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

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

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

D2
c

x

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
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