On a form of degree $d$ in $2d+1$ variables ($d\geq 4$)

0.16 MB·

by  Manoj Verma

ON A FORM OF DEGREE d IN 2d+1 VARIABLES (d 4)
≥
MANOJVERMA
4
1 Abstract. Fork≥2,wederiveanasymptoticformulaforthenumberofzerosoftheforms
0 Qki=1(x22i−1+x22i)+Qki=1(x22k+2i−1+x22k+2i)−x24kk+1
2 and
n x1Qki=1(x22i+x22i+1)+x2k+2Qki=1(x22k+2i+1+x22k+2i+2)−x24kk++13
a inthebox1≤xi≤P usingthecirclemethod.
J

9 Mathematics SubjectClassification2010: Primary11D45;Secondary11D85, 11P55.

]
T
N
.
h 1. Introduction
t
a
m
The problems of representation of positive integers and of non-trivial representation of
[
zero by forms with integral coefficients have attracted attention of many mathematicians

1 since the beginning of the 20th century. Most of the work in this area has been carried
v
out using the circle method. The square-root barrier implies that the circle method is
6
unlikely to yield an asymptotic formula in case of a form of degree d with less than 2d+1
6
3 variables. In 1962 Birch, Davenport and Lewis [1] found an asymptotic formula for the
2 number representations of zero in a box x P by the cubic forms in seven variables of
i
. | | ≤
1 the type
0 N (x ,x ,x )+N (x ,x ,x )+a x3
4 1 1 2 3 2 4 5 6 7 7
1 where N and N are norm forms of some cubic field extensions of Q and a is a nonzero
1 2 7
: integer. Incidentally, whereas an asymptotic formula for the number of representations
v
i of an integer as a sum of eight cubes has been established by Vaughan [4], an asymptotic
X
formula in the case of a sum of seven cubes seems to be out of reach as of now. However,
r
a a lower bound of the expected order of magnitude has been established by Vaughan [5].
A few more families of cubic forms in seven variables have been dealt with. However, for
d 4 no example of an asymptotic formula for the number of representations of either
≥
zero or of large positive integers by a form of degree d in 2d+1 variables seems to exist in
the literature. In this paper, for each d 4, we establish an asymptotic formula for the
≥
numberof zeros of aparticular form of degree din 2d+1variables ina suitableexpanding
box using the circle method. It turns out that our proof goes through in cases d = 1,2,3
as well and retrieves previously known results in those cases. For d even, say d = 2k,
k 1, we look at the number zeros of the form
≥
k k
f(x) = f(x ,...,x )= (x2 +x2 )+ (x2 +x2 ) x2k
1 4k+1 2i−1 2i 2k+2i−1 2k+2i − 4k+1
i=1 i=1
Y Y
while for d odd, say d= 2k+1, k 0, we look at the number of zeros of the form
≥
k k
f(x) = f(x ,...,x )= x (x2 +x2 )+x (x2 +x2 ) x2k+1
1 4k+3 1 2i 2i+1 2k+2 2k+2i+1 2k+2i+2 − 4k+3
i=1 i=1
Y Y
1
2 MANOJVERMA
in a box 1 x P. When convenient we shall abbreviate and write
i
≤ ≤
f(x) = f(x ,...,x )= f (x ,...,x )+f (x ,...,x ) xd (1.1)
1 2d+1 d 1 d d d+1 2d − 2d+1
where
k (x2 +x2 ) if d is even, d= 2k, k 1
fd(x1,...,xd) = x i=1k 2(ix−21 +x22i ) if d is odd, d = 2k+1,≥k 0 .
(cid:26) Q1 i=1 2i 2i+1 ≥
We could define fd = fd(x1,.Q..,xd) recursively: f1(x1) = x1, f2(x1,x2) = x21 +x22, and
f (x ,...,x ) = (x2 +x2) f (x ,...,x ) for d 2.
d 1 d d−1 d · d−2 1 d−2 ≥
Taking x = x = x = x = ... = 1, x = x = x = x =
d−2 2d−2 d−4 2d−4 d−3 2d−3 d−5 2d−5
... = 0 and applying the four square theorem we see that this form represents every
integer infinitely many times; our interest lies in establishing the asymptotic formula for
the number of representations of zero with 1 x P. Only minor changes are required
i
≤ ≤
to prove the expected asymptotic formula for the number of representations of a large
positive integer N by this form (or by the form obtained by changing the sign in the
front of xd2d+1 to +) with the variables restricted so that 1 ≤ xi ≤ ⌊Nd1⌋. −
Note that if d 3 then the form f is singular (e.g., consider the point (1, 0, 0, ..., 0))
≥
(soDeligne’s estimate for complete exponential sumsis notapplicable) andits h-invariant
is 3. (The h-invariant of a form F, denoted h(F), is defined as the smallest integer h such
that F can be written in the form
F(x) = A (x)B (x)+...+A (x)B (x)
1 1 h h
where A (x) and B (x) are forms of positive degrees having rational coefficients.)
i i
Inwhatfollows, ε can take any positive real valuein any statement in which itappears,
e(α) = e2πiα, the symbols and O have their usualmeanings with the implicit constants
≪
depending at most on ε and d. If A is a set, #A denotes the cardinality of A. If n,m
are positive integers, d(n) denotes the number of positive divisors of n and d (n) denotes
m
the number of (ordered) factorizations of n with m positive integral factors. For a real
numberx, x and x denote thegreatest integer less than or equal to x andthe smallest
⌊ ⌋ ⌈ ⌉
integer greater than or equal to x respectively.
Theorem 1. The number of zeros R(0;P) of the form f in (1.1) with 1 x P satisfies
i
≤ ≤
R(0;P) = Pd+1SJ +O(Pd+1−1+5·12d−1+ε)
where
∞ q
a
S = q−2d−1S(q,a) with S(q,a) = e f(z) ,
q
q=1 a=1 zmodq (cid:18) (cid:19)
X X X
(a,q)=1
and
∞
J = e(γf(ξ))dξ dγ;
Z−∞ Z[0,1]2d+1 !
S and J are positive.
Remark. One can also consider the problem of finding an asymptotic formula for the
number of zeros of the form f inside the box x P but then the contribution from
i
| | ≤
those “degenerate” zeros of f that lie inside the box x P and also in the set
i
| | ≤
U = x Z2d+1 :f (x ,...,x ) = f (x ,...,x )= x = 0
d 1 d d d+1 2d 2d+1
{ ∈ }
would be 22d−2P2d−2+O (P2d−3) in case d is odd and 22d−6d2P2d−4+O (P2d−5) in case
d d
d is even and thus the contribution from the zeros of f with at least one of the x equal to
i
zero would be Pd+2 if d 5 and cannot be determined precisely via the circle method.
≫ ≥
ON A FORM OF DEGREE d IN 2d+1 VARIABLES (d≥4) 3
On the other hand, the number of zeros of f with x P but none of the x equal to
i i
| | ≤
zero is 22d+1 times the number of zeros with 1 x P in case d is even, and is 22d−1
i
≤ ≤
times the sum of the number of zeros with 1 x P of the three forms
i
≤ ≤
f (x ,...,x )+f (x ,...,x ) xd ,
d 1 d d d+1 2d − 2d+1
f (x ,...,x )+f (x ,...,x ) xd
− d 1 d d d+1 2d − 2d+1
and
f (x ,...,x ) f (x ,...,x ) xd ;
d 1 d − d d+1 2d − 2d+1
theorem 1 applies to the latter two forms as well.
2. The Minor Arcs
Let P be a large positive integer, I = 1,2,...,P , B = I2d+1 and
{ }
F(α) = e(αf(x)). (2.1)
x∈B
X
Then the number of zeros of the form f inside the box 1 x P is
i
≤ ≤
1 1
R(0;P) = F(α)dα = F (α)2F (α)dα (2.2)
1 2
Z0 Z0
where
F (α) = e(αf (x ,...,x ))
1 d 1 d
x1,.X..,xd∈I
and
F (α) = e( αxd).
2
−
x∈I
X
Let δ be a sufficiently small positive real number to be specified later. For 1 q
Pδ, 1 a q, (a,q) = 1, define the major arcs M(q,a) by ≤ ≤
≤ ≤
a
M(q,a) = α : α Pδ−d
− q ≤
(cid:26) (cid:12) (cid:12) (cid:27)
(cid:12) (cid:12)
and let M be the union of the M(q,a). The(cid:12)set m =(cid:12) (Pδ−d,1+Pδ−d] M forms the minor
(cid:12) (cid:12) \
arcs.
Lemma 2.1. We have
1
F (α) 2dα Pd+ε.
1
| | ≪
Z0
Proof. We assume that d is even, d = 2k, k 1; the proof is similar in case d is odd. We
≥
have
1
F (α)2dα = a(n)2
1
| |
Z0 0≤nX≤2kPd
where
a(n)= # (x ,...,x ) Id : f (x ,...,x )= n for n 0.
1 d d 1 d
{ ∈ } ≥
Clearly a(0) = 1 and a(n) = Pd. For n 1, in any solution of k (x2 +
0≤n≤2kPd ≥ i=1 2i−1
x2 ) = n we must have x2 +x2 = n (1 i k) for some factorization n = n ...n
2i P 2i−1 2i i ≤ ≤ Q 1 k
of n with k (positive) factors. For a positive integer r, the number of solutions of the
4 MANOJVERMA
equation s2+t2 = r in non-negative integers s,t is 8d(r) rε. Thus, for 1 n 2kPd
≤ ≪ ≤ ≤
we have
a(n) nε...nε = d (n)nε nεnε P2dε.
≪ 1 k k ≪ ≪
n1.X..nk=n
Hence
1
F (α)2dα = 1+ a(n)2 1+P2dε a(n) Pd+2dε. (cid:3)
1
| | ≪ ≪
Z0 1≤nX≤2kPd 1≤nX≤2kPd
Lemma 2.2. We have
d+1+ε− δ
F(α)dα P 2d−1.
m ≪
Z
Proof. For α m, by Dirichlet’s theorem on Diophantine approximation, there exist a,q
∈
with (a,q) = 1,q Pd−δ and α a q−1Pδ−d < q−2. Since α m (Pδ−d,1 Pδ−d),
≤ − q ≤ ∈ ⊆ −
we must have 1 a < q and(cid:12) q >(cid:12)Pδ (otherwise α would be in M(q,a)). By Weyl’s
≤ (cid:12) (cid:12)
inequality (See [2] or [6]), (cid:12) (cid:12)
1+ε− δ
F2(α) P 2d−1.
≪
This, with (2.2) and lemma 2.1 gives
F(α)dα = F2(α)F (α)dα sup F (α) F (α)2dα
(cid:12)Zm (cid:12) (cid:12)Zm 1 2 (cid:12) ≤ (cid:18)α∈m| 2 |(cid:19)(cid:18)Zm| 1 | (cid:19)
(cid:12) (cid:12) (cid:12) (cid:12)1
(cid:12)(cid:12) (cid:12)(cid:12) (cid:12)(cid:12) sup F2(α) (cid:12)(cid:12) F1(α)2dα Pd+1+ε−2dδ−1. (cid:3)
≤ (cid:18)α∈m| |(cid:19)(cid:18)Z0 | | (cid:19) ≪
3. The Major Arcs
Todealwiththemajorarcsweneedtoobtainanapproximationtothegeneratingfunction
F(α), (2.1), in terms of the auxiliary functions
a
S(q,a) = e f(z) (3.1)
q
zmodq (cid:18) (cid:19)
X
and
I(β) = e(βf(ξ))dξ. (3.2)
Z[0,P]2d+1
Lemma 3.1. For α M(q,a), writing α = (a/q)+β, we have
∈
F(α) = q−2d−1S(q,a)I(β)+O(P2d+2δ).
ON A FORM OF DEGREE d IN 2d+1 VARIABLES (d≥4) 5
Proof. We have
a a
F +β = e +β f(x)
q q
(cid:18) (cid:19) x∈B (cid:18)(cid:18) (cid:19) (cid:19)
X
a
= e +β f(qy+z)
q
zmodq y:qy+z∈B (cid:18)(cid:18) (cid:19) (cid:19)
X X
a
= e f(qy+z) e(βf(qy+z))
q
zmodq y:qy+z∈B (cid:18) (cid:19)
X X
a
= q−4k−1 e f(z) q2d+1e(βf(qy+z)).
q
zmodq (cid:18) (cid:19)y:qy+z∈B
X X
The inner sum is a Riemann sum of the function e(βf(ξ)) over a 2d + 1-dimensional
cuboid which depends on z. We want to replace it by the integral over the 2d + 1-
dimensional cube [0,P]2d+1 and estimate the error in doing so. Firstly, any first-order
partial derivative of e(βf(ξ)) is of the form 2πiβg(ξ)e(βf(ξ)) where g(ξ) is a degree
d 1 form (in 2d+1 real variables) whose coefficients are bounded and which itself is,
−
therefore, O(Pd−1) on the cube [0,P]2d+1. Therefore, e(βf(ξ)) does not vary by more
that O(β qPd−1) on a cube of side q. There are q2d+1 terms in the outer sum and
| |
(P/q)2d+1 terms in each inner sum so the error in replacing the sum by the integral
≪
over the corresponding cuboid is O(β qPd−1q2d+1(P/q)2d+1) = O(β qP3d). Secondly,
| | | |
each side of this cuboid depends on z but is within length q of the sides of [0,P]2d+1.
Since the integrand is bounded by 1 in absolute value, the error in replacing the region
of integration by [0,P]2d+1 is O(qP2d). Since we have q Pδ and β Pδ−d, the total
≤ | | ≤
error is O(P2d+2δ)+O(P2d+δ) = O(P2d+2δ). Hence,
a a
F +β = q−2d−1 e f(z) I(β)+O(P2d+2δ)
q q
(cid:18) (cid:19) zmXodq (cid:18) (cid:19)(cid:16) (cid:17)
which gives the lemma. (cid:3)
Lemma 3.2. We have
F(α)dα = Pd+1S(Pδ)J(Pδ)+O(Pd+5δ)
M
Z
where
q
S(Q) = q−2d−1S(q,a) (3.3)
q≤Q a=1
X X
(a,q)=1
and
J(µ) = e(γf(ξ))dξ dγ. (3.4)
Z|γ|<µ Z[0,1]2d+1 !
Proof. Note that M(q,a) = [(a/q) Pδ−d,(a/q)+Pδ−d] has length 2Pδ−d. The change
−
of variable α= (a/q)+β and lemma 3.1 give
Pδ−d
F(α)dα = q−2d−1S(q,a) I(β)dβ +O(Pd+3δ).
ZM(q,a) Z−Pδ−d
6 MANOJVERMA
Summing over 1 q Pδ, 1 a q, (a,q) = 1, we get
≤ ≤ ≤ ≤
Pδ−d
F(α)dα = S(Pδ) I(β)dβ +O(Pd+5δ).
ZM Z−Pδ−d
Applying the change of variable γ = Pdβ, the integral in the above equation becomes
Pδ
P−d I(P−dγ)dγ.
Z−Pδ
Since f is a homogeneous function of degree d, the change of variable ξ′ = P−1ξ gives
(P−dγ) = e(γP−df(ξ))dξ = e(γf(P−1ξ))dξ
Z[0,P]2d+1 Z[0,P]2d+1
= P2d+1 e(γf(ξ′))dξ′
Z[0,1]2d+1
which completes the proof of the lemma. (cid:3)
4. The Singular Integral
Note that the inner integral appearingin the definition of J(µ), (3.4), is boundedtrivially
by 1. Applying the change of variable ξ′ = γ1/2kξ we see that
e(γf(ξ))dξ = γ−2dd+1 e(f(ξ′))dξ′.
1
Z[0,1]2d+1 Zγd[0,1]2d+1
We think of the last integral as an iterated integral and apply Lemma 7.3 of Vaughan [6]
to each iteration to get the bound
e(γf(ξ))dξ min(1,γ−2dd+1). (4.1)
Z[0,1]2d+1 ≪
Lemma 4.1. For µ 1,
≥
J(µ) = J +O(µ−d+d1)
where
∞
J = e(γf(ξ))dξ dγ >0.
Z−∞ Z[0,1]2d+1 !
Proof. The convergence of J(µ) to J as µ at the rate asserted follows from (4.1).
→ ∞
To show that J > 0, we imitate [2, Chapter 4]. Writing α for (α ,...,α ) and α for
1 1 d 2
(α ,...,α ),
d+1 2d
λ
J = lim e(γ(f (α )+f (α ) αd )dα dα dα dγ.
λ→∞Z−λ Z[0,1]2d+1 d 1 d 2 − 2d+1 1 2 2d+1!
Since 01e(−γαd2d+1)dα2d+1 = 01 d1αd1−1e(−γα)dα and −λλe(γµ)dγ = sin(2πλµ)/πµ for
any real numbers λ and µ,
R R R
J = d1 λl→im∞Z[0,1]2d+1α1d−1(cid:18)sin(π2π(fλd((fαd1()α+1)f+d(fαd2()α−2)α−)α))(cid:19)dα1dα2dα.
ON A FORM OF DEGREE d IN 2d+1 VARIABLES (d≥4) 7
Putting β = f (α )+f (α ) α so that α = f (α )+f (α ) β,
d 1 d 2 − d 1 d 2 −
1 2⌊d/2⌋+1 sin(2πλβ)
J = lim φ(β) dβ
dλ→∞Z−1 πβ
where
φ(β) = d1 Z{β<fd(α1)+fd(α2)<β+1}∩[0,1]2d(fd(α1)+fd(α2)−β)d1−1dα1dα2.
Fourier’s integral theorem for a finite interval is applicable and gives
J = φ(0) = (fd(α1)+fd(α2))d1−1dα1dα2.
Z{0<fd(α1)+fd(α2)<1}∩[0,1]2d
Since the region of integration is a non-empty open set and the integrand is positive
throughout the region of integration, J > 0. (cid:3)
5. The Singular Series
From the definition of S(Q), (3.3),
q
S(Q) = S(q) = q−2d−1 S (q,a)2S (q,a) (5.1)
1 2
q≤Q q≤Q a=1
X X X
(a,q)=1
where
q q
a a
S (q,a) = e f (x ,...,x ) , S (q,a) = e xd
1 d 1 d 2
q −q
x1,.X..,xd=1 (cid:18) (cid:19) Xx=1 (cid:18) (cid:19)
and
q
S(q) = q−2d−1 S (q,a)2S (q,a).
1 2
a=1
X
(a,q)=1
S(q) is a multiplicative function of q. In fact, if (q ,q ) = (a,q q ) = 1 and we choose
1 2 1 2
a ,a so that a q +a q = 1 then for i= 1,2
1 2 2 1 1 2
S (q q ,a) = S (q q ,a q +a q ) = S (q ,a )S (q ,a ). (5.2)
i 1 2 i 1 2 1 2 2 1 i 1 1 i 2 2
This follows from the fact that as r and r run through a complete (or reduced) set of
1 2
residues modulo q and q respectively, r q +r q runs through a complete (or reduced)
1 2 2 1 1 2
set of residues modulo q q . Thus it suffices to study S (q,a) when q is a prime power
1 2 i
and (a,q) = 1. For a prime power pl and p ∤ a we have
pl pl auy2 pl auz2
S (pl,a) = N (p,l,u) e e . (5.3)
1 d−2  pl pl 
u=1 y=1 (cid:18) (cid:19)z=1 (cid:18) (cid:19)
X X X
 
where
N (p,l,u) = # (x ,...,x ): 1 x ,...,x pl,f (x ,...,x ) u(mod pl) .
d−2 1 d−2 1 d−2 d−2 1 d−2
{ ≤ ≤ ≡ }
For a fixed u with pj u, writing u = pju and dividing the sum over y (or z) into pj
1
k
subsums by grouping pl−j consecutive terms together we get
pl auy2 pl auz2 pj pl−j au (pl−jv +v )2 pl−j au v2
e = e = e 1 1 2 = pj e 1 2 .
pl pl pl−j pl−j
Xy=1 (cid:18) (cid:19) Xz=1 (cid:18) (cid:19) vX1=1vX2=1 (cid:18) (cid:19) vX2=1 (cid:18) (cid:19)
8 MANOJVERMA
The sum over v is a quadratic Gauss sum modulo pl−j and its value is well known (e.g.,
2
see [3, Chapter 9]). We find that
p2l if j = l
pl auy2 pl auz2  pl+j if p ≡ 1(mod 4)
e e = ( 1)l+jpl+j if p 3(mod 4) = G(p,l,j) ,
Xy=1 (cid:18) pl (cid:19)Xz=1 (cid:18) pl (cid:19)  0− if p =≡ 2, j =l−1
2 √ 1 2l+j if p = 2, j l 2
 · − · ≤ −

say, depends on p,l and j but not on a or u1. Thus, from (5.3),
l−1
S (pl,a) = p2lM (p,l,l)+ M∗ (p,l,j)G(p,l,j) (5.4)
1 d−2 d−2
j=0
X
where, for d 3, l 1, 0 j l,
≥ ≥ ≤ ≤
M (p,l,j) := # (x ,...,x ): 1 x ,...,x pl, pj f (x ,...,x ) .
d−2 1 d−2 1 d−2 d−2 1 d−2
{ ≤ ≤ | }
and, for d 3, l 1, 0 j l 1,
≥ ≥ ≤ ≤ −
M∗ (p,l,j) := # (x ,...,x ) :1 x ,...,x pl, pj f (x ,...,x )
d−2 { 1 d−2 ≤ 1 d−2 ≤ k d−2 1 d−2 }
= M (p,l,j) M (p,l,j +1).
d−2 d−2
−
The equation (5.4) also holds for d= 1 and d= 2 if we define
M (p,l,j) =0 for 0 j l, M∗ (p,l,j) = 0 for 0 j l 1,
−1 ≤ ≤ −1 ≤ ≤ −
1 if j = 0 1 if j = 0
f = 1, M (p,l,j) = and M∗(p,l,j) = .
0 0 0 if 1 j l 0 0 if 1 j l 1
(cid:26) ≤ ≤ (cid:26) ≤ ≤ −
It is clear that
M (p,l,j) = pl−j for 0 j l and M∗(p,l,j) = pl−j(1 p−1) for 0 j l 1.
1 ≤ ≤ 1 − ≤ ≤ −
The treatment of the singular series is trivial in cases d = 1,2,3 now; we dispose them
off. When d = 1 we have
S (pl,a) = 0; S (q,a) = S (q,a) =S(q) = 1 if q = 1, 0 if q > 1; S= 1. (5.5)
1 1 2
When d = 2 we have
S1(pl,a) = G(p,l,0); S1(q,a) 2q; S2(q,a) 2q; S(q) 4√2q−23. (5.6)
| | ≤ | | ≤ | | ≤
When d = 3 we have p
S1(pl,a) < (l+1)p2l; S1(q,a) < q2d(q); S2(q,a) q2/3; S(q) q−43+ε. (5.7)
| | | | | ≪ | | ≪
To deal with the cases d 4, we first calculate M (p,l,j) = # (w,x) : 1 w,x
2
≥ { ≤ ≤
pl, pj w2 + x2 for 1 j l. We do so by dividing the set (w,x) : 1 w,x
| } ≤ ≤ { ≤ ≤
pl, pj w2+x2 according to the highest power of p dividing w. Thus, for 1 j l,
| } ≤ ≤
M2(p,l,j) = # (w,x) :1 w,x pl, p⌊j+21⌋ w, pj w2+x2
{ ≤ ≤ | | }
⌊j−1⌋
2
+ # (w,x) : 1 w,x pl, pm w, pj w2+x2
{ ≤ ≤ k | }
m=0
X
⌊j−1⌋
2
= p2l−2⌊j+21⌋+ # (w′,x′): 1 w′,x′ pl−m, p ∤w′, pj−2m w′2+x′2 .
{ ≤ ≤ | }
m=0
X
Since 1 has two square roots modulo pj−2m if p 1 (mod 4), one square root modulo
− ≡
pj−2m if pj−2m = 2, and no square root in other cases, the mth term in the sum equals
ON A FORM OF DEGREE d IN 2d+1 VARIABLES (d≥4) 9
pl−m(1 1) 2 p(l−m)−(j−2m) = 2p2l−j(1 1) if p 1 (mod 4), 22l−j−1 if pj−2m = 2 and
− p · · − p ≡
zero otherwise. Thus, for 1 j l,
≤ ≤
p2l−2⌊j+21⌋+2 j+1 p2l−j(1 1) if p 1(mod 4)
⌊ 2 ⌋ − p ≡
M2(p,l,j) =  p2l−2⌊j+21⌋ if p 3(mod 4) .
 22l−j if p =≡ 2


Hence, for 0 j l 1,
≤ ≤ −
(j +1)p2l−j(1 1)2 if p 1(mod 4)
− p ≡
p2l−j(1 1 ) if p 3(mod 4) and j is even
M∗(p,l,j) =  − p2 ≡ .
2
 0 if p 3(mod 4) and j is odd
 22l−j−1 if p =≡ 2



Thus for 0 j l 1 we have M∗(2,l,j) = 22l−1−j, M∗(p,l,j) < p2l−j if p 3(mod 4),
≤ ≤ − 2 2 ≡
and M∗(p,l,j) < (j +1)p2l−j for all p. Also, M (p,l,l) pl if p = 2 or p 3(mod 4),
2 2 ≤ ≡
and M (p,l,l) < (l+1)pl for all p. To estimate S (q,a) we consider the cases of even d
2 1
and odd d separately.
Now suppose that d is even and d 4, say d = 2k and k 2. Then, for 0 j l 1,
≥ ≥ ≤ ≤ −
k−1
M∗ (p,l,j) = M∗(p,l,j )
2k−2 2 i
0≤Xji≤l−1 Yi=1
j1+...+jk−1=j
Thus
k−1
M∗ (2,l,j) = 22l−1−ji
2k−2
0≤Xji≤l−1 Yi=1
j1+...+jk−1=j
= 2(k−1)(2l−1)−j
0≤Xji≤l−1
j1+...+jk−1=j
lk−12(k−1)(2l−1)−j lk−122kl−2l−1−j.
≤ ≤
If p 3(mod 4), then
≡
k−1
M∗ (p,l,j) < p2l−ji
2k−2
0≤Xji≤l−1 Yi=1
j1+...+jk−1=j
= p2(k−1)l−j
0≤Xji≤l−1
j1+...+jk−1=j
lk−1p2(k−1)l−j = lk−1p2kl−2l−j.
≤
10 MANOJVERMA
If p 1(mod 4), then
≡
k−1
M∗ (p,l,j) < (j +1)p2l−ji
2k−2 i
0≤Xji≤l−1 Yi=1
j1+...+jk−1=j
k−1
= p2(k−1)l−j (j +1)
i
0≤Xji≤l−1 Yi=1
j1+...+jk−1=j
< lk−1p2(k−1)l−j 1
0≤jXi≤l−1
j1+...+jk−1=j
< lk−1lk−1p2(k−1)l−j = l2k−2p2kl−2l−j.
Dividing the (2k 2)-tuples (x ,...,x ) with pl f (x ,...,x ) into the ones
1 2k−2 2k−2 1 2k−2
− |
with one of the factors x2 +x2 divisible by pl and the ones with none of the factors
2i−1 2i
x2 +x2 divisible by pl we see that for any prime p
2i−1 2i
k−1
k 1
M (p,l,l) − M (p,l,l) (p2l)k−2+ M∗(p,l,j )
2k−2 ≤ 1 2 · 2 i
(cid:18) (cid:19) 0≤Xji≤l−1 Yi=1
j1+...+jk−1≥l
< (k 1)(l+1)pl p2kl−4l+l2k−2p2kl−2l−l
− ·
< (l+1)2k−2p2kl−3l.
(The last inequality follows since (k 1)(l+1)+l2k−2 (k 1)(2l)+l2k−2 = 2k−2 l+
− ≤ − 1 ·
l2k−2 < (l+1)2k−2.) Thus from (5.4),
(cid:0) (cid:1)
S (pl,a) < (l+1)(l+1)2k−2p2kl−l = (pl)2k−1d(pl)2k−1.
1
| |
Thus from (5.2) for even d 4 and (a,q) = 1,
≥
S (q,a) q2k−1d(q)2k−1 q2k−1+ε = qd−1+ε. (5.8)
1
| | ≤ ≪
Now supposethat d is odd and d 5, say d= 2k+1 and k 2. Note that there is one
≥ ≥
value of x modulo pl that is divisible by pl while for 0 j l 1 there are pl−j(1 p−1)
1
≤ ≤ − −
values of x modulo pl that are exactly divisible by pj. Hence, for 0 j l 1,
1
≤ ≤ −
k−1
M∗ (p,l,j) = pl−j0(1 p−1) M∗(p,l,j ).
2k−1 − 2 i
0≤Xji≤l−1 Yi=1
j0+...+jk−1=j
Thus
k−1
M∗ (2,l,j) = 2l−j0(1 2−1) 22l−1−ji
2k−1 −
0≤Xji≤l−1 Yi=1
j0+...+jk−1=j
1
= 2l+(k−1)(2l−1)−j
2 ·
0≤Xji≤l−1
j0+...+jk−1=j
lk2l+2kl−2l−(k−1)−j−1 lk22kl−l−j−2.
≤ ≤