 p2s + 1 X + + 2s = ζ(qs)ζ(4s) 1+ p7s − p5s p − 1 m=6 q q pms p ps + 2

Y

p4s

= ζ(4s)Lq (s).

Similarly, if q ≡ ±43, ±53 (mod 120), then         5 4 6 5 + = + = 0. q q q q

Hence

Kq (s) :=

ζ(qs)

Y p

=

Lq (s) . ζ(4s)

!    q−1  X m+1 m 1 1− + + q q pms (p2s − 1)2 (p2s − 1)2 m=6 p4s

1

The proof is complete.



We now are in a position to prove Theorem 1 in the case q ≡ ±5 (mod 24). Assume first that q ≡ ±19, ±29 (mod 120) and let ℓq (n) be the n-th coefficient of the Dirichlet series Lq (s). From Lemma 4, λq ⋆ 1 = ℓq ⋆ a4 ⋆ a−1 2 and therefore r  r   X X X X x x 1 = . ℓq (d)L (λq ⋆ 1) (n) = ℓq (d) M 2 m d d 1/4 n6x

d6x

d6x

m6(x/d)

Since L(z) ≪ zδc (z) for some c > 0 X

n6x

r X |ℓq (d)|  x  √ δc d d d6x   r  X X x q (d)|  |ℓ√ δc ≪ x1/2  + d √ √ d d6 x x x

(λq ⋆ 1) (n) ≪ x1/2

The Dirichlet series Lq (s) := consequently

P∞

n=1 ℓq (n)n

X d6z

and by partial summation

−s

is absolutely convergent in the half-plane σ >

1 5,

|ℓq (d)| ≪q,ε z 1/5+ε

X |ℓq (d)| √ ≪q,ε z −3/10+ε . d d>z

We infer that X

n6x

    (λq ⋆ 1) (n) ≪ x1/2 δc x1/4 + x7/20+ε ≪ x1/2 δc x1/4 .

Now suppose that the Riemann hypothesis is true. By [1], which is a refinement of [9], we know that M (z) ≪ε z 1/2 ω(z). The method of [9, 1] may be adapted to the function L yielding L(z) ≪ε z 1/2 ω(z) log z. e

Observe that, for any a > 2, ε > 0 and z > ee p  p  log z exp log z (log log z)a 6 exp log z (log log z)a+ε

WHEN THE NUMBER OF DIVISORS IS A QUADRATIC RESIDUE

7

so that L(z) ≪ε z 1/2 ω(z) and hence X

n6x

(λq ⋆ 1) (n) ≪ x1/4

r X |ℓq (d)|  x  √
X |ℓq (d)| √
≪ x1/4 ω x ω ≪ x1/4 ω x 1/4 1/4 d d d d6x

d6x

achieving the proof in that case. The case q = 5 is similar but simpler since λ5 ⋆ 1 = a5 ⋆ a−1 2 by (3). Finally, when q ≡ ±43, ±53 (mod 120), we proceed as above. Let νq (n) be the n-th coefficient −1 of the Dirichlet series Lq (s). Then λq ⋆ 1 = νq ⋆ a−1 4 ⋆ a2 from Lemma 4, so that r   X X X x 1 (λq ⋆ 1) (n) = µ(m)M νq (d) 2 m d 1/4 n6x

d6x

m6(x/d)

and estimating trivially yields X X |νq (d)| √ (λq ⋆ 1) (n) ≪ x1/2 d n6x d6x

X

m6(x/d)1/4

1 δc m2



1 m2

r  x d

and we complete the proof as in the previous case.



Remark 5. Let us stress that a bound of the shape X (λq ⋆ 1) (n) ≪ x1/4+ε n6x

for all x sufficiently large and small ε > 0, is a necessary and sufficient condition the Riemann hyPfor ∞ pothesis. Indeed, if this estimate holds, then by partial summation the series n=1 (λq ⋆ 1) (n)n−s is absolutely convergent in the half-plane σ > 14 . Consequently, the function Kq (s)ζ(2s)−1 is analytic in this half-plane. In particular, ζ(2s) does not vanish in this half-plane, implying the Riemann hypothesis, proving the necessary condition, the sufficiency being established above. 5. A short interval result for the case q = 5 5.1. Introduction. This section deals with sums of the shape X (λ5 ⋆ 1) (n) x 0 such that |f ′ (x)| ≍ λ1 . Then R(f, N, δ) ≪ N λ1 + N δ + δλ−1 1 + 1.

(4)

This result is essentially a consequence of the mean value theorem. The second tool is [4, Theorem 7] with k = 3. Lemma 8. Let s
∈ Q∗ \ {±2, ±1} and X > 0 such that N 6 X 1/s . Then there exists a constant c3 := c3 (s) ∈ 0, 41 depending only on s such that, if N 2 δ 6 c3

(5)

then R



X , N, δ ns



≪ XN 3−s


1/7

+ δ XN 59−s


1/21

.

Our last result relies the short sum of λ5 ⋆ 1 to a problem of counting integer points near a smooth curve. Lemma 9. Let 1 6 y 6 x. Then X (λ5 ⋆ 1) (n) ≪ max

(16y 2 x−1 )1/5