Waring’s Problem

Lectured by V. R. Neale

Easter Term 2012

Course description

This course is intended as an introduction to one of the central topics of additive numbertheory, namely Waring’s problem and its solution using the Hardy-Littlewood circle method.

The course is tentatively expected to cover the following topics.

Introduction to Waring’s problem. The application of the Hardy-Littlewood circle methodto the solution of Waring’s problem. Weyl’s inequality and Hua’s lemma. The applicationof Weyl’s inequality to Diophantine approximation. Improved bounds for Waring’s problemfor higher powers. Applications of the circle method to related problems.

There are a number of references for Waring’s problem and the circle method, including thebooks listed below.

References

1. H. Davenport, Analytic methods for Diophantine equations and Diophantine inequali-

ties, Cambridge Mathematical Library, Cambridge University Press, Cambridge, 2005.

2. H. Iwaniec and E. Kowalski, Analytic Number Theory, Colloquium Publications 53,American Mathematical Society, Providence, Rhode Island, 2004.

3. R.C. Vaughan, The Hardy-Littlewood Method, Second edition, Cambridge Tracts inMathematics 125, Cambridge University Press, Cambridge, 1997.

4. Ben Green’s lecture notes for Part III Additive number theory, Chapter 3

Last updated: Mon 28th May, 2012

In progress! So please let me know of any corrections: glt1000@cam.ac.uk

IntroductionLecture 1

Our starting point is the following famous result of Lagrange.

Lagrange. Every positive integer can be written as the sum of four squares. (Note that weinclude 0 as a square.)

This is potentially a surprising result, but once one believes it one might naturally ask whethersomething similar holds for cubes, and so on. In 1770, Waring asserted (in Latin) that everypositive integer is a sum of 4 squares, or 9 cubes, or 19 biquadrates, and so on.

↑ ↑fourth powers ???

This gave rise to Waring’s problem, which is to prove the following.

Conjecture (Waring’s Problem). For each k > 2, there exists s = s(k) such that everypositive integer can be written as a sum of s kth powers.

This was first proved by Hilbert in 1909. In the 1920s, Hardy and Littlewood gave a newproof using what is now known as the Hardy-Littlewood circle method. Their proof gave anasymptotic formula for the number of ways to write a large N as a sum of s kth powers.

Since then, there has been a lot of work on applying the circle method to other problems.One notable example is Vinogradov’s proof that every sufficiently large odd integer is a sumof three primes. His slight variant on the circle method is the basis of our approach in thiscourse.

The first key idea is to count the solutions in order to deduce the existence of at least one.We are interested in

∣∣{(x1, . . ., xs) ∈ Ns : N = xk1 + · · ·+ xk

s

}∣∣.

We are interested in the smallest possible s; this is still a topic of research. In fact, there aretwo relevant quantities.

Definition. Let g(k) be the smallest s such that every positive integer is a sum of s kth

powers. Let G(k) be the smallest s such that every sufficiently large positive integer isa sum of s kth powers.

The value of g(k) is almost known for all k, but this is for boring reasons. For example,3k − 1 is difficult to write as a sum of kth powers because the only available summands are1 and 2k. This forces g(k) to be artificially large (at least for some k).

The value of G(k) is more interesting and much less well understood. We know that G(2) = 4and that G(4) = 16, and otherwise we have upper and lower bounds that do not meet.

Our aim here is to obtain an upper bound for G(k), but it will not be very good.

We will count solutions using an integral. We have

∣∣{(x1, . . ., xs) ∈ Ns : N = xk1 + · · ·+ xk

s

}∣∣ =∫ 1

0

(N1/k∑

n=0

e2πiθnk

)s

e−2πiθN dθ.

1

We are asking for an asymptotic formula of the form

S(N)Nsk−1 + o(N

sk − 1).

For large enough N the main term will dominate and this will allow us to deduce the existenceof solutions.

Let’s look at the main term. A heuristic argument (thinking about the number of kth

powers available for representing numbers up to N) suggests that Nsk−1 is the right order of

magnitude.

The ‘singular series’ S(N) records the fact that some numbers can be represented in moreways than others. It turns out that ‘local behaviour’ (behaviour modulo small numbers) isthe important thing. We shall spend some time analysing S(N) in order to show that it isbounded below by a positive constant, uniformly in N .

Before that, we must estimate the integral to arrive at the asymptotic formula. We muststudy the sum. There are two types of behaviour.

The trivial bound for the sum is N1/k. This is our benchmark for comparison.

Examples.

• If θ = 1/3 then θnk just depends on n modulo 3. We do not expectmuch cancellation in the sum; its size will be similar to N1/k. This istrue for θ a rational with small denominator, and, moreover, for θ veryclose to a rational with small denominator.

-q @Iq@Iq-q @Iq@Iqq

• If θ = 1/√

2 then {θnk} does not seem to favour one direction over another. Weexpect lots of cancellation in the sum; its size will be much smaller than N1/k.This is true for θ not close to a rational with small denominator.

We shall therefore split the range of integration into two types of point:

• the ‘major arcs’ (θ close to a rational with small denominator) – these will give themain term, plus an error; and

• the ‘minor arcs’ (the rest) – these will contribute only to the error term.

We shall analyse the major and minor arcs separately, starting with the minor arcs.

Motto. ‘If it looks difficult, then approximate!’

Notation.Lecture 2

• For real θ, we write e(θ) = e2πiθ.

• We are interested in writing N as a sum of kth powers, so we should concentrateon kth powers up to N . We write [N1/k] for {n : 0 6 n 6 ⌊N1/k⌋}.

↑ ↑slightly non-standard floor function

Let S = {nk : 0 6 n 6 N1/k} = {nk : n ∈ [N1/k]}.• Write 1X for an indicator function. For example, 1S denotes the indicator function

of the set S, so 1S(x) =

{1 if x ∈ S0 if x /∈ S

2

• We write 1S for the Fourier transform of 1S : define

1S(θ) =∑

n∈S

e(nθ) =∑

n∈[N1/k]

e(θnk).

• For functions f and g, write f = O(g), or f ≪ g, if there is a positive constant Csuch that |f(x)| 6 Cg(x) for all x (“f grows at most as fast as g”.) We can addsubscripts to indicate dependence on other variables.

For example,

q/2∑

n=1

1

n= O(log q), and log x = Oε(x

ε) for any ε > 0, also written as

log x≪ε xε.

We write f = o(g) iff(x)

g(x)→ 0 as x→∞. (“f grows more slowly than g”.)

For example,1

x= o(1), and N

sk−1−δ = o(N

sk−1) for any δ > 0.

• We write ‖λ‖ for the distance from λ to the nearest integer. That is, ‖λ‖ =min{|λ− n| : n ∈ Z}.

We are interested in counting solutions.

Lemma 1. #{(x1, . . ., xs) ∈ [N1/k]s : N = xk

1 + · · ·xks

}=

∫ 1

0

1S(θ)se(−Nθ) dθ.

Proof. We can rewrite the left-hand side as

LHS = #{(x1, . . ., xs) ∈ [N1/k]s : N = xk

1 + · · ·xks

}

=∑

x1,...,xs∈[N1/k]

1{N=xk1+···+xk

s}

= 1S ∗ · · · ∗ 1S︸ ︷︷ ︸s times

(N) (convolution)

=

∫ 1

0

1S∗· · ·∗1S(θ)e(−Nθ) dθ (by Fourier inversion)

=

∫ 1

0

1S(θ)se(−Nθ) dθ2

Remark. Alternatively, we could rewrite the right-hand side:

∫ 1

0

1S(θ)se(−Nθ) dθ =∑

x1,...,xs∈[N1/k]

∫ 1

0

e(θ(xk

1 + · · ·xks −N)

)dθ,

and then we could use the fact that

∫ 1

0

e(mθ) dθ =

{1 if m = 00 if m ∈ Z \ {0}

We need to define the major and minor arcs. Remember that the major arcs are shortintervals around rationals with small denominator – these contribute the most to the integral.

Definition. We define the major arc around the point a/q to be the set

Ma,q = {θ ∈ [0, 1) : ‖θ − a/q‖ 6 T/N}.↑

points very close to 1 are in the same major arc as points very close to 0

3

We define the major arcs M to be M =⋃

16q6Q

⋃

a mod q(a,q)=1

Ma,q

and the minor arcs m to be m = [0, 1) \M.

Here, Q and T are parameters that we shall choose later; we shall obtain variousconstraints on them. (Think of them both as Nγ for some small γ > 0.)

Remark. There is some flexibility in the choice of major and minor arcs; they did not haveto be exactly as we have defined them. A good rule of thumb when using the circlemethod is to develop techniques for dealing with the major arcs, and then to make themajor arcs as large as those techniques allow, since the minor arcs tend to be the moreproblematic part.

We want to know that the major arcs are disjoint, because when we estimate the integral wewant to sum the estimates for the integrals over each separate major arc.

Lemma 2. Suppose that Q−2 > 4N−1T . Then the major arcs are disjoint.

Proof. We have Ma,q = {θ ∈ [0, 1) : ‖θ− a/q‖ 6 T/N}, so each major arc has width 2T/N .The distance between the centres of the distinct major arcs Ma,q and Ma′,q′ is

∥∥∥∥a

q− a′

q′

∥∥∥∥ >1

qq′>

1

Q2since 1 6 q, q′ 6 Q.

So as long as Q−2 > 4N−1T , we are done. 2

↑constant is not important

Plan. Minor arcs, then major arcs, then singular series.

We shall take the ‘basic route’ for now, since this gives strategies that are applicable to otherproblems. We shall then prove some ‘upgrades’ that lead to better results.

The Minor Arcs

Our task in this section is to prove that

∫

m

1S(θ)se(−Nθ) dθ = o(Nsk−1).

We shall in fact prove that

∫

m

|1S(θ)|s dθ = o(Nsk−1) for large enough s.

We do this by obtaining an upper bound for∣∣1S(θ)

∣∣ =∣∣∑

n∈[N1/k] e(θnk)∣∣ for θ ∈ m.

We shall use Weyl’s inequality, which is a quantitative statement that if θ is not close to arational with small denominator then θnk is pretty uniformly distributed modulo 1.

Theorem 3 (Weyl’s Inequality). Let f(x) = αxk + αk−1xk−1 + . . . + α1x + α0 be a real

polynomial of degree k with leading coefficient α.

Suppose that α has a rational approximation a/q, where q is positive, a and q are

coprime, and∣∣∣α− a

q

∣∣∣ 61

q2.

4

Then for any P and any ε > 0 we have

∣∣∣∣∣

P∑

x=1

e(f(x)

)∣∣∣∣∣≪ε P 1+ε

(P−1/K + q−1/K +

(P k

q

)−1/K)

,

where K = 2k−1 and the implied constant depends only on ε and k.

Remarks.

(i) We shall think of k as fixed throughout the course, so we do not record dependenceof implicit constants on k.

(ii) We shall apply Weyl’s inequality with P = N1/k.

(iii) The trivial bound is P . If P δ 6 q 6 P k−δ for some δ > 0, then we get animprovement over the trivial bound. Indeed, we have

∣∣∣∣∣

P∑

x=1

e(f(x)

)∣∣∣∣∣≪ε P 1+ε

(P−1/K + P−δ/K + P−δ/K

)

If q is small, then we do not expect a saving over the trivial bound, because in thiscase α is close to a rational with small denominator and we then expect major arcbehaviour.

Proof. We start by studying |∑x e(λx)|, where λ is fixed and the sum is over x lying in aninterval of length at most P .

This is a geometric progression: we have trivial bound from the triangle inequality

ւ∣∣∣∣∑

x

e(λx)

∣∣∣∣ 62

|1− e(λ)| ≪1

‖λ‖↑

standard useful fact:

����

�� TT

This is not useful if λ is an integer, or is very close to an integer, but in that case wecan use the trivial bound of P .

The key idea is that repeated differentiation of a polynomial leads to a linear polyno-mial.

We saw last time that for fixed λ, if x ranges over an interval of length at most P , thenLecture 3 ∣∣∑x e(λx)

∣∣≪ min(P, ‖λ‖−1).

Idea: use induction on k, and repeated ‘differentiation’.

It will be convenient to allow sums to be over intervals of length at most P . Define

Sk(f) =

P2∑

x=P1+1

e(f(x)

),

where the subscript k records the degree of the polynomial f , and the only importantthing about P1 and P2 is that P2 − P1 6 P . We do not record the precise values.

5

We have

∣∣∣∣∣

P2∑

x=P1+1

e(f(x)

)∣∣∣∣∣

2

=

P2∑

x1,x2=P1+1

e(f(x2)− f(x1)

)(complex conjugates – useful idea)

= P2 − P1︸ ︷︷ ︸x2=x1

+ 2Re∑

x1<x2

e(f(x2)− f(x1)

)

↑makes no difference for now

Write y = x2−x1, so 1 6 y 6 P2−P1, and f(x2)−f(x1) = f(y+x1)−f(x1) = ∆yf(x1).(‘Differentiation’ – ∆yf is a polynomial of degree k − 1.)

So x ranges over a subinterval of (P1, P2),

depending on y (it might be empty)

↓

|Sk(f)|2 6 P + 2Re

P∑

y=1

∑

x

∆ye(f(x)

)

6 P + 2

P∑

y=1

∣∣Sk−1(∆yf)∣∣

↑since ∆yf has degree k − 1

Similarly,

∣∣Sk−1(∆yf)∣∣2 6 P + 2

P∑

z=1

∣∣Sk−2(∆y,zf)∣∣

↑range of summation depends on y and z

We want to use induction, so we want to use our bound for |Sk−1(∆yf)|2 in our boundfor |Sk(f)|2. We use Cauchy-Schwarz. We have (absorbing constants into the ≪)

|Sk(f)|4 ≪ P 2 +

( P∑

y=1

|Sk−1(∆yf)|)2

+ P

P∑

y=1

|Sk−1(∆yf)|

≪ P 2 + P

P∑

y=1

|Sk−1(∆yf)|2

≪ P 2 + PP∑

y=1

(P +

P∑

z=1

|Sk−2(∆y,zf)|)

≪ P 3 + P

P∑

y,z=1

|Sk−2(∆y,zf)|

Continuing inductively, we get

|Sk(f)|2n ≪ P 2n−1 + P 2n−n−1P∑

y1=1

· · ·P∑

yn=1

|Sk−n(∆y1,...,ynf)|

Note that ∆y1,...,yk−1f(x) = k!αy1. . .yk−1x + β for some β. (We expect a linear poly-

nomial after differentiating k − 1 times, so just have to check the coefficient of x.)

6

We use our bound with n = k − 1, and our observation from the start of the proof, toget

|Sk(f)|2k−1 ≪ P 2k−1−1 + P 2k−1−kP∑

y1=1

· · ·P∑

yk−1=1

min(P, ‖k!αy1. . .yk−1‖−1

)

To consider the sum, collect together terms for which k!y1. . .yk−1 has the same value;call this value m. There are at most d(m) possibilities for each yi, and d(m) ≪ε mε,so our inequality becomes ↑ ↑

#divisors standard bound

|Sk(f)|K ≪ε PK−1 + PK−k+εk!P k−1∑

m=1

min(P, ‖αm‖−1

)

We recall that α is close to a/q. Split the range of summation into blocks of q consecutiveterms (the last block may have fewer terms).

There are ≪ P k−1

q+ 1 blocks. The sum over the block M,M + 1, . . .,M + q − 1 is

q−1∑

m=0

min(P, ‖α(M + m)‖−1

).

We have α(M + m) = αM +a

qm + O

(1

q

), since

∣∣∣α− a

q

∣∣∣ 61

q2and 0 6 m < q.

As m ranges from 0 to q− 1, the term am also ranges over 0 to q− 1 modulo q, since aand q are coprime. Write r ≡ am mod q, and write b for the integer nearest to qαM .

Then the sum becomes αM , and an O(1/q) error

↓q−1∑

r=0

min

(P,

∥∥∥∥b

q+

r

q+ O

(1

q

)∥∥∥∥−1)

.

Some values of r will contribute P , but there are at most O(1) (i.e., constant) of them.

The remaining values of r contribute

∥∥∥∥b + r

q+ O

(1

q

)∥∥∥∥≫u

q, where 1 6 u 6 q/2.

So the sum over this block is ≪ P +

q/2∑

u=1

q

u≪ P + q log q.

Thus, using our earlier bound,

|Sk(f)|K ≪ε PK−1 + PK−k−ε

(#blocks︷ ︸︸ ︷

P k−1

q+ 1

)from each block︷ ︸︸ ︷(P + q log q)

≪ε PK+ε

(P−1 + q−1 +

(P k

q

)−1)

(If q > P k then the bound we are aiming for would be worse than the trivial bound,so we may assume that q 6 P k.) 2

7

We can use this to bound∣∣1S(θ)

∣∣ when θ ∈ m.

Lemma 4. Suppose that Q ≫ Nγ and T ≫ Nγ for some γ > 0. If θ ∈ m, then∣∣1S(θ)

∣∣ ≪N1/k−γ′

, where γ′ > 0 depends on γ.

Proof. By Dirichlet (straightforward pigeonhole argument), there is q with 1 6 q 6 N/T

such that∣∣∣θ − a

q

∣∣∣ 61

q

T

N– and so, in particular,

∣∣∣θ − a

q

∣∣∣ 61

q2.

If q 6 Q, then θ is in the major arc Ma,q. But we are considering θ ∈ m, so we mayassume that Q < q 6 N/T .

Apply Weyl’s inequality to the polynomial θxk, with P = N1/k. We get

∣∣∣1S(θ)∣∣∣ =

∣∣∣∣∣∑

n∈[N1/k]

e(θnk)

∣∣∣∣∣

≪ε N1k +ε

(N−1/kK + q−1/K +

(N

q

)−1/K)

, where K = 2k−1

≪ε N1k (1−1/K)+ε + N1/k+εQ−1/K + N1/k+εT−1/K

≪ N1k−γ′

for some γ′ > 0, since Q≫ Nγ and T ≫ Nγ . 2

Our next task is to show that, for large enough s, the contribution from the minor arcs isnegligibly small. Our approach here does not give a good bound for s – it just shows thatthere is some s. Later in the course, we shall work harder to get a good bound.

We now conclude our study of the minor arcs (for now).Lecture 4

Proposition 5 (The contribution from the minor arcs). Suppose that Q ≫ Nγ andT ≫ Nγ for some small γ > 0. Then for sufficiently large s (depending on k and γ),we have ∫

m

1S(θ)se(−Nθ) dθ = o(Nsk−1) ← (error term)

Proof. Our bound on |1S(θ)| in Lemma 4 was only a small saving over the trivial bound.

We can save a little more by looking at∫ 1

0|1S(θ)|2 dθ.

We have

∫ 1

0

|1S(θ)|2 dθ =∑

x1,x2∈[N1/k]

∫ 1

0

e(θ(xk

2 − xk1))dθ = N1/k

↑ ↑zero only

if x1 = x2

big saving

over N2( 1k−γ′)

(Parseval’s identity in Fourier analysis.)

8

Then we have∣∣∣∣∫

m

1S(θ)se(−Nθ) dθ

∣∣∣∣ 6

∫

m

|1S(θ)|s dθ

6 supθ∈m

|1S(θ)|s−2

∫ 1

0

|1S(θ)|2 dθ

≪ N ( 1k−γ′)(s−2)N1/k

ր տmay as well extend the

range of integration

standard strategy:

pull out most and use

Parseval on what’s left

So we want to ensure that

(1

k− γ′

)(s− 2) +

1

k<

s

k− 1 ⇐⇒ −γ′s− 2

k+ 2γ′ +

1

k< −1

⇐⇒ sγ′ > 2γ′ − 1

k+ 1

⇐⇒ s > 2 +1

γ′

(1− 1

k

)

↑γ′ depends on γ

So as long as s is large enough, we obtain the required bound. 2

The Major Arcs

We want to study 1S(θ) where θ ∈ M. Recall that θ ∈ Ma,q precisely when θ = a/q + twhere |t| 6 T/N . So we want to estimate (“whenever it looks difficult, approximate”)

1S(a/q + t) =∑

n∈[N1/k]

e

((a

q+ t

)nk

).

We shall start by looking at θ = a/q. We shall want to work with subintervals of [N1/k], sothat we can treat tnk as being approximately constant.

We need one small piece of notation. Define

S(a, q) = En(mod q) e

(a

qnk

)=

1

q

q−1∑

n=0

e

(a

qnk

)

The trivial bound for |S(a, q)| 6 1, so this is our benchmark for comparison.

Lemma 6. Let I ⊆ [N1/k] be an interval. Then

∑

n∈I

e

(a

qnk

)= |I|S(a, q) + O(q).

Proof. Idea: e(

aq nk

)depends only on the congruence class of n modulo q.

9

For r ∈ {0, 1, . . ., q − 1}, let Ir = {n ∈ I : n ≡ r (mod q)}.

Then |Ir| =|I|q

+ O(1). We have

∑

n∈I

e

(a

qnk

)=

q−1∑

r=0

∑

n∈Ir

e

(a

qnk

)=

q−1∑

r=0

|Ir|e(

a

qrk

)

=

q−1∑

r=0

|I|q

e

(a

qrk

)+ O(q) = |I|S(a, q) + O(q)

2

We want to approximate tnk as constant on short intervals, and then to replace a sum by anintegral. First we prove a preliminary lemma about approximating sums by integrals.

Lemma 7. Let f : R→ C be a differentiable function. Then

∣∣∣∣∣

J∑

j=1

f(j)−∫ J

0

f(x) dx

∣∣∣∣∣ ≪ J maxy∈[0,J]

|f ′(y)|+ maxy∈[0,J]

|f(y)|

(A crude bound – we are not going to work hard to prove it.)

Proof. Note that for y ∈ [j − 12 , j + 1

2 ], we have

|f(y)− f(j)| 6 maxx∈[j− 1

2,j+ 1

2]|f ′(x)|,

by the mean value theorem.

We split [0, J ] into intervals of length 1. We get

∣∣∣∣∣

J∑

j=1

f(j)−∫ J

0

f(x) dx

∣∣∣∣∣ 6 J maxy∈[0,J]

|f ′(y)|+ maxy∈[0,J]

|f(y)|

↑allow for endpoints 2

We can now estimate 1S

(a

q+ t

).

Lemma 8. Let θ =a

q+ t be in the major arc Ma,q. Then

1S(θ) = S(a, q)

∫ N1/k

0

e(tyk) dy + O(N1/2kQ1/2T 1/2

)

︸ ︷︷ ︸just what turns up in the proof

Proof. Idea:

• split [N1/k] into intervals on which tnk doesn’t vary much

• approximate tnk by a constant on each interval

• approximate resulting sum by an integral

Let M be a parameter to be chosen later. (The integer nearest to N1/2kQ1/2T−1/2 willturn out to be a good choice.)

10

For 1 6 j 6

⌊N1/k

M

⌋, define

Ij ={(j − 1)M + 1, (j − 1)M + 2, . . ., (j − 1)M + M

}∩ [N1/k].

Then each interval Ij has length M , except possibly the last (which may be shorter).(The single point n = 0 contributes a negligible amount.)

Choose nj = jM to be the representative of interval Ij .

We want to approximate tnk by tnkj on Ij . For n ∈ Ij , we have

|tnk − tnkj | = |t| |n− nj | (nk−1 + . . . + nk−1

j ) ≪ N−1TMNk−1

k ≪ MN−1/kT

The error in a single term is∣∣∣∣e((a

q+ t)nk

)− e

(a

qnk + tnk

j

)∣∣∣∣ =∣∣1− e

(tnk

j − tnk)∣∣

≪ |tnkj − tnk| using |1− e(λ)| ≪ |λ|

≪ MN−1/kT

So the error term when we sum over Ij is not too big:

∑

n∈Ij

e

((a

q+ t

)nk

)=

∑

n∈Ij

e

(a

qnk

j + tnkj

)+ O

(|Ij |MN−1/kT

)

= e(tnkj )|Ij |S(a, q) + O(Q) + O(M2N−1/kT )

(using Lemma 6). So the Fourier coefficient becomes

1S

(a

q+ t

)=

⌊N1/k/M⌋∑

j=1

e(tnkj )|Ij |S(a, q) + O(M−1N1/kQ) + O(M−1N1/kM2N−1/kT )

= S(a, q)M

⌊N1/k/M⌋∑

j=1

e(tnkj ) + O(M) + O(M−1N1/kQ) + O(MT )

↑last Ij may be shorter

(Absorb the O(M) into the O(MT ).)

We now use Lemma 7 to approximate the sum by an integral, using f(x) = e(txkMk).(Recall that nj = jM .) We have

∣∣∣∣∣∣

⌊N1/k/M⌋∑

j=1

e(tjkMk)−∫ ⌊N1/k/M⌋

0

e(txkMk) dx

∣∣∣∣∣∣

≪ M−1N1/k |t|MkM−(k−1)N1k (k−1) + 1

≪ TN−1N + 1 ≪ T

So

1S

(a

q+ t

)= S(a, q)M

∫ ⌊N1/k/M⌋

0

e(txkMk) dx + O(MT ) + O(M−1N1/kQ).

11

Substitute y = Mx in the integral to get

1S(θ) = S(a, q)

∫ N1/k

0

e(tyk) dy + O(MT ) + O(M−1N1/kQ)

Standard strategy: choose parameter(s) to make the error terms approximately equal.

So choose M to make MT and M−1N1/kQ approximately equal.

M being close to N1/2kQ1/2T−1/2 will be good, and then both error terms becomeO(N1/2kQ1/2T 1/2). 2

Our next task is to estimate the contribution from the major arcs. We need some preliminaryLecture 5

lemmas.

Lemma 9. For any real θ, we have

∣∣∣∣∫ 1

0

e(θxk) dθ

∣∣∣∣≪ min(1, |θ|−1/k

).

Proof. The trivial bound gives

∣∣∣∣∫ 1

0

e(θxk) dθ

∣∣∣∣ 6 1 for all θ.

We can concentrate on θ > 1.

We have

∣∣∣∣∫ 1

0

e(θxk) dθ

∣∣∣∣ =1

θ1/k

∫ θ1/k

0

e(yk) dy, via y = θ1/kx.

So it suffices to prove that∣∣∣∫X

0e(yk) dy

∣∣∣≪ 1 for all X > 1. We have

∣∣∣∣∫ X

0

e(yk) dy

∣∣∣∣ 6

∣∣∣∣∫ 1

0

e(yk) dy

∣∣∣∣+∣∣∣∣∫ X

1

e(yk) dy

∣∣∣∣ 6 1 +

∣∣∣∣∫ X

1

e(yk) dy

∣∣∣∣.

But∫ X

1

e(yk) dy =

∫ X

1

1

kyk−1kyk−1e(yk) dy

integrate by parts:

u = 1/kyk−1, dv = kyk−1e(yk)

=

[1

kyk−12πie(yk)

]X

1

+k − 1

k2πi

∫ X

1

1

yke(yk)dy

So ∣∣∣∣∫ 1

0

e(θxk) dθ

∣∣∣∣ ≪1

Xk−1+ 1 +

∫ X

1

1

ykdy ≪ 1

2

We need to compute a certain integral. We do the calculation now, and see later where itarises.

Lemma 10. For s > 2k + 1, we have

∫ ∞

−∞

(∫ 1

0

e(τxk)dx

)s

e(−τ) dτ =Γ(1 + 1/k)s

Γ(s/k),

where Γ is the Gamma function.

Proof. Idea:

12

• Show that the integral is limε→012ε

∣∣{x ∈ [0, 1]s : |xk1 + · · ·+ xk

s − 1| 6 ε}∣∣

(Lebesgue measure)

• Compute this limit.

We shall introduce a useful cut-off (not proved here)∫ ∞

−∞

sin(2πτε)

πτe(τy)dτ = 1|y|6ε

We need some estimates:

sin(2πτε)

2πτε= 1 + O(τε) and

∣∣∣∣sin(2πτε)

τε

∣∣∣∣ 61

|τε|

Also,∣∣∣∣∫

[0,1]se(τ(xk

1 + · · ·+ xks − 1)

)dx1 · · · dxs

∣∣∣∣ =

∣∣∣∣∫ 1

0

e(τxk)dx

∣∣∣∣s

≪ min(1, |τ |−s/k

)

by Lemma 9.

For any ε ∈ (0, 1), we have∫ ∞

−∞

∫

[0,1]se(τ(xk

1 + · · ·+ xks − 1)

)dx1 · · · dxsdτ

=

∫ ∞

−∞

sin(2πτε)

2πτε

∫

[0,1]se(τ(xk

1 + · · ·+ xks − 1)

)dx1 · · · dxsdτ

+O

(∫ ε−1

0

τε

∣∣∣∣∣

∫

[0,1]se(τ(xk

1 + · · ·+ xks − 1)

)dx1 · · · dxs

∣∣∣∣∣ dτ

+

∫ ∞

ε−1

(1− 1

τε

) ∣∣∣∣∣

∫

[0,1]se(τ(xk

1 + · · ·+ xks − 1)

)dx1 · · · dxs

∣∣∣∣∣ dτ

)

=1

2ε

∫

[0,1]s

∫ ∞

−∞

sin(2πτε)

πτe(τ(xk

1 + · · ·+ xks − 1)

)dτdx1 · · · dxs

+O

(∫ 1

0

τεdτ +

∫ ε−1

1

τετ−s/kdτ +

∫ ∞

ε−1

(1− 1

τε

)τ−s/kdτ

)

=1

2ε

∫

[0,1]s1|xk

1+···+xk

s−1|6εdx1. . .dxs + O(ε)

↑using s > 2k + 1

So ∫ ∞

−∞

(∫ 1

0

e(τxk)dx

)s

e(−τ)dτ = limε→0

1

2ε

∫

[0,1]s1|xk

1+···+xk

s−1|6εdx1. . .dxs

Taking the limit inside, integrating over xs and making substitutions show that this is

1

ks

∫ 1

0

· · ·∫ 1

00<y1+...+ys−1<1

[y1. . .ys−1

(1− (y1 + . . . + ys−1)

)] 1k−1

dy1. . .dys−1

and a standard calculation shows that this isΓ(1 + 1/k)s

Γ(s/k). 2

13

We need an easy consequence of Weyl’s inequality.

Lemma 11. Let q be a natural number and let a be coprime to q.

Then |S(a, q)| ≪ε q−1K +ε, where K = 2k−1.

Remark. Later in the course, we shall see that in fact |S(a, q)| ≪ q−1/k, but proving thiswill need rather more work.

Proof. We apply Weyl’s inequality (Theorem 3) with α = a/q, f(x) = (a/q)nk, and P = q.This gives

∣∣∣∣∣

q∑

n=1

e

(a

qnk

)∣∣∣∣∣ ≪ε q1+ε(q−1/K + q−1/K + (qk−1)−1/K

)≪ε q1− 1

K +ε

2

We can now, at last, estimate the contribution from the major arcs.

Proposition 12 (The contribution from the major arcs). Suppose that Q and T sat-isfy the following conditions:

• Q→∞ and T →∞ as N →∞• Q−2 > 4TN−1 (so the major arcs are disjoint)

• Q1/2T 1/2 6 N1/2k (∗) to control error term when raising to sth power

• N−1/2kQ5/2T 3/2 = O(N−γ′

) for some γ′ > 0

• Q2T 1− sk = O(N−γ′

) for some γ′ > 0

(For example, we could take Q = Nγ and T = Nγ for some suitable γ > 0.)

Then, for s > 2k + 1, the integral over the major arcs is

∫

M

1S(θ)se(−Nθ) dθ =Γ(1 + 1/k)s

Γ(s/k)N

sk−1

∞∑

q=1

∑

a (mod q)(a,q)=1

S(a, q)se

(−aN

q

)+ o(N

sk−1)

Proof. Recall from Lemma 8 that if θ =a

q+ t ∈Ma,q then

1S(θ) = S(a, q)

∫ N1/k

0

e(tyk)dy + O(N1/2kQ1/2T 1/2)

We raise this to the sth power, using our assumption (∗) that Q1/2T 1/2 6 N1/2k tocontrol the error term:

1S(θ)s = S(a, q)s

(∫ N1/k

0

e(tyk)dy

)s

+ O(Ns−1

k N1/2kQ1/2T 1/2)

14

Integrating over Ma,q, we get

∫

Ma,q

1S(θ)se(−Nθ) dθ =

∫

|t|6TN−1

S(a, q)s

(∫ N1/k

0

e(tyk)dy

)s

e

((a

q+ t

)N

)dt

+ O

(N

sk− 1

2k Q1/2T 1/2 TN−1︸ ︷︷ ︸length of interval

)

= S(a, q)se

(−a

qN

)∫

|t|6TN−1

(∫ N1/k

0

e(tyk)dy

)s

e(−Nt) dt

+ O(N

sk−1− 1

2k Q1/2T 3/2)

Since the major arcs are disjoint (Lemma 2), we can simply sum. We have

∫

M

1S(θ)se(−Nθ) dθ

=

Q∑

q=1

∑

a (mod q)(a,q)=1

S(a, q)se

(−aN

q

)∫

|t|6TN−1

(∫ N1/k

0

e(tyk)dy

)s

e(−Nt) dt

+ O(N

sk−1− 1

2k Q5/2

︸︷︷︸≪ Q2 terms in the sum

T 3/2)

We make two substitutions in the integrals.Lecture 6

∫

|t|6TN−1

(∫ N1/k

0

e(tyk)dy

)s

e(−Nt) dt

(x = y/N1/k) = Ns/k

∫

|t|6TN−1

(∫ 1

0

e(tNxk)dx

)s

e(−Nt) dt

(τ = Nt) = Nsk−1

∫

|τ |6T

(∫ 1

0

e(τxk)dx

)s

e(−τ) dτ

We have two remaining tasks

• extend the range of integration over τ to R

• extend the range of summation over q to N

The error in the integral if we extend the range of integration is

∣∣∣∣∣

∫

|τ |>T

(∫ 1

0

e(τxk)dx

)s

e(−τ) dτ

∣∣∣∣∣ 6

∫

|τ |>T

∣∣∣∣∫ 1

0

e(τxk)dx

∣∣∣∣s

dτ

≪∫

|τ |>T

|τ |−s/kdτ using Lemma 9

≪ T 1− sk since s > k

15

So the overall error in making this change is at most

≪Q∑

q=1

∑

a (mod q)(a,q)=1

T 1− sk N

sk−1 ≪ N

sk−1Q2T 1− s

k

Incorporating the value of the integral (which we found in Lemma 10), we have

∫

M

1S(θ)se(−Nθ) dθ =Γ(1 + 1/k)s

Γ(s/k)N

sk−1

Q∑

q=1

∑

a (mod q)(a,q)=1

S(a, q)se

(−aN

q

)

+ O(Nsk−1Q2T 1− s

k ) + O(Nsk− 1

2k−1Q5/2T 3/2)

The error incurred in extending the range of summation over q is∣∣∣∣∣∣∣∣

Γ(1 + 1/k)s

Γ(s/k)N

sk−1

∑

q>Q

∑

a (mod q)(a,q)=1

S(a, q)se

(−aN

q

)∣∣∣∣∣∣∣∣

≪ Nsk−1

∑

q>Q

∑

a (mod q)(a,q)=1

|S(a, q)|s ← want a better boundthan the trivial bound

≪ε Nsk−1

∑

q>Q

∑

a (mod q)(a,q)=1

q−sK +ε using our bound from Lemma 11

(K = 2k−1)

≪ε Nsk−1

∑

q>Q

q1− sK +ε

≪ Nsk−1Q1− s

2K since s > 2k + 1

So we have

∫

M

1S(θ)se(−Nθ) dθ =Γ(1 + 1/k)s

Γ(s/k)N

sk−1

∞∑

q=1

∑

a (mod q)(a,q)=1

S(a, q)se

(−aN

q

)

+ O(Nsk−1Q1− s

2K ) + O(Nsk−1Q2T 1− s

k )

+ O(Nsk− 1

2k−1Q5/2T 3/2)

Our assumptions on Q and T show that these error terms are all o(Nsk−1). 2

Definition. The singular series is

S(N) =Γ(1 + 1/k)s

Γ(s/k)

∞∑

q=1

∑

a (mod q)(a,q)=1

S(a, q)se

(−aN

q

).

Theorem 13. For sufficiently large s (depending on k), we have

#{(x1, . . ., xs) : xk

1 + · · ·+ xks = N

}= S(N)N

sk−1 + o(N

sk−1).

Proof. This follows immediately from Propositions 5 and 12. 2

16

The Singular Series

Our task in this section is to prove that S(N) is bounded below by a positive constant,uniformly in N . That is, there is some c > 0 not depending on N such that S(N) > c.

We write S(N) = β∞β(N), where

β∞ =Γ(1 + 1/k)s

Γ(s/k)and β(N) =

∞∑

q=1

∑

a (mod q)(a,q)=1

S(a, q)se

(−aN

q

)

Our work on Lemma 10 shows that β∞ reflects the ‘number’ of solutions to the equationxk

1 + · · · + xks = 1 (and if we incorporate the N

sk−1 term then we are thinking about the

equation xk1 + · · ·+ xk

s = N).

It turns out that β(N) reflects the number of solutions to the congruences nk1 · · · + nk

s ≡ N(mod q) as q varies. If nk

1 · · · + nks = N then we certainly have a solution to the congru-

ence modulo q for all q. What is less obvious is that if we have (many) solutions to thesecongruences, then there is a solution to the equation – roughly speaking, this is a sort of‘local-to-global principle’.

Here is an outline plan for this section:

• Show that we can write β(N) as a product∏

p βp(N), where the local factor βp(N)

reflects the number of solutions to the congruence nk1 · · · + nk

s ≡ N modulo powers ofthe prime p.

• Show that almost all of the βp(N) are very close to 1.

• Show that βp(N) > 0 for all p.

• Deduce that β(N) (and so S(N)) is bounded below by a positive constant, uniformlyin N .

We can show that β(N) is related to counting solutions to congruences.Lecture 7

Lemma 14. We have

β(N) = limQ→∞

En1,...,ns (mod Q)Q 10 (mod Q)

(nk

1 · · ·+ nks −N

),

where the limit is taken over a sequence of Q tending to infinity in such a way thatvp(Q) → ∞ for each prime p (for example, Q(n) = n!), (and this Q is nothing to dowith the major arcs).

17

Proof. For fixed Q, the right hand side is

En1,...,ns (mod Q)Q 10 (mod Q)

(nk

1 + · · ·+ nks −N

)

= En1,...,ns (mod Q)

∑

u (mod Q)

e

(u(nk

1 + · · ·+ nks −N)

Q

)(Fourier inversion)

=∑

u (mod Q)

e

(−uN

Q

)[En (mod Q)e

(unk

Q

)]sSay r = (u,Q),u = ar, Q = qr

=∑

q|Q

∑

16a6q(a,q)=1

e

(−aN

q

)[1

Q

Q∑

n=1

e

(ank

q

)]s

=∑

q|Q

∑

16a6q(a,q)=1

e

(−aN

q

)[Q/q

Q

q∑

n=1

e

(ank

q

)]s

=∑

q|Q

∑

a (mod q)(a,q)=1

e

(−aN

q

)S(a, q)s

Taking the limit over Q as described, this becomes

∞∑

q=1

∑

a (mod q)(a,q)=1

e

(−aN

q

)S(a, q)s = β(N).

2

We want to write this as a product, essentially using the Chinese Remainder Theorem. Foreach prime p and each r > 0, define

M(pr) = En1,...,ns (mod pr)pr 10 (mod pr)

(nk

1 + · · ·+ nks −N

),

a weighted count of the number of solutions to the congruence nk1 + · · ·+ nk

s ≡ N (mod pr).Note that M(p0) = 1 for all p.

Define the local factor βp(N) = limr→∞ M(pr) – we shall check that this limit exists, andthat the βp(N) really are the local factors.

Lemma 15.

(i) For s > 2k + 1 and for each prime p, the limit in the definition of βp(N) exists.Moreover, we have βp(N) = 1 + Oε

(p1− s

K +ε), where K = 2k−1.

“Almost all βp(N) are close to 1.”

(ii) For s > 2k + 1, there is some p0 such that

1

26∏

p>p0

βp(N) 63

2,

i.e. the infinite product converges.

18

Proof. We can express M(pr) using an appropriate exponential sum:

M(pr) = En1,...,ns (mod pr)pr 10 (mod pr)

(nk

1 + · · ·+ nks −N

),

= En1,...,ns (mod pr)

∑

u (mod pr)

e

(u(nk

1 + · · ·+ nks −N)

pr

)

=∑

u (mod pr)

e

(−uN

pr

)[En (mod pr)e

(unk

pr

)]s

=∑

u (mod pr)

e

(−uN

pr

)S(u, pr)s

We have M(pr) = 1 +

r−1∑

j=0

[M(pj+1)−M(pj)

], since M(p0) = 1.

So we want to show that the summands are small enough that this converges. We have

M(pj+1)−M(pj) =∑

u (mod pj+1)

e

(− uN

pj+1

)S(u, pj+1)s

−∑

u (mod pj)

e

(−upN

pj+1

)

1

pj

∑

n (mod pj)

e

(upnk

pj+1

)

s

=∑

u (mod pj+1)

e

(− uN

pj+1

)S(u, pj+1)s

−∑

v (mod pj+1)p|v

e

(− vN

pj+1

)S(v, pj+1)s

=∑

u (mod pj+1)(u,p)=1

e

(− uN

pj+1

)S(u, pj+1)s

We can use our bound from Lemma 11: for u coprime to p, we have∣∣S(u, pj+1)

∣∣ ≪ε

p(j+1)(− 1K +ε), where K = 2k−1. So

∣∣M(pj+1)−M(pj)∣∣ 6

∑

u (mod pj+1)(u,p)=1

∣∣S(u, pj+1)∣∣s

≪ε pj+1p(j+1)(− sK +ε)

≪ε p(j+1)(1− sK +ε)

So the partial sumsr−1∑

j=0

(M(pj+1) −M(pj)

)are bounded above by a convergent geo-

metric series, since s > 2k + 1, and so limr→∞

M(pr) exists.

19

It is now easy to bound∣∣M(pr)− 1

∣∣. We have

∣∣M(pr)− 1∣∣ 6

r−1∑

j=0

∣∣M(pj+1)−M(pj)∣∣

≪ε

r−1∑

j=0

p(j+1)(1− sK +ε)

≪ε p1− sK +ε 1

1− p1− sK +ε

≪ε p1− sK +ε since p > 2

uniformly in r, so βp(N) = 1 + Oε

(p1− s

K +ε).

For (ii), the bound on the infinite product follows immediately, since s > 2k + 1. 2

We still need to prove that β(N) =∏

p

βp(N). Recall that

β(N) =∞∑

q=1

∑

a (mod q)(a,q)=1

e

(−aN

q

)S(a, q)s

and, by our above work,

βp(N) =

∞∑

j=0

∑

u (mod pj)(u,p)=1

e

(−uN

pj

)S(u, pj)s

It is now natural to define A(q) =∑

a(mod q)(a,q)=1

e

(−aN

q

)S(a, q)s.

Note that in particular, A(pj+1) = M(pj+1)−M(pj).

Then β(N) =

∞∑

q=1

A(q) and βp(N) =

∞∑

j=0

A(pj), and this latter sum is absolutely convergent.

(Idea: prove that A is multiplicative, and then done.)

We want to think about multiplicativity.

Lemma 16. The sum S(a, q) is multiplicative, in the sense that if q1 and q2 are coprimenatural numbers, then S(a1, q1)S(a2, q2) = S(a1q2 + a2q1, q1q2) for any a1 coprime toq1, and any a2 coprime to q2.

Remark. This will be helpful when proving an improved bound on |S(a, q)|.

Proof. For each n modulo q1q2, there is a unique pair (n1 modulo q1, n2 modulo q2) suchLecture 8

that

n

q1q2≡ n1

q1+

n2

q2(mod 1) (Chinese remainder theorem)

Then we have

20

S(a1q2 + a2q1, q1q2) = En (mod q1q2)e

((a1q2 + a2q1)n

k

q1q2

)

= En1 (mod q1)n2 (mod q2)

e

((a1q2 + a2q1)(q1q2)

k−1

(n1

q1+

n2

q2

)k)

= En1 (mod q1)n2 (mod q2)

e

((a1q2 + a2q1)

(qk−12 nk

1

q1+

qk−11 nk

2

q2

))

= En1 (mod q1)n2 (mod q2)

e

(a1(q2n1)

k

q1+

a2(q1n2)k

q2

)

= En1 (mod q1)e

(a1n

k1

q1

)En2 (mod q2)e

(a2n

k2

q2

)(q1, q2 coprime)

= S(a1, q1)S(a2, q2) 2

We can use this to show that A, given by A(q) =∑

a(mod q)(a,q)=1

e

(−aN

q

)S(a, q)s, is multiplicative.

Lemma 17. Let q1 and q2 be coprime natural numbers. Then A(q1)A(q2) = A(q1q2) – thatis, A is multiplicative.

Proof. Write f(a, q) for the summand in the definition of A(q), so f(a, q) = S(a, q)se(−aN

q

).

Then, by Lemma 16, we have f(a1, q1)f(a2, q2) = f(a1q2 + a2q1, q1q2) whenever a1 iscoprime to q1 and a2 is coprime to q2. We have a bijection (Z/q1Z)× × (Z/q2Z)× ∼=(Z/q1q2Z)×, and explicitly the pair (a1, a2) corresponds to a1q2 + a2q1. 2

Corollary 18. The βp(N) are the local factors for this problem, in the sense that β(N) =∏p βp(N).

Proof. This follows immediately from the multiplicativity of A and the absolute convergenceof the series for βp(N). 2

Now that we have written β(N) as a product of local factors, we can show that it is boundedbelow by a positive constant, uniformly in N . We already know that for large primes p thefactors βp(N) behave well (we showed that there is p0 such that

∏p>p0

βp(N) >12 ). We

need to show that individually each βp(N) cannot be too small. (We need only check thisfor small p, but it will be no extra work to do it for all p.)

To do this, we remember that βp(N) was defined as the limit of M(pr) as r → ∞, whereM(pr) is a weighted count of the number of solutions to a certain congruence. So we wantto show that this congruence has many solutions.

Plan.

• Find some R such that if mk ≡ c (mod pR) has a solution (where c is a fixedinteger not divisible by p), then for r > R the congruence nk ≡ c (mod pr) alsohas a solution

• Show that if we have a1, . . ., as such that ak1 + · · ·+ ak

s ≡ N (mod pR) and a1 6≡ 0(mod p) then for each r > R there are many solutions to nk

1 + · · · + nks ≡ N

(mod pr)

• Show that there is a solution to ak1 + · · ·+ ak

s ≡ N (mod pR)

21

We start with the first of these. It turns out to be a good idea to distinguish between twocases: p odd and p = 2.

Lemma 19.

(i) Let p be an odd prime and write k = pvk0 where p and k0 are coprime. Let c bean integer not divisible by p. If there is m such that mk ≡ c (mod pv+1), then foreach r > v + 1 there is an n such that nk ≡ c (mod pr).

(ii) Write k = 2vk0 where k0 is odd. Let c be an odd integer. If there is m suchthat mk ≡ c (mod 2v+2), then for each r > v + 2 there is an n such that nk ≡ c(mod 2r)

Proof.

(i) We have m such that mk ≡ c (mod pv+1). Take r > v + 1. The multiplicativegroup (Z/prZ)× is cyclic of order pr−1(p− 1), say with generator g. Note that gis also a generator for (Z/pv+1Z)× since r > v +1. Say c = gγ , m = gα, and wantn = gβ .

We have αk ≡ γ (mod pv(p− 1)), that is, αpvk0 ≡ γ (mod pv(p− 1)). So we cancertainly find β such that βpvk0 ≡ γ (mod pr−1(p− 1)).

(ii) The group (Z/2rZ)× has order 2r−1, so every element has order dividing 2r−1.

If k is odd and xk ≡ 1 (mod 2r), then x ≡ 1 (mod 2r), so as x ranges over allvalues so does xk, so we can certainly solve nk ≡ c (mod 2r).

If k is even, then we have k = 2vk0 where k0 and v > 1. Then if x is odd, wehave xk ≡ (x2)2

v−1k0 ≡ 1 (mod 4), so the kth powers lie in {y : y ≡ 1 (mod 4)} ⊂(Z/2rZ)×. This is cyclic of order 2r−2, generated by 5. The argument is the sameas in (i). 2

We can now show that if we can solve the starting congruence then there are many solutionsto subsequent ones.

Lemma 20.

(i) Let p be an odd prime and write k = pvk0 where p and k0 are coprime. Supposethat we have a1, . . ., as such that ak

1 + · · · + aks ≡ N (mod pv+1), and suppose

moreover that a1 6≡ 0 (mod p). Then for each r > v + 1 there are at leastp(r−v−1)(s−1) solutions to nk

1 + · · ·nks ≡ N (mod pr).

(ii) Write k = 2vk0 where k0 is odd. Suppose that we have a1, . . ., as such thatak1 + · · · + ak

s ≡ N (mod 2v+2), and suppose moreover that a1 is odd. Then foreach r > v + 2 there are at least 2(r−v−2)(s−1) solutions to nk

1 + · · · + nks ≡ N

(mod 2r).

Proof.

(i) Choose n2, . . ., ns modulo pr such that ni ≡ ai (mod pv+1) for 2 6 i 6 s. Thereare p(r−v−1)(s−1) ways of doing this.

By construction, ak1 ≡ N −nk

2 − · · ·−nks (mod pv+1) and we know that this has a

solution with a1 6≡ 0 (mod p), so by Lemma 19 there is some n1 modulo pr withnk

1 ≡ N − nk2 − · · · − nk

s (mod pr).

(ii) Very similar. 2

We are trying to prove that β(N) is bounded below by a positive constant, uniformly inLecture 9

N , and the remaining step of the argument is to show that there are solutions to the ‘basecongruences’.

22

Lemma 21. Let p be a prime, and write k = pvk0 where p and k0 are coprime. For s > 4k,there exist a1, . . ., as such that

{ak1 + · · ·+ ak

s ≡ N (mod pv+1) if p odd

ak1 + · · ·+ ak

s ≡ N (mod 2v+2) if p = 2

}

and a1 6≡ 0 (mod p).

In fact, if k is odd then our argument will show that s > 2k + 1 suffices.

Remark. For k = 4, p = 2 we really do need s = 16, and similarly for other powers of 2.There are many ways to prove the result; our approach here is a mix of Davenport’sbook and Green’s lecture notes.

Proof. If N is coprime to p, then the condition a1 6≡ 0 (mod p) will be satisfied automat-ically. If N is not coprime to p, we instead look at N − 1, and then include 1k as asummand – so we need to allow s to be one larger for this.

Case 1: p = 2

If k is odd, then (as in the proof of Lemma 19) we can solve ak1 ≡ N (mod 2v+2)

and we are done. We can look at 0 < N < 2v+2, and by choosing ai from {0, 1}we can obtain all such N using at most 2v+2 − 1 summands. But 2v+2 6 4k, sos > 4k will do.

Case 2: p is odd

Let A = {nk : n ∈ Z/pv+1Z} and let A× = {nk : n ∈ (Z/pv+1Z)×}. We areinterested in the sumsets ℓA = {a1+· · ·+aℓ : ai ∈ A}, and we want mA = Z/pv+1Zfor some m.

Note that these are invariant under multiplication by elements of A×.

Define an equivalence relation on Z/pv+1Z by: x1 ∼ x2 ⇐⇒ x1 ≡ hx2 (mod pv+1)for some h ∈ A×. Then each ℓA is a union of equivalence classes, and ℓA ⊆ (ℓ+1)A.

How big are the equivalence classes? One is {0}. If x 6≡ 0 (mod pv+1), thenh1x ≡ h2x (mod pv+1) implies that p | h1 − h2, so the number of elements in theequivalence class of x is at least p−1

(k,p−1) , the number of kth powers modulo p.

So for each ℓ, we have either ℓA = (ℓ + 1)A or |(ℓ + 1)A| > |ℓA|+ p−1(k,p−1) .

If ℓA = (ℓ + 1)A for some ℓ, then ℓA = Z/pv+1Z as 1 ∈ A.

If this does not occur for ℓ = 0, 1, . . .,m− 1, then m(p−1)(k,p−1) 6 |mA| 6 pv+1.

So m < 2k. So s > 2k + 1 will suffice. 2

We can collect together the recent results to prove

Lemma 22. For s > 4k, there are constants cp > 0 such that βp(N) > cp > 0 for all N .

23

Proof. By Lemmas 21 and 20, for odd p and r > v + 1, we have

M(pr) = p−rs∑

n1,...,ns (mod pr)

pr 10 (mod pr)(nk1 + · · ·+ nk

s −N)

> p−r(s−1)p(r−v−1)(s−1)

> p(v+1)(1−s)

So βp(N) > p(v+1)(1−s).

Very similarly, we get β2(N) > p(v+2)(1−s). 2

Proposition 23 (singular series). For s > 2k + 1, there is a positive constant c such thatβ(N) > c > 0 for all N .

Proof. This follow immediately from:

• Lemma 15∏

p>p0βp(N) >

12

• Corollary 18 β(N) =∏

p βp(N)

• Corollary 22.2

We have now proved could in principle find an explicit value

↓Theorem 24. Let k > 2 be an integer. Then there is s = s(k) such that

#{(x1, . . ., xs) : xk

1 + · · ·+ xks = N

}= S(N)N

sk−1 + o(N

sk−1)

and there is a constant c > 0 such that S(N) > c > 0 for all N .

Corollary 25. Let k > 2 be an integer. Then there is s = s(k) such that every sufficientlylarge integer can be written as a sum of s kth powers.

We now start to think about improvements to the argument.

At the moment, our bound on s is weak because our argument for the minor arcs led to aweak bound. We shall prove Hua’s lemma, which will allow us to improve this considerably.

We shall first improve our bound on |S(a, q)| – we shall show that if a and q are coprime,then |S(a, q)| ≪ q−1/k (where, as usual, the implied constant may depend on k).

We proved in Lemma 16 that S(a, q) is multiplicative, so we shall concentrate on finding abound for |S(a, pj)| for prime powers pj . We start by looking just at primes, and we shallthen use that to help with prime powers.

Lemma 26. Let p be a prime and let a be coprime to p. Let δ = (k, p−1). Then |S(a, p)| 6(δ − 1)p−1/2. ← good!

Proof. Let g be a generator for (Z/pZ)×. Define a multiplicative character χ : Z/pZ → Cby

χ(gj) = e(j/δ)

χ(0) = 0

24

Then

1 + χ(gn) + · · ·+ χ(gn(δ−1)) =

δ−1∑

j=0

e

(nj

δ

)= δ 1δ|n = δ 1gn is a kth power mod p

So

pS(a, p) =

p−1∑

m=0

e

(amk

p

)

= 1 +

p−1∑

n=1

[1 + χ(gn) + · · ·+ χ

(gn(δ−1)

)]e

(agn

p

)

=∑

x (mod p)

[1 + χ(x) + · · ·+ χ(xδ−1)

]e

(ax

p

)

=

δ−1∑

j=0

∑

x (mod p)

χ(xj)e

(ax

p

)

The term with j = 0 is∑

x(mod p)

e

(ax

p

)= 0, so we concentrate on 1 6 j 6 δ − 1.

We have

∣∣∣∣∣∣

∑

x(mod p)

χ(xj)e

(ax

p

)∣∣∣∣∣∣

2

=∑

x,y (mod p)

χ(xj)χ(yj)e

(a

p(x− y)

).

We can ignore terms with y = 0, since χ(0) = 0.

If y 6= 0, we look at χ((xy−1)j

)and put x ≡ yz (mod p). We have

∣∣∣∣∣∣

∑

x (mod p)

χ(xj)e

(ax

p

)∣∣∣∣∣∣

2

=∑

z (mod p)

χ(zj)∑

y (mod p)y 6=0

e

(a

py(z − 1)

)

︸ ︷︷ ︸p−1 if p|z−1−1 if p∤z−1

= (p− 1)χ(1)−∑

z 6=1

χ(zj)

= pχ(1)︸︷︷︸1

−∑

z

χ(zj)

︸ ︷︷ ︸0

= p

So |S(a, p)| 6 p−1(δ − 1)p1/2. 2

Last time we saw that if p is prime and a is coprime to p, and if δ = (k, p − 1), thenLecture 10

|S(a, p)| 6 (δ − 1)p−1/2. Today we look at S(a, pj).

25

Lemma 27. Let p be prime and let a be coprime to p.

(i) If p ∤ k and 2 6 r 6 k, then S(a, pr) = p−1.

(ii) If r > k, then S(a, pr) = p−1S(a, pr−k), unless p = k = 2.↑

don’t need p ∤ k here

If p = k = 2, then this holds for r > 4, and S(a, 8) = 12e(

a8

).

Proof. We look at r > 2, and have S(a, pr) =1

pr

pr−1∑

n=0

e

(ank

pr

).

Write k = pvk0 where p and k0 are coprime.

We can make a substitution to reduce to a smaller power of p. Write n = pr−v−1x + y,where 0 6 x 6 pv+1 − 1 and 0 6 y 6 pr−v−1 − 1. We are interested in

nk = (pr−v−1x + y)k = yk + kpr−v−1xyk−1 +

k∑

j=2

(k

j

)(pr−v−1x)jyk−j

︸ ︷︷ ︸want this to disappear

modulo p. If v = 0 then each term in the sum over j is divisible by p2(r−1) and so bypr as r > 2. So

nk ≡ yk + k0pr−1xyk−1 (mod pr)

↑k = k0 for v = 0

If v > 1 and r > k then we have r > k > pv > v + 2 unless p = k = 2, so we look atr > v + 3. (If p = k = 2 and r = 3 then it is an easy calculation to find S(a, pr).)

What is the power of p dividing(kj

)?

Certainly pv | k. Also, vp(j!) =∑

m>1

⌊j

pm

⌋< j, so v

((k

j

))> v − (j − 1).

So the jth term in the sum is divisible by pℓ, where ℓ > v − (j − 1) + j(r − v − 1) =r − r + v − j + 1 + j(r − v − 1) = r + (r − v − 1)︸ ︷︷ ︸

>2

(j − 1) − j︸︷︷︸>2

> r.

So nk ≡ yk + k0pr−1xyk−1 (mod pr) in all cases. This gives

S(a, pr) =1

pr

pr−v−1−1∑

y=0

pv+1−1∑

x=0

e

(ayk

pr+

ak0pr−1xyk−1

pr

)

=1

pr

pr−v−1−1∑

y=0

e

(ayk

pr

) pv+1−1∑

x=0

e

(ak0p

r−1xyk−1

p

)

︸ ︷︷ ︸=0 if p∤y

=pv+1 if p|y

(let y = pz) =pv+1

pr

pr−v−2−1∑

z=0

e

(azk

pr−k

)

26

If v = 0 and 2 6 r 6 k, then each summand is 1 and we get

S(a, pr) =pv+1

prpr−v−2 =

1

p.

If r > k, then the summand depends only on the congruence class of z modulo pr−k,so

S(a, pr) =pv+1

pr

pr−v−2

pr−k

pr−k−1∑

z=0

e

(azk

pr−k

)=

1

pS(a, pr−k).

2

We can now prove the promised bound on |S(a, q)|.Lecture 11

Proposition 28. Let q > 2 be an integer and let a be coprime to q. Then |S(a, q)| ≪ q−1/k.

Proof. Write T (a, q) = q1/kS(a, q), so we want to prove that |T (a, q)| ≪ 1. For now, lookat k > 3; we shall return to k = 2 later.

Let q = pr1

1 . . .prnn . Then, by the multiplicativity of S (Lemma 16), we have T (a, q) =∏n

j=1 T (aj , prj

j ) for some suitable aj , each coprime to the corresponding pj .

If rj > k + 1 then by Lemma 27 we have T (aj , prj

j ) = T (aj , prj−kj ), and so we may

assume that rj 6 k for all j.

The contribution to |T (a, q)| from primes dividing k is certainly O(1), so we may ignorethese primes.

If pj ∤ k and rj > 2, then by Lemma 27 we have |T (aj , prj

j )| = p−1j p

rj/kj 6 1.

If rj = 1, then by Lemma 26 we have |T (aj , pj)| 6 kp−1/2j p

1/kj 6 kp

−1/6j for k > 3.

If pj > k6, then |T (aj , pj)| 6 1, and there are at most O(1) primes p with p 6 k6.

So |T (a, q)| ≪ 1.

For k = 2, if q1 is odd then |S(a1, q1)| = q−1/21 (squaring argument very similar to one

we have seen before). An arbitrary q can be written as 2rq1 where q1 is odd. ThenT (a, q) = T (a1, q1)T (a2, 2

r).

As above, Lemma 27 allows us to assume that r 6 3, and it is then easy to check that|T (a2, 2

r)| ≪ 1. 2

Consequences of Proposition 28

We can use this bound wherever we used our previous bound (Lemma 11). In particular:

• Proposition 12: the estimate for the contribution from the major arcs holds for s >

2k + 1 (rather than 2k + 1)

• Lemma 15: the limit in the definition of βp(N) exists and the infinite product convergesfor s > 2k + 1 (rather than 2k + 1)

• Proposition 23: the lower bound on the singular series holds for

{s > 2k + 1 if k odds > 4k if k even

27

We now return to the minor arcs, to show that their contribution is negligible for a widerrange of s. Weyl’s inequality (Theorem 3) gave |1S(θ)| = O(N

1k−γ′

) for θ in the minor arcsm (this was Lemma 4).

We got an extra saving in Proposition 5 using Parseval’s identity:

∫ 1

0

∣∣1S(θ)∣∣2 dθ = #

{(x1, x2) ∈ [N1/k]2 : xk

1 = xk2

}= N1/k

Hua’s lemma gives a bound on higher moments

∫ 1

0

∣∣1S(θ)∣∣2r

dθ.

Lemma 29 (Hua’s lemma). For r ∈ {1, 2, . . ., k}, we have

∫ 1

0

∣∣1S(θ)∣∣2r

dθ ≪ε N1k (2r−r+ε) for any ε > 0

Proof. Let P = N1/k and let Ir =

∫ 1

0

∣∣1S(θ)∣∣2r

dθ. We use induction on r.

For r = 1, we have I1 =

∫ 1

0

∣∣1S(θ)∣∣2 dθ = P .

Suppose that the result holds for some r 6 k − 1.

In the proof of Weyl’s inequality (Theorem 3), we proved, using repeated squaring andCauchy-Schwarz, that

∣∣1S(θ)∣∣2r

≪ P 2r−1 + P 2r−r−1Re

(P∑

y1=1

· · ·P∑

yr=1

Sk−r

(∆y1,...,yr

f))

whereSk−r

(∆y1,...,yr

f)

=∑

x

e(θ∆y1,...,yr

(xk))

↑ ↑differencing sum over a subinterval of

(0, P ] depending on y1, . . ., yr

Multiply by∣∣1S(θ)

∣∣2r

and integrate, to obtain

Ir+1 =

∫ 1

0

∣∣1S(θ)∣∣2r+1

dθ

≪ P 2r−1Ir + P 2r−r−1P∑

y1=1

· · ·P∑

yr=1

Re

(∫ 1

0

Sk−r∆y1,...,yrf∣∣1S(θ)

∣∣2r

dθ

)

We consider the integral, which is

∫ 1

0

∑

x

e(θ∆y1,...,yr

(xk)) ∑

16u1,...,u2r−16P

16v1,...,v2r−16P

e(θ(uk

1 + · · ·+ uk2r−1 − vk

1 − · · · − vk2r−1

))dθ

This counts solutions to

∆y1,...,yr(xk) + uk

1 + · · ·+ uk2r−1 − vk

1 − · · · − vk2r−1 = 0 (∗)

where y1, . . ., yr, x, u1, . . ., u2r−1 , v1, . . ., v2r−1 run from 1 to P .

28

Now ∆y1,...,yr(xk) is an integer divisible by y1, . . ., yr, and it is a strictly increasing

function of x (since r < k). So for a fixed value, there are at most Oε(Pε) values of

y1, . . ., yr and at most one x giving this value.

SoIr+1 ≪ P 2r−1Ir + P 2r−r−1M,

where M counts solutions to (∗) and

M ≪ P 2r

P ε1 .

ր ↑ ↑ui, vj yi x

So Ir+1 ≪ε P 2r−1P 2r−r+ε + P 2r−r−1P 2r+ε ≪ε P 2r+1−(r+1)+ε, as required. 2

Consequences of Lemma 29

• Proposition 5: for s > 2k +1, the contribution from the minor arcs is negligible. Indeed,

∣∣∣∣∫

m

1S(θ)se(−Nθ) dθ

∣∣∣∣ 6

∫

m

∣∣1S(θ)∣∣s dθ

6 supθ∈m

∣∣1S(θ)∣∣s−2k

∫ 1

0

∣∣1S(θ)∣∣2k

dθ

≪ε N ( 1k−γ′)(s−2k)

︸ ︷︷ ︸Weyl

N1k (2k−k+ε)

︸ ︷︷ ︸Hua

and since s > 2k + 1, this is o(Nsk−1).

• Theorem 24: the asymptotic formula holds for s > 2k + 1

• Corollary 25: if s > 2k +1, then every sufficiently large integer can be written as a sumof s kth powers.

A Waring-type problemLecture 12

Rather than trying to write numbers as sums of kth powers, we try to write them as sums ofcertain ‘bracket quadratics’ (or ‘generalised quadratics’):

N = n1⌊n1

√2⌋+ · · ·+ ns⌊ns

√2⌋

↑floor function

These bracket quadratics grow approximately quadratically, but localy they behave ratherdifferently from the squares. For example, modulo q it turns out that n and ⌊n

√2⌋ behave

roughly independently and uniformly: n⌊n√

2⌋ behaves like nm modulo q. In particular,every congruence class can be achieved by a bracket quadratic, so there are no local reasonswhy a number is not a sum of bracket quadratics.

Density considerations (there aren’t enough bracket quadratic) show that we cannot hope towrite every number as a sum of two bracket quadratics, but there is no reason to rule outthe possibility that every sufficiently large integer is a sum of three bracket quadratics.

29

Write A = {n⌊n√

2⌋ : 0 6 n 6√

N}. As with Waring’s problem, we are interested in

1A(θ) =∑

n∈[√

N ]

e(θn⌊n

√2⌋)

.

In Waring’s problem, one can use Weyl’s inequality:

• either {θn2} is approximately equidistributed in R/Z (circle)— minor arc behaviour, θ not close to a rational with small denominator

• or {θn2} is approximately equidistributed at points— major arc behaviour, θ close to a rational with small denominator

We might expect {θn⌊n√

2⌋} not to be equidistributed if θ is close to a rational with smalldenominator. But we need more major arcs. It turns out that

n√

2⌊n√

2⌋ =1

2

(2n2 + ⌊n

√2⌋2 − {n

√2}2)

,

↑fractional part

and so we need major arcs around points of the forma + b

√2

q.

Green and Tao have produced useful results on the quantitative distribution of certain orbitsin nilmanifolds (building on qualitative work by many other people).

Very roughly speaking, a nilmanifold is like a non-abelian torus. In the abelian case, we haveR/Z and more generally Rn/Zn.

Perhaps the simplest non-abelian example is the Heisenberg nilmanifold G/Γ, where

G =

1 R R

1 R1

and Γ =

1 Z Z

1 Z1

.

It turns out that an element

1 x y

1 z1

Γ in G/Γ can have representative

1 {x} {y − x⌊z⌋}

1 {z}1

.

In this way, one can use nilmanifolds to obtain bracket quadratics.

Using the Green-Tao work, it turns out that there are three cases.

1. The nilsequence is approximately equidistributed over thewhole nilmanifold – minor arc behaviour, θ not close to(a + b

√2)/q for small q

2. The nilsequence is equidistributed in a nilsubmanifold – minorarc behaviour, θ not very close to (a + b

√2)/q for small q

3. The nilsequence is approximately equidistributed on points –major arc behaviour, θ very close to (a + b

√2)/q for small q

30

One can analyse this carefully, estimating the contribution from the major arcs either usingnilsequences or directly as we did for Waring’s problem, to obtain the following result: forsufficiently large s, we have

#{n1, . . ., ns ∈ [N1/2] : N = n1⌊n1

√2⌋+ · · ·+ ns⌊ns

√2⌋}

= S(N)Ns2−1 + o(N

s2−1)

where the singular series is

S(N) = 2−s/4 Γ(3/2)s

Γ(s/2)

∞∑

q=1

∞∑

b=−∞

∑

a (mod q)(a,b,q)=1

S(a, b, q)s

(∫ 1

0

e

(− b

2qx2

)dx

)s

e

(−(

a + b√

2

q

)N

),

︸ ︷︷ ︸β∞

︸ ︷︷ ︸β(N)

where S(a, b, q) = En,m(mod 2q)e

(anm

q+

b

2q(2n2 + m2)

).

We can, with some work, interpret β(N) as counting solutions. We might expect to need thecongruences

n1⌊n1

√2⌋+ · · ·+ ns⌊ns

√2⌋ ≡ N (mod q) .

Now n1⌊n1

√2⌋ (mod q) is not very nice, but it is approximately like n1m1, so we might look

at n1m1 + · · ·+ nsms ≡ N (mod q).

But this is not enough. If n1⌊n1

√2⌋ + · · ·ns⌊ns

√2⌋ = N , then certainly the congruence

modulo q holds for all q, but also we have

√2

q

(n1⌊n1

√2⌋+ · · ·+ ns⌊ns

√2⌋)≡ n√

2

q(mod 1) .

Why does this matter here when it did not for the squares? We had n2√

2 is very well-behavedmodulo 1, whereas n

√2⌊n√

2⌋ is not. Using our identity for n√

2⌊n√

2⌋ and approximating,we find that β(N) approximately counts solutions to

n1m1 + · · ·+ nsms ≡ N (mod q)

1

2q

(2n2

1 + · · ·+ 2n2s + m2

1 + · · ·+ m2s − x2

1 − · · · − x2s

) ∼= N√

2

q(mod 1) ,

↑approximate, as congruencesmodulo 1 are rather delicate

where n1, . . ., ns,m1, . . .,ms range modulo q, and x1, . . ., xs range modulo 1.

We then try to write β(N) as a singular product. This needs care, as one does not quite havethe independence one would hope for, but one can prove that

β(N) =

∫ 1

0

· · ·∫ 1

0

χη(−x21 − · · · − x2

s − 2N√

2)︸ ︷︷ ︸mod 1 contribution

∏

p

βp(N) dx1 · · · dxs

where χη is a smoothed indicator function of an interval of width η around integers, and

βp(N) are local factors similar to these for Waring’s problem, but with pairs of congruences.

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