In this section we prove that the estimate (A) from Proposition 3.4 holds for the sequence{aφ,ψ,n}ndened in (3.30) withθ1= 1/6.
Proposition 3.6. Let an=aφ,ψ,n, where aφ,ψ,n is dened as in (3.30), and letAd(X)be dened as in (3.25). Then for all >0 and allX ≥2, we have
Ad(X)X
5 6+.
Proof. Recall that
Ad(X) =
X
Norm(n)≤X
n≡0 modd
an.
Since the sequence an is supported on odd ideals n, we see thatAd(X) = 0 unless dis odd. Hence we may assume without loss of generality that dis an
odd ideal. Let
R(X) :=D4(X) =(u, v)∈ D ∪ε2D ∪ε4D ∪ε6D:u2−2v2≤X . (3.33)
By Proposition 3.5 and denition (3.30), we have
Ad(X) = X (u,v)∈R(X) u+v√2≡0 modd [u+v√2]φ,ψ, where [u+v√2]φ,ψ is dened as in (3.29).
We now reformulate the congruence condition u+v√2 ≡ 0 modd. Propo-
sition 3.5 implies that there is an element d1+d2 √
2∈ D which generates d.
Then the congruence above is equivalent to saying that there exist integerse1
ande2 such thatu+v √
2 = (d1+d2 √
2)(e1+e2 √
2), i.e. such that
u=d1e1+ 2d2e2
and
v=d2e1+d1e2.
In other words,(u, v)is in the image of the linear transformation
Ld:= d1 2d2 d2 d1 :Z2→Z2
Figure 3.2: The regionR(X)and the lattice pointsR(d, X)
of determinant
D:= Norm(d) =d21−2d22.
Hence we dene
R(d, X) :={(u, v)∈ R(X) : (u, v)∈Image(Ld)} (depicted in Figure 3.2), and we rewrite the sum Ad(X)as
Ad(X) =
X
(u,v)∈R(d,X)
[u+v√2]φ,ψ.
Using the fact that|[u+v√2]φ,ψ| ≤1, we obtain the trivial bound
|Ad(X)| ≤ X (u,v)∈R(d,X) 1 = X L−d1R(X)∩Z2 1. (3.34) Sinced1+d2 √
2∈ D, we have the inequalities
d2 1
2 ≤D≤d 2 1,
which implies thatdiam(L−d1)D−1/2. Hence Lemma 3.14 gives
|Ad(X)| ≤a4XD−1+O(D−
1
2X12 + 1)XD−1+X12D−12 + 1, (3.35)
where the implied constant is absolute. This estimate will be useful when D
is large compared toX.
Next we split the sum Ad(X)into8·16sums where the congruence classes of
uandv modulo16are xed, sayu≡u0mod 16andv≡v0mod 16for some
congruence classes u0 and v0 modulo16withu0 invertible modulo16. Foru
andv satisfying these congruences, we have
[u+v√2]φ,ψ=δ(u0, v0)
v
u
,
where δ(u0, v0) ∈ {±1} depends only on the congruence classes u0 and v0
modulo16. Hence it remains to give estimates for sums of the type
Ad(u0, v0, X) := X (u,v)∈R(u0,v0,d,X) v u , where R(u0, v0,d, X) :={(u, v)∈ R(d, X) : (u, v)≡(u0, v0) mod 16}.
Splitting the sum according to the value ofu, we obtain Ad(u0, v0, X) = X 0≤u≤R1(X) u≡u0mod 16 Au,d(v0, X), (3.36) where Au,d(v0, X) := X v∈Iu (u,v)∈Ld(Z2) v≡v0mod 16 v u . Here R1(X) = sup{u∈R: (u, v)∈ R(X)} X 1 2
and Iu is an interval (or a union of 2 disjoint intervals) of size ≤ 2R2(X),
where
R2(X) = sup{|v| ∈R: (u, v)∈ R(X)} X 1 2.
We now unwind the condition(u, v)∈Ld(Z2), i.e. that (u, v)is in the image
ofLd. Consider the system of equations inxandy:
(
u=d1x+ 2d2y
v=d2x+d1y.
Let d:= gcd(d1, d2)and writed1=dd01, d2=dd02. Recall thatd and so also
d1 is odd, so thatd= gcd(d1,2d2). If the system (3.37) has a solution overZ,
then dmust divide u. This means that Ad(u0, v0, X) = X 0≤u≤R1(X) u≡u0mod 16 u≡0 modd Au,d(v0, X).
Now suppose u≡0 modd, and letxu, yu∈Zbe such that
u=d1xu+ 2d2yu.
Then all solutions (x, y)∈Z2to the rst equation in (3.37) are given by
(x, y) = (xu−2d02k, yu+d01k), k∈Z.
Hence
v=d2(xu−2d02k) +d1(yu+d01k) =d2xu+d1yu+Dk/d,
which means that (3.37) has a solution overZif and only if
v≡d2xu+d1yumodD/d.
Note thatDis odd, so thatD/dand16are coprime. Letvube the congruence
class modulo16D/dsuch that
(
vu≡d2xu+d1yumodD/d
vu≡v0mod 16.
Thus we have proved that ifu≡0 modd, then Au,d(v0, X) = X v∈Iu v≡vumod 16D/d v u .
Let eu = gcd(vu,16D/d), write 16D/d = eudu, vu = euvu0, and perform a
change of variables v=euv0, so that
Au,d(v0, X) = eu u X v0∈I0 u v0≡v0umoddu v0 u ,
where Iu0 =Iu/eu. Since gcd(v0u, du) = 1, we can now detect the congruence
conditionv0≡vu0 modduvia Dirichlet characters modulo du. In other words,
Au,d(v0, X) = 1 ϕ(du) eu u χ(v0 u) X χmoddu X v0∈I0 u χ(v0) v0 u , (3.38)
where v0
u denotes the multiplicative inverse of v0u modulo du. Let χ be a
Dirichlet character modulo du. If the character
v07→χ(v0)
v0 u
is trivial, thenu=f g2for somef dividingd
u (and therefore dividing16D/d)
and some integerg. The number of suchu≤R1(X)is ≤τ(16D/d)R1(X)12 DX14.
In this case we use the trivial bound
X v0∈I0 u χ(v0) v0 u #Iu0 ≤#IuX 1 2,
where the implied constant in is absolute. Hence the contribution of such
utoAd(u0, v0, X)is
DX
3
4. (3.39)
On the other hand, if the character
v07→χ(v0)
v0
u
is not trivial, its conductor is at most
16Du/dDX12,
and so the Polya-Vinogradov inequality gives the estimate
X v0∈I0 u χ(v0) v0 u D 1 2X 1 4+.
Combining this with (3.36), (3.38), and (3.39), we have proved the bound
Ad(X)D
1
2X34+. (3.40)
We use (3.40) forD < X1/6 and (3.35) forD≥X1/6 to obtain
Ad(X)X
5 6+.