6 AMADOU DIOGO BARRY AND STEPHANE
´
R. LOUBOUTIN
and
Mc(d, δ, x) =
c2d
4
(1
1
c4
)
(
X2
+ O(X)
)
+
c2d
2
(
X2 log(c2)
+ O(X)
)

1
2c4
(
(c4

1)X2
+ O(X)
)
+ O(X)

d
log(c2)
2c2
(
c4X2
+ O(X)
)
+c2dX2
(
c2X
X
log(u/X)
u
du + O(
1
X
)
)
+O(X2)
=
2c2d(log2 c)X2
+ O(dX) +
O(X2),
which proves the Lemma.
3. Proof of Theorem 5
Set
0 λn(m) =
d|m
e−π(m/d−d)2/n.
Then,
S4(p) =
χ∈Xp
+ a,b,c,d
χ(ab)¯(cd)e−π(a2+b2+c2+d2)/p
χ
=

a,b,c,d
e−π(a2+b2+c2+d2)/p
+
p 1
2

a,b,c,d
cd≡±ab (mod p)
e−π(a2+b2+c2+d2)/p


a
e−πa2/p
4
+
p 1
2
p−1
j=1
a,b
ab≡j (mod p)
e−π(a2+b2)/p
2


0
e−πu2/pdu
4
+
p 1
2
p−1
j=1
m≥1
m≡j (mod p)
a|m
e−π(a2+m2/a2)/p
2
=
−p2/16
+
p 1
2
p−1
j=1
m≥1
m≡j (mod p)
λp(m)e−2πm/p
2

−p2/16
+
p 1
2
p−1
j=1
λp(j)2e−4πj/p

−p2/16
+
p 1
2
e−4π
p−1
j=1
λp(j)2
=
−p2/16
+
p 1
2
e−4πΛ(p).
By Proposition 7, it follows that
S4(p)
p2
log p.
6
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