Abstract. We discuss the relationship between various additive problems concerning squares.

PROGRESSIONS AND SUMSETS OF SQUARES

JAVIER CILLERUELO AND ANDREW GRANVILLE

Abstract. We discuss the relationship between various additive problems concerning squares.

1. Squares in arithmetic progression

Let σ(k) denote the maximum of the number of squares in a +b, . . . , a + kb as we vary over positive integers a and b. Erd˝os conjectured that σ(k) = o(k) which Szemer´edi [30] elegantly proved as follows: If there are more than δk squares amongst the integers a +b, . . . , a +kb (where k is sufficiently large) then there exists four indices 1 ≤ i

< i

<

i

< i

≤ k in arithmetic progression such that each a + i

b is a square, by Szemer´edi’s theorem. But then the a + i

b are four squares in arithmetic progression, contradicting a result of Fermat. This result can be extended to any given field L which is a finite extension of the rational numbers: From Faltings’ theorem we know that there are only finitely many six term arithmetic progressions of squares in L, so from Szemer´edi’s theorem we again deduce that there are o

(k) squares of elements of L in any k term arithmetic progression of numbers in L. (Xavier Xarles [31] recently proved that are never six squares in arithmetic progression in Z[ √

d] for any d.)

In his seminal paper Trigonometric series with gaps [27] Rudin stated the following conjecture:

Conjecture 1. σ(k) = O(k

).

It may be that the most squares appear in the arithmetic progression −1 + 24i, 1 ≤ i ≤ k once k ≥ 8 yielding that σ(k) =

q

8

3

k + O(1). Conjecture 1 evidently implies the following slightly weaker version:

Conjecture 2. For any ε > 0 we have σ(k) = O(k

).

Bombieri, Granville and Pintz [4] proved that σ(k) = O(k

), and recently Bombieri and Zannier [5] have proved that σ(k) = O(k

).

2. Rudin’s approach

Let e(θ) := e

throughout. The following well-known conjecture was discussed by Rudin (see the end of section 4.6 in [27]):

2000 Mathematics Subject Classification:11N36.

The idea of this paper came out while the first author was visiting the Centre de Recherches Math´ematiques of Montreal and he is extremely grateful for their hospitality.

1

Conjecture 3. For any 2 ≤ p < 4 there exists a constant C

such that, for any trigonometric polynomial f (θ) = P

k

a

e(k

θ) we have

(2.1) kf k

≤ C

kf k

.

Here, as usual, we define kfk

:= R

0

|f(t)|

dt for a trigonometric polynomial f . Con- jecture 3 says that the set of squares is a Λ(p)-set for any 2 ≤ p < 4, where E is a Λ(p)-set if there exists a constant C

such that (2.1) holds for any f of the form f (θ) = P

nk∈E

a

e(n

θ) (a so-called E−polynomial). By H¨older’s inequality we have, for r < s < t,

(2.2) kf k

≤ kfk

kf k

;

taking r = 2 we see that if E is a Λ(t)-set then it is a Λ(s)-set for all s ≤ t.

Let r(n) denote the number of representations of n as the sum of two squares (of positive integers). Taking f (θ) = P

1≤k≤x

e(k

θ), we deduce that kf k

= x, whereas kf k

= P

n

#{1 ≤ k, ` ≤ x : n = k

+ `

}

≥ P

n≤x²

r(n)

³ x

log x; so we see that (2.1) does not hold in general for p = 4.

Conjecture 3 has not been proved for any p > 2, though Rudin [27] has proved the following theorem.

Theorem 1. If E is a Λ(p)-set then any arithmetic progression of N terms contains

¿ N

elements of E. In particular, if Conjecture 3 holds for p then σ(k) = O(k

).

Proof: We use Fej´er’s kernel κ

(θ) := P

|j|≤N

(1 −

) e(jθ). Note that kκ

k

= 1 and kκ

k

= P

|j|≤N

(1 −

)

¿ N so, by (2.2) with r = 1 < s = q < t = 2 we have kκ

k

¿ 1

N

so that kκ

k

¿ N

where

+

= 1.

Suppose that n

, n

. . . , n

are the elements of E which lie in the arithmetic progres- sion a + ib, 1 ≤ i ≤ N. If n

= a + ib for some i, 1 ≤ i ≤ N then n

= a + mb + jb where m = [(N +1)/2] and |j| ≤ N/2; and so 1 −

≥

. Therefore, for g(θ) := P

1≤`≤σ

e(n

θ),

we have Z

0

g(−θ)e((a + bm)θ)κ

(bθ)dθ ≥ σ 2 . On the other hand, we have kgk

≤ C

kgk

¿ √

σ since E is a Λ(p)-set and g is an E-polynomial. Therefore, by H¨older’s inequality,

(2.3)

¯ ¯

¯ ¯ Z

0

g (−θ)e((a + bm)θ)κ

(bθ)dθ

¯ ¯

¯ ¯ ≤ kgk

kκ

k

¿ √ σN

and the result follows by combining the last two displayed equations.

It is known that Conjecture 3 is true for polynomials f (θ) = P

k≤N

e(k

θ) and Antonio C´ordoba [18] proved that Conjecture 3 also holds for polynomials f (θ) = P

k≤N

a

e(k

θ) when the coefficients a

are positive real numbers and non-increasing.

3. Sumsets of squares For a given finite set of integers E let f

(θ) = P

k∈E

e(kθ). Mei-Chu Chang [11]

conjectured that for any ε > 0 we have

kf

k

¿

kf

k

for any finite set of squares E. As kf

k

= P

n

r

(n) where r

(n) is the number of representations of n as a sum of two elements of E , her conjecture is equivalent to:

Conjecture 4 (Mei-Chu Chang). For any ε > 0 we have that

(3.1) kf

k

= X

n

r

(n) ¿ |E|

= kf

k

for any finite set E of squares.

We saw above that P

n

r

(n) À |E|

log |E| in the special case E = {1

, . . . , k

}, so conjecture 4 is sharp, in the sense one cannot entirely remove the ε.

Trivially we have kf

k

= X

n

r

(n) ≤ max r

(n) X

n

r

(n) ≤ |E| · |E|

= |E|

for any set E; it is surprisingly difficult to improve this estimate when E is a set of squares. The best such result is due to Mei-Chu Chang [11] who proved that

X

n

r

(n) ¿ |E|

/ log

|E|

for any set E of squares. Assuming a major conjecture of arithmetic geometry we can improve Chang’s result, in a proof reminiscent of that in [4].

Theorem 2. Assume the Bombieri-Lang conjecture. Then, for any set E of squares, and any set of integers A, we have

X

n

r

(n) ¿ |A|

|E|

+ |A||E|.

In particular X

n

r

(n) ¿ |E|

.

Proof: One consequence of [10] is that there exists an integer κ, such that if the Bombieri- Lang conjecture is true then for any polynomial f (x) ∈ Z[x] of degree six which does not have repeated roots, there are no more than κ rational numbers m for which f (m) is a square. For any given set of four elements a

, . . . , a

∈ A consider all integers n for which there exist b

, . . . , b

∈ E with n = a

+ b

= · · · = a

+ b

. Evidently f (b

) = (b

b

b

)

where f(x) = (x

+ a

− a

)(x

+ a

− a

)(x

+ a

− a

), and so there cannot be more than κ such integers n. Therefore,

X

n

µ r

(n) 4

¶

= X

n

#{a

, . . . , a

∈ E : ∃b

, . . . , b

∈ E, with n = a

+b

, i = 1, . . . , 4}

= X

a1,...,a4∈A

#{n : ∃b

, . . . , b

∈ E, with n = a

+ b

, i = 1, . . . , 4} ≤ κ µ |A|

4

¶ . We have P

n

r

(n) = |E||A|; and so P

n

r

(n)

¿ P

n

¡

E+A(n) 4

¢ + P

n

r

(n) ¿

|A|

+ |A||E|. Therefore the result follows from Holder’s inequality since X

n

r

(n) ≤ Ã X

n

r

(n)

!

à X

n

r

(n)

!

¿ |A|

|E|

+ |A||E|.

The final form of Theorem 2 was inspired by email correspondance with Jozsef Soly- mosi, who also suggested the following:

Theorem 3. Assume the Bombieri-Lang conjecture. There exists a constant c > 0 such that for any finite set E of squares we have r

(n) ≤ c p

|E + E| for all integers n.

Proof: Suppose, on the contrary, that r

(n) > p

κ|E + E| for some integer n, where κ is the constant in the Bombieri-Lang conjecture. Let A := {a : a +a

= n, a, a

∈ E}, so that

max

r

(m) ≥ 1

|A + A|

X

m

r

(m) = |A|

|A + A| = r

(n)

|A + A| ≥ κ|E + E|

|E + E | = κ.

Now for each a for which m − a ∈ A ⊂ E, we have m − a, n − a, n − m + a ∈ E, and therefore an integral point on the curve y

= (x

− m)(x

−n)(x

+ n − m), contradicting the Bombieri-Lang conjecture.

By the Cauchy-Schwarz inequality we have

|E|

|A|

= ³X

r

(n)

´

≤ |E + A| X

r

(n) ¿ |E + A|(|A|

|E|

+ |A||E|), so we can deduce from Theorem 2 that

|E + A| À |A| + |E| min{|A|, |E|

},

still under the assumption of the Bombieri-Lang conjecture. In particular,

|E + A| À |E||A|

when |A| ¿ |E|

which is sharp, and also

(3.2) |E + E | À |E|

.

This leads us to the following conjecture.

Conjecture 5 (Ruzsa). For all ε > 0 we have

|E + E| À |E|

, for every finite set, E, of squares of integers.

Theorem 4. Conjecture 4 implies Conjecture 5 with the same ε.

Proof: By the Cauchy-Schwarz inequality we have

|E|

= ( X

n

r

(n))

≤ |E + E| · X

n

r

(n) and the result follows.

Theorem 5. Conjecture 5 implies Conjecture 2 with ε

= ε

/(4 − 2ε

).

Proof: If E is a set of squares which is a subset of an arithmetic progression P of length k then E + E ⊂ P + P . From conjecture 5 we deduce that

|E|

¿ |E + E| ≤ |P + P | = 2k − 1

and the result follows.

In particular, applying this result to (3.2) shows that the Bombieri-Lang conjecture implies σ(k) ¿ k

. However one can easily do better than this: Suppose that there are σ

squares amongst a + ib, 1 ≤ i ≤ k which are ≡ r (mod s); that is the squares amongst a + rb + jsb, 0 ≤ j ≤ [k/s]. This gives rise to ¡

r,s

6

¢ rational points on the set of curves y

= x Q

j=1

(x + n

) for 0 ≤ n

< n

< · · · < n

≤ [k/s]. Summing over all r (mod s) and all s > σ/10 we get

k

σ

À

µ 10k/σ 5

¶

À X

s>σ/10

X

r (mods)

µ σ

6

¶

À X

s>σ/10

s

µ [σ/s]

6

¶ À σ

and we obtain σ(k) ¿ k

. This is all irrelevant since stronger upper bounds were proved unconditionally in [4] and [5].

An affine cube of dimension d in Z is a set of integers {b

+ P

i∈I

b

: I ⊂ {1, . . . , d}}

for non-zero integers b

, . . . , b

. In [28], Solymosi states

Conjecture 6 (Solymosi). There exists an integer d > 0 such that there is no affine cube of dimension d of distinct squares.

This was asked earlier as a question by Brown, Erd˝os and Freedman in [9], and Hegyv´ari and S´ark¨ozy [21] have proved that an affine cube of squares, all ≤ n, has dimension ≤ 48(log n)

.

This conjecture follows from the Bombieri-Lang conjecture for if there were an affine cube of dimension d then for any x

∈ {b

+ P

i∈I

b

: I ⊂ {3, . . . , d}} we have that x

+ b

, x

+ b

, x

+ b

+ b

are also squares, in which case there are ≥ 2

integers x for which f(x) = (x

+ b

)(x

+ b

)(x

+ b

+ b

) is also square; and so 2

≤ κ, as in the proof of theorem 2.

In [28], Solymosi gives a beautiful proof that for any set of real numbers A, if |A + A| ¿

|A|

then A contains many affine cubes of dimension d. Therefore we deduce a weak version of Ruzsa’s conjecture from Solymosi’s conjecture:

Theorem 6. Conjecture 6 implies that there exists δ > 0 for which |E + E| À |E|

, for any set E of squares.

The Erd˝os-Szemer´edi conjecture states that for any set of integers A we have

|A + A| + |A ¦ A| À

|A|

.

In fact they gave a stronger version, reminiscent of the Balog-Szemer´edi-Gowers theo- rem:

Conjecture 7 (Erd˝os-Szemer´edi). If A is a finite set of integers and G ⊂ A × A with

|G| À |A|

then

(3.3) |{a + b : (a, b) ∈ G}| + |{ab : (a, b) ∈ G}| À

|G|

.

Mei-Chu Chang [11] proved that a little more than Conjecture 7 implies Conjecture 4:

Theorem 7. If (3.3) holds whenever |G| ≥

|A| then Conjecture 4 holds.

Proof: Let B be a set of k non-negative integers and E = {b

: b ∈ B}. Define G

:= {(a

, a

) : ∃b, b

∈ B with a

= b + b

, a

= b − b

, and b

− b

∈ M } where M ⊂ E − E; and so A

:= {a

, a

: (a

, a

) ∈ G

} ⊂ (B + B ) ∪ (B − B ). Therefore

|A

| ≤ 2|G

|.

Since {a + a

: (a, a

) ∈ G} and {a − a

: (a, a

) ∈ G} are subsets of {2b : b ∈ B}, they have ≤ k elements; and {aa

: (a, a

) ∈ G} ⊂ M . Therefore (3.3) implies that

|G

|

¿

|M | + k. Since, trivially, |G

| ≤ k

we have P

m∈M

r

(m) = |G

| ¿ k

(|M | + k).

Now let M be the set of integers m for which r

(m) ≥ k

, so that P

m∈M

r

(m) ≥ k

|M | and hence P

m∈M

r

(m) ¿ k

by combining the last two equations. There- fore, as r

(m) ≤ k,

kf

k

= X

m

r

(m)

≤ X

m∈E−E

k

+ k X

m∈M

r

(m) ¿ k

.

She also proves a further, and stronger, result along similar lines:

Conjecture 8 (Mei-Chu Chang). If A is a finite set of integers and G ⊂ A × A then (3.4) |{a + b : (a, b) ∈ G}| · |{a − b : (a, b) ∈ G}| · |{ab : (a, b) ∈ G}| À

|G|

. Theorem 8. Conjecture 8 holds if and only if Conjecture 4 holds.

Proof: Assume Conjecture 8 and define B, A and G

as in the proof of theorem 7, so that ( P

m∈M

r

(m))

= |G

|

¿ k

|M|. We partition E − E into the sets M

:= {m : 2

≤ r

(m) < 2

} for j = 1, 2, . . . , J := [log(2k)/ log 2]; then (2

|M

|)

≤ ( P

m∈Mj

r

(m))

¿ k

|M

| so that P

m∈Mj

r

(m)

≤ 2

|M

| ¿ k

. Hence

kf

k

= X

m

r

(m)

< X

j

X

m∈Mj

r

(m)

¿ Jk

¿ k

, as desired.

Now assume Conjecture 4 and let G

:= {(a, b) ∈ G : ab = n}. Then |G|

= ( P

n

|G

|)

≤ |{ab : (a, b) ∈ G}| · P

n

|G

|

, while X

n

|G

|

= Z

0

¯ ¯

¯ ¯

¯ ¯ X

(a,b)∈G

e(4abt)

¯ ¯

¯ ¯

¯ ¯

2

dt = Z

0

¯ ¯

¯ ¯

¯ ¯ X

(a,b)∈G

e((a + b)

t)e(−(a − b)

t)

¯ ¯

¯ ¯

¯ ¯

2

dt

which, letting E

:= {r

: r = a ± b, (a, b) ∈ G}, is

≤ Z

0

¯ ¯

¯ ¯

¯ ¯ X

r²∈E+

e(r

t) X

s²∈E−

e(−s

t)

¯ ¯

¯ ¯

¯ ¯

2

dt ≤ kf

k

kf

k

by the Cauchy-Schwarz inequality. Since kf

k

¿ |{a ± b : (a, b) ∈ G}| · |G|

by

Conjecture 4, our result follows by combining the above information.

4. Solutions of a quadratic congruence in short intervals

We begin with a connection between additive combinatorics and the Chinese Remain- der Theorem. Suppose that n = rs with (r, s) = 1; and that for given sets of residues Ω(r) ⊂ Z/rZ and Ω(s) ⊂ Z/sZ we have Ω(n) ⊂ Z/nZ in the sense that m ∈ Ω(n) if and only if there exists u ∈ Ω(r) and v ∈ Ω(s) such that m ≡ u (mod r) and m ≡ v (mod s). When (r, n/r) = 1 consider the map which embeds Z/rZ → Z/nZ by tak- ing u (mod r) and replacing it by U (mod n) for which U ≡ u (mod r) and U ≡ 0 (mod n/r); we denote the image of Ω(r) by Ω(r, n) under this map. The key remark, which follows immediately from the definitions, is that

Ω(n) = Ω(r, n) + Ω(s, n).

Thus if n = p

. . . p

where the primes p

are distinct then

Ω(n) = Ω(p

, n) + Ω(p

, n) + · · · + Ω(p

, n).

Particularly interesting is where Ω

(n) is the set of solutions m (mod n) to f (m) ≡ 0 (mod n), for given f (x) ∈ Z[x]. Can there be many elements of Ω

(n) in a short interval when f has degree two? A priori this seems unlikely since the elements of the Ω(r, n) are so well spread out, that is the distance between any pair of elements is ≥ n/r because they are all divisible by n/r.

The next theorem involves the distribution of the elements of Ω(n) in the simplest non-trivial case, in which each Ω(p

) has just two elements, namely {0, 1}, so that Ω(n) is the set of solutions of x(x − 1) ≡ 0 (mod n) (see also [3]).

Theorem 9. Let Ω(n) be the set of solutions of x(x − 1) ≡ 0 (mod n). Then (1) Ω(n) has an element in the interval (1, n/k + 1).

(2) For any ε > 0 there exists n = p

. . . p

such that Ω(n) ∩ (1, (

− ε)n] = ∅.

(3) For any ε > 0 there exists n = p

. . . p

such that if x ∈ Ω(n) then |x| < εn.

Proof. Let Ω(p

, n) = {0, x

} where x

≡ 1 (mod p

) and x

≡ 0 (mod p

) for any i 6= j. Then Ω(n) = {0, x

}+· · ·+{0, x

}. Let s

= n and s

be the least positive residue of x

+ · · · + x

(mod n) for r = 1, . . . , k so that s

= 1. By the pigeonhole principle, there exists 0 ≤ l < m ≤ k such that s

and s

lie in the same interval (jn/k, (j +1)n/k], and so |s

−s

| < n/k with s

−s

≡ x

+· · ·+x

(mod n) ∈ Ω(n). If s

−s

> 1 then we are done. Now s

− s

≡ 1 (mod n) but is not = 1, so s

− s

6= 0, 1. Thus we must consider when s

− s

< 0. In this case x

+ · · · + x

+ x

+ · · · + x

(mod n) ∈ Ω(n) and is ≡ s

− (s

− s

) ≡ 1 − (s

− s

), and the result follows.

To prove (2) let us take k − 1 primes p

, . . . , p

> k, and integers a

= [p

/k] for j = 1, . . . , k − 1. Let P = p

· · · p

and determine r (mod P ) by the Chinese Remainder Theorem satisfying ra

(P/p

) ≡ 1 (mod p

) for j = 1, . . . , k − 1. Now let p

be a prime

≡ r (mod P ), and let a

the least positive integer satisfying a

P ≡ 1 (mod p

). Let n = p

· · · p

so that x

= a

n/p

for i = 1, . . . , k. Now n/k ≥ x

> n/k − n/p

> 0 for i = 1, . . . , k − 1 and since x

+ · · · + x

≡ 1 (mod n) with 1 ≤ x

< n we deduce that x

+ · · · + x

= n + 1 and therefore 1 + n/k ≤ x

< 1 + n/k + n P

i=1

1/p

. Now elements of Ω(n) are of the form P

i∈I

x

and we have | P

i∈I

x

− n|I |/k| ≤ 1 + 2n P

i=1

1/p

,

and this is < εn provided p

> 2k/ε for each p

. Finally, since the cases I = ∅ and

I = {1, . . . , k} correspond to the cases x = 0 and x = 1 respectively, we have that any

other element is greater than (1/k − ε)n.

To prove (3) we mimic the proof of (2) but now choosing non-zero integers a

satisfying

|

| <

for j = 1, . . . , k − 1. This implies that |a

/p

| < ε/2 and then | P

i∈I

x

| < εn.

In the other direction, we give a lower bound for the length of intervals containing k elements of Ω(n).

Theorem 10. Let integer d ≥ 2 be given, and suppose that for each prime power q we are given a set of residues Ω(q) ⊂ (Z/qZ) which contains no more than d elements.

Let Ω(n) be determined for all integers n using the Chinese Remainder Theorem, as described at the start of this section. Then, for any k ≥ d, there are no more than k integers x ∈ Ω(n) in any interval of length n

, where α

(k) =

> 0 with 0 < ε

(k) =

+ O(

).

Proof. Let x

, . . . , x

elements of Ω(n) such that x

< · · · < x

< x

+ n

. Let q a prime power dividing n. Each x

belongs to one of the d classes (mod q) in Ω(q). Write r

, . . . , r

to denote the number of these x

belonging to each class. Then Q

1≤i<j≤k+1

(x

− x

) is a multiple of q

(

). The minimum of P

i=1

¡

i

2

¢ under the restriction P

i

r

= k+1 is d ¡

2

¢ +rs where r, s are determined by k+1 = rd+s, 0 ≤ s < d.

Finally

n

(

) > Y

1≤i<j≤k+1

(x

− x

) > n

(

)

and we get a contradiction, by taking α

(k) = (d ¡

2

¢ + rs)/ ¡

2

¢ . The next theorem is an easy consequence of the proof above.

Theorem 11. If x

< · · · < x

are solutions to the equation x

≡ a (mod b), then x

− x

> b

, where ` is the largest odd integer ≤ k.

First proof. For any maximal prime power q dividing b, (a, q) must be an square so we can write x

= y

Q

q

(a, q)

with y

≡ a

(mod q

) where q

= q/(a, q) and (a

, q

) = 1.

Let Ω(q

) be the solutions of y

≡ a

(mod q

). Now, since (a

, q

) = 1 we have that

|Ω(q

)| ≤ 2 and we can apply theorem 10 to obtain that x

− x

= (y

− y

) Y

q

(a, q)

≥ Ã Y

q

q/(a, q)

!

Y

q

(a, q)

≥ ( Y

q

q)

. Now, notice that α

(k − 1) = 1/2 − 1/(2l) where l is the largest odd number ≤ k.

Second proof. Write x

= a+r

b where r

= 1 < r

< · · · < r

(if necessary, by replacing a in the hypothesis by x

− b). Consider the k-by-k Vandermonde matrix V with (i, j)th entry x

. The row with i = 1 + 2I has jth entry (a + r

b)

; by subtracting suitable multiples of the rows 1 + 2`, ` < I , we obtain a matrix V

with the same determinant where the (2I + 1, j) entry is now (r

b)

. Similarly the row with i = 2I + 2 has jth entry x

(a + r

b)

; by subtracting suitable multiples of the rows 2 + 2`, ` < I, we obtain a matrix V

with the same determinant where the (2I + 2, j) entry is now x

(r

b)

. Finally we arrive at a matrix W by dividing out b

from rows 2I + 1 and 2I + 2 for all I. Then the determinant of V , which is Q

1≤i<j≤k

(x

− x

), equals b

times the

determinant of W , which is also an integer, and the result follows.

The advantage of this new proof is that if we can get non-trivial lower bounds on the determinant of W then we can improve Theorem 11. We note that W has (2I + 1, j) entry r

, and (2I + 2, j) entry x

r

.

Remark: Taking k = ` to be the smallest odd integer ≥

, then we can split our interval into two pieces to deduce from Theorem 11 a weak version of Conjecture 9:

There are no more than

solutions x to the equation x

≡ a (mod b) in any interval of length b

. From this it follows that the number of solutions x to the equation x

≡ a (mod b) in any interval of length L is

¿ 1 + log L log

³

1 +

´ .

This result, with ‘1/2’ replaced by ‘1/d’, was proved for the roots of any degree d polynomial mod b by Konyagin and Steger in [23].

A slightly improvement on the theorem above would have interesting consequences.

Conjecture 9. There exists a constant N such that there are no more than N solutions 0 < x

< x

< · · · < x

< x

+ b

to the equation x

≡ a (mod b), for any given a and b.

Theorem 12. Conjecture 9 implies Conjecture 1.

Proof. Suppose that there are ` À k

squares amongst a + b, a + 2b, . . . , a + kb, which we will denote x

< x

< · · · < x

. By conjecture 9 we have x

− x

≥ [(` − 1)/N ]b

, whereas (k − 1)b ≥ (x

+ x

)(x

− x

) ≥ (x

− x

)

. Therefore [(` − 1)/N ]

≤ (k − 1) which implies that ` ≤ N (1 + √

k − 1).

Conjecture 9 would follow easily from theorem 10 if we could get the exponent 1/2 for some k, instead of 1/2 − ε

(k). Conjecture 9 can be strengthened and generalized as follows:

Conjecture 10. Let integer d ≥ 1 be given, and suppose that for each prime power q we are given a set of residues Ω(q) ⊂ (Z/qZ) which contains no more than d elements.

Ω(b) is determined for all integers b using the Chinese Remainder Theorem, as described at the start of this section. Then, for any ε > 0 there exists a constant N (d, ε) such that for any integer b there are no more than N (d, ε) integers n, 0 ≤ n < b

with n ∈ Ω(b).

In theorem 10 we proved such a result with the exponent ‘1 − ε’ replaced by ‘1/d − ε’.

We strongly believe Conjecture 10 with ‘1−ε’ replaced by ‘1/d’, analogous to Conjecture 9. In a 1995 email to the second author, Bjorn Poonen asked Conjecture 10 with ‘1 − ε’

replaced by ‘1/2’ for d = 4; his interest lies in the fact that this would imply the uniform boundedness conjecture for rational preperiodic points of quadratic polynomials (see [25]).

Conjecture 10 does not cover the case Ω

(b) = {m (mod b) : f (m) ≡ 0 (mod n)}

for all monic polynomials f of degree d since, for example, the polynomial (x − a)

≡ 0

(mod p

) has got p

solutions (mod p

), rather than d. One may avoid this

difficulty by restricting attention to squarefree moduli (as in a conjecture posed by

Croot [19]); or, to be less restrictive, note that if f (x) has more than d solutions

(mod p

) then f must have a repeated root mod p, so that p divides the discriminant of f :

Conjecture 11. Fix integer d ≥ 2. For any ε > 0 there exists a constant N (d, ε) such that for any monic f (x) ∈ Z[x] there are no more than N (d, ε) integers n, 0 ≤ n < b

, with f (n) ≡ 0 (mod b) for any integer b such that if p

divides b then p does not divide the discriminant of f.

5. Lattice points on circles

Conjecture 12. There exists δ > 0 and integer m > 0 such that if a

+ b

= n with a

, b

> 0 and a

≡ a

(mod q) for i = 1, . . . , m then q = O(n

).

Theorem 13. Conjecture 12 implies Conjecture 1

Proof. Suppose that x

< · · · < x

are distinct squares belonging to the arithmetic progression a + b, a + 2b, . . . , a + kb with (a, b) = 1, where r > √

8lk, with l sufficiently large > m. We may assume that (a, b) = 1 and that b is even. There are r

sums x

+ x

each of which takes one of the values 2a + 2b, 2a + 3b, . . . , 2a + 2kb, and so one of these values, say n, is taken ≥ r

/(2k − 1) > 4l times. Therefore we can write n = r

+ s

for j = 1, 2, . . . , 4l for distinct pairs (r

, s

), and let v

= r

+ is

. Note that n ≡ 2 (mod 8).

Let Π = Q

1≤i<j≤4l

(v

− v

) 6= 0. We will prove that |Π| ≥ b

(

)(n/2)(

), by considering the powers of the prime divisors of b and n which divide Π. Note that (n/2, b) = 1.

Suppose p

kb where p is a prime, and select w (mod p

) so that w

≡ a (mod p

).

Note that each r

, s

≡ w or − w (mod p

): We partition the v

into four subsets J

, J

, J

, J

depending on the value of (r

(mod p

), s

(mod p

)). Note then that p

divides v

− v

if v

, v

belong to the same subset, and so p

to the power P

i

¡

i| 2

¢ > 4 ¡

2

¢ divides Π.

Now let p be an odd prime with p

kn. If p ≡ 3 (mod 4) then p

must divide each r

and s

so that then p

(

) divides Π. If p ≡ 1 (mod 4) let us suppose π is a prime in Z[i] dividing p; then π

¯ π

divides v

for some 0 ≤ e

≤ e. If e

≤ e

we deduce that π

π ¯

divides v

− v

. We now partition the values of j into sets J

, . . . , J

depending on the value of e

. The power of π dividing Π is thus P

i=0

i P

g=i+1

|J

||J

| + P

i=0

(e − i) ¡

i| 2

¢ , and the power of ¯ π dividing Π is thus P

i=0

(e −g ) P

i=0

|J

||J

|+ P

i=0

(e−i) ¡

i| 2

¢ . It is easy to show that P

0≤i<g≤e

(i +e − g)m

m

+e P

i=0

¡

i

2

¢ , under the conditions that P

i

m

is fixed and each m

≥ 0, is minimized when m

= m

, m

= · · · = m

= 0.

Therefore the power of π plus the power of ¯ π dividing Π is ≥ 2e ¡

2

¢ . Finally |r

−r

|, |s

− s

| ≤ (x

−x

)/(x

+x

) ≤ (k −1)b/(2 √

a + b), and so |v

−v

|

≤ (k−1)

b

/(2(a+b)), giving that |Π| ≤ (k

b

/(2(a+b)))

(

). Putting these all together with the fact that n > 2(a + b), we obtain 2

(a + b)

≤ k

b

. However this implies that n ≤ 2k(a + b) ≤ 2

k

b

¿ k

n

¿ k

n

, for l sufficiently large; and therefore a + b < n ¿ k

.

Let u

, . . . u

be the distinct integers in [1, b/2] for which each u

≡ a (mod b), so that

d ³ 2

, by the Chinese Remainder Theorem, where ω(b) denotes the number of prime

factors of b. The number of x

≡ u

(mod b/2) is ≤ 1 + ((a + kb)

− a

)/(b/2) ≤

1 + 2(k/b)

; and thus r ¿ 2

+ k

2

/b

. This is ¿ k

provided ω(b) ¿ log k,

which happens when b ¿ k

by the prime number theorem. The result follows.

Here is a flowchart of the relationships between the conjectures above:

10.

If |Ω(q)| ≤ d, for any prime power q|b then |Ω(b) ∩ [0, b

]| < C(d, ε)

®¶

11.

If f(x) ∈ Z[x], is monic, degree d and p

|b = ⇒ p - disc(f ) then f (n) ≡ 0 (mod b) has no more than

N (d, ε) solutions 0 ≤ n ≤ b

+3

9.

∃ m such that x

≡ r (mod q), 1 ≤ i ≤ m = ⇒ max |x

− x

| > q

®¶

3.(Rudin)

For any 2 ≤ p < 4 ∃C

such that if f(θ) = P

k

a

e(k

θ) then kf k

≤ C

kf k

®¶

1.(Rudin)

σ(k) ¿ k

⁺³

2.(Rudin)

σ(k) ¿ k

12.

∃δ > 0, ∃m such that a

+ b

= n, a

≡ a

(mod q)

1 ≤ i ≤ m = ⇒ q = O(n

)

KS

5.(Ruzsa)

If E ⊂ squares,

|E + E| À |E|

KS

8.(Mei Chu−Chang)

|{a + b : (a, b) ∈ G}|×

|{a − b : (a, b) ∈ G}|×

|{ab : (a, b) ∈ G}| À

|G|

ks +3

4. (Mei Chu−Chang)

If E ⊂ squares, P

m

r

(m) ¿ |E|

KS

7.(Erdos−Szemeredi−Chang)

If G ⊂ A × A, and |G| ≥ |A|/2 then

|{a + b : (a, b) ∈ G}|+

|{ab : (a, b) ∈ G}| À

|G|

KS

Dependencies: ε

= ε

/(4 − 2ε

), ε

= 2/p − 1/2, ε

= 3ε

/2, ε

= 3ε

/4, ε

= ε

, ε

= 4ε

.

If E ⊂ squares and A ⊂ Z, P

m

r

(m) ¿ |A|

|E |

+ |A||E| ⁺³

If E ⊂ squares and A ⊂ Z,

|E + A| À

 



 

|E||A| if |A| ¿ |E|

|E|

if |E|

¿ |A| ¿ |E|

|A| if |E|

¿ |A|.

®¶

Bombieri-Lang Conjecture

KS &. V V V V V V V V V V V V V V V V V V V V V V V V V V V V

V V V V V V V V V V V V V V V V V V V V V V V V V V V V

®¶®¶

If E ⊂ squares,

|E + E| À |E|

6.(Solymosi)

∃d > 0 such that there is no affine cube of dimension d of distinct squares

®¶

If E ⊂ squares, r

(n) ¿ p

|E + E|

If E ⊂ squares, ∃²

> 0 such that |E + E| À |E|

Conjecture 13. For any α < 1/2, there exists a constant C

such that for any N we have

#{(a, b), a

+ b

= n, N ≤ |b| < N + n

} ≤ C

. A special case of interest is where N = 0:

(5.1) #{(a, b), a

+ b

= n, |b| < n

} ≤ C

.

Heath-Brown pointed out that one has to be careful in making an analogous conjecture in higher dimension as the following example shows: Select integer r which has many representations as the sum of two squares; for example, if r is the product of k distinct primes that are ≡ 1 (mod 4) then r has 2

such representations. Now let N be an arbitrarily large integer and consider the set of representations of n = N

+ r as the sum of three squares. Evidently we have ≥ 2

such representations in an interval whose size, which depends only on k, is independent of n. However, one can get around this kind of example in formulating the analogy to conjecture 13 in 3-dimensions, since all of these solutions live in a fixed hyperplane. Thus we may be able to get a uniform bound on the number of such lattice points in a small box, no more than three of which belong to the same hyperplane.

It is simple to prove (5.1) for any α ≤ 1/4 (and Conjecture 13 for α ≤ 1/4 with N ¿ n

), but we cannot prove (5.1) for any α > 1/4. Conjecture 13 and the special case 5.1 are equivalent to the following conjectures respectively:

Conjecture 14. The number of lattice points {(x, y) ∈ Z

: x

+ y

= R

} in an arc

of length R

is bounded uniformly in R.