4 AMADOU DIOGO BARRY AND STEPHANE

´

R. LOUBOUTIN

Since

1≤mx

d1|m

1

c

√

m≤d1≤c

√

m

1 = Sc(x) = O(x)

we may assume that d2 = d1, or even that d2 d1. Since

δ|D1

δ|D2

µ(δ) =

1 if gcd(D1, D2) = 1

0 otherwise,

we have

(2) Tc(x) = 2Uc(x) + O(x),

where

Uc(x) =

1≤mx

d1|m

1

c

√

m≤d1≤c

√

m

d2|m

1

c

√

m≤d2d1

1

=

dc

√

x

D1

c

√

x

d

D2D1

gcd(D1,D2)=1

m=kdD1D2x

d2Di 2/c2≤m≤c2d2Di 2,

i∈{1,2}

1

=

dc

√

x

δ

c

√

x

d

µ(δ)

D1

c

√

x

dδ

D2D1

m=kdδ2D1D2x

d2δ2Di 2/c2≤m≤c2d2δ2Di 2,

i∈{1,2}

1

(write di = dDi with gcd(D1, D2) = 1 and m = kdD1D2, then change Di into δDi

and m into kdδ2D1D2).

Hence,

(3) Uc(x) =

dc

√

x

δ

c

√

x

d

µ(δ)Mc(d, δ, x),

where

Mc(d, δ, x) =

D1

c

√

x

dδ

D2D1

Nc(d, δ, D1, D2, x),

with Nc(d, δ, D1, D2, x) being the number of positive integers k such that

(4) k

x

dδ2D1D2

and

1

c

√

kdδ2D1D2 ≤ dδDi ≤ c

√

kdδ2D1D2 for i ∈ {1, 2}, i.e. such that

(5)

dD1

c2D2

≤ k ≤

c2dD2

D1

,

which implies

D1/c2

≤ D2.

To begin with, Uc(x) is not too large. Indeed, by (5), we have

d/c2

≤ k ≤

c2d

and

Nc(d, δ, D1, D2, x) ≤ (c2 − 1/c2 + 1)d. Therefore,

Mc(d, δ, x) ≤

(c2

−

1/c2

+ 1)d

D1

c

√

x

dδ

D2

c

√

x

dδ

1 ≤

(c4

+

c2

− 1)

x

dδ2

and Uc(x) ≤

∑

dc

√

x

∑

δc

√

x/d

Mc(d, δ, x) x log x. Thus, by (2), we have

Tc(x) x log x.

To prove our Proposition, we use (2), (3) and the following Lemma 9.

4

´

R. LOUBOUTIN

Since

1≤mx

d1|m

1

c

√

m≤d1≤c

√

m

1 = Sc(x) = O(x)

we may assume that d2 = d1, or even that d2 d1. Since

δ|D1

δ|D2

µ(δ) =

1 if gcd(D1, D2) = 1

0 otherwise,

we have

(2) Tc(x) = 2Uc(x) + O(x),

where

Uc(x) =

1≤mx

d1|m

1

c

√

m≤d1≤c

√

m

d2|m

1

c

√

m≤d2d1

1

=

dc

√

x

D1

c

√

x

d

D2D1

gcd(D1,D2)=1

m=kdD1D2x

d2Di 2/c2≤m≤c2d2Di 2,

i∈{1,2}

1

=

dc

√

x

δ

c

√

x

d

µ(δ)

D1

c

√

x

dδ

D2D1

m=kdδ2D1D2x

d2δ2Di 2/c2≤m≤c2d2δ2Di 2,

i∈{1,2}

1

(write di = dDi with gcd(D1, D2) = 1 and m = kdD1D2, then change Di into δDi

and m into kdδ2D1D2).

Hence,

(3) Uc(x) =

dc

√

x

δ

c

√

x

d

µ(δ)Mc(d, δ, x),

where

Mc(d, δ, x) =

D1

c

√

x

dδ

D2D1

Nc(d, δ, D1, D2, x),

with Nc(d, δ, D1, D2, x) being the number of positive integers k such that

(4) k

x

dδ2D1D2

and

1

c

√

kdδ2D1D2 ≤ dδDi ≤ c

√

kdδ2D1D2 for i ∈ {1, 2}, i.e. such that

(5)

dD1

c2D2

≤ k ≤

c2dD2

D1

,

which implies

D1/c2

≤ D2.

To begin with, Uc(x) is not too large. Indeed, by (5), we have

d/c2

≤ k ≤

c2d

and

Nc(d, δ, D1, D2, x) ≤ (c2 − 1/c2 + 1)d. Therefore,

Mc(d, δ, x) ≤

(c2

−

1/c2

+ 1)d

D1

c

√

x

dδ

D2

c

√

x

dδ

1 ≤

(c4

+

c2

− 1)

x

dδ2

and Uc(x) ≤

∑

dc

√

x

∑

δc

√

x/d

Mc(d, δ, x) x log x. Thus, by (2), we have

Tc(x) x log x.

To prove our Proposition, we use (2), (3) and the following Lemma 9.

4