On the Sum of Dilations of a Set

We show that for any relatively prime integers $1\leq p<q$ and for any finite $A \subset \mathbb{Z}$ one has $$|p \cdot A + q \cdot A | \geq (p + q) |A| - (pq)^{(p+q-3)(p+q) + 1}.$$


Introduction
Let A and B be finite sets of real numbers.The sumset and the product set of A and B are defined by For d > 0 the dilation of A by d is defined by while for any real number x, the translation of A by x is defined by The Erdős-Szemerédi sum-product conjecture [8] claims that for any finite subset of the positive integers, either the sumset A + A or the product set A • A must be big, more precisely max{|A + A|, |A • A|} ≫ |A| 2−ǫ  for any ǫ > 0. The best result in this direction, due to Solymosi [11], is the above bound with the weaker exponent 4/3 − ǫ.Another realization of this phenomenon is if an expression of sets uses both addition and multiplication, then it produces a big set.For example, both come from the method of Elekes [6] using the Szemerédi-Trotter incidence geometry, see the book of Tao and Vu [12].Improving the exponent in these bounds is a challenging problem, and 2 − ǫ is certainly expected.Such a problem seems easier for more variables, for example is proved by the first author [1].Changing the role of addition and multiplication in most of these expressions does not change the results or our expectation dramatically.However, the two variable expressions A • (1 + A) and p • A + q • A are exceptions, because translation seriously alters multiplicative behavior, while dilation seems rather harmless.It is a beautiful consequence of the incidence geometry that A • (1 + A) is big; for example, Jones and Roche-Newton [9] proved On the other hand, let q > p ≥ 1 be relatively prime integers and for X = {1, 2, . . ., |X|}, obviously p • X + q • X ⊂ {p + q, . . ., (p + q)|X|}, that is Bukh [2] proved that for coprime integers λ 1 , . . ., λ k one has Our main result says |p • A + q • A| ≥ (p + q)|A| − C p,q .Cilleruelo, Hamidoune and Serra showed this for p = 1 and q prime and these results were extended by Du, Cao, and Sun [5] when q is a prime power or the product of two primes.Hamidoune and J. Rué [7] solved the case when p = 2 and q prime and these results were extended by Ljujic [10] to p = 2 and q the power of an odd prime or the product of two odd primes.While it is certainly feasible that for some C that only depends on λ 1 , . . ., λ k , our method seems to only handle the case where k ≤ 2. We will confine ourselves to subsets of the integers.Transforming our result to the case when p, q and the set A are from rationals is an obvious task.As Sean Eberhard pointed out, keeping p and q integers and extending our result to where A is a subset of the real numbers follows from, say, Lemma 5.25 in [12].Our main result is as follows.
Theorem 1.1.For any relatively prime integers 1 ≤ p < q and for any finite A ⊂ Z one has Note that the multiplicative constant p + q is best possible as the above example (1) shows.On the other hand we do not attempt to get the best additive constant, and improvements in that respect are very probable.For example, our method gives q2 (q−2)(q+1) to the case p = 1, or simply q! to the case p = 1, q is a prime.We cannot even decide the true size of the best additive constant.The following example shows that, even in the special case p = 1, the additive constant is not polynomial in q.It also suggests that possibly a better constant can be proved for all sufficiently large sets A, avoiding such pathological cases.
For t a positive integer and 0 ≤ a < q 2 also an integer, let If we set a = ⌊ √ q⌋ and t = ⌊log 2 √ q⌋, it is easy to see that |A| = a t+1 and |A + q . The authors are thankful to Imre Ruzsa for drawing their attention to the present problem, as well as for pointing out this example.

Preliminaries
For nonempty finite subsets of the real numbers by observing that We extend this argument in the following lemma.
Lemma 2.1.Let A be a nonempty subset of the real numbers and q > p ≥ 1.Then Proof.Let A = {a 1 , . . ., a n } where a 1 < a 2 < . . .< a n .Then For each 1 ≤ i ≤ n − 1 there are three elements in the previous list, pa i + qa i < pa i+1 + qa i < pa i + qa i+1 , and one more element It is worth noting that this gives |A + 2 • A| ≥ 3|A| − 2, which settles Theorem 1.1 in the case p = 1, q = 2.This is best possible by (1).
The main purpose of this section is to give a short proof of |A + 3 • A| ≥ 4|A| − 4 in order to introduce some of the main ideas of the proof of the general theorem.This was proved by Cilleruelo, Silva, and Vinuesa [4] using different methods and they were able to classify the sets where equality holds.The method we use requires that A is a subset of the integers, and can be extended to a general lower bound for all cases with q ≥ 3, though we leave this extension to the interested reader.The bound 4|A| − 4 is also best possible, as is follows from the next construction.Let 0 ≤ d < q ≤ n be integers and let X := {i + xq : 0 ≤ i ≤ d, 0 ≤ x < n}.Note that |X| = (d + 1)n.It is easy to check that X + q • X is precisely the set of integers in the interval [0, . . ., (q + 1)(d + (n − 1)q)] that are equivalent to one of {0, . . ., d} modulo q.It follows that |X + q • X| = (q + 1)|X| − (d + 1)(q − d), and the optimal choice d = ⌊(q − 1)/2⌋ gives |X + q • X| = (q + 1)|X| − ⌊ q+1 2 ⌋⌈ q+1 2 ⌉.We would like to remark here one can use 1.1 and the methods of [3] to show for |A| ≥ (q − 1) 2 q q 2 −q+1 , one has |A + q • A| ≥ (q + 1)|A| − ⌊ q+1 2 ⌋⌈ q+1 2 ⌉.Furthermore, their methods can be used to show X is the unique set, up to an affine transformation, where equality holds.Translation and dilation do not change |A + 3 • A|, we may translate A so that 0 is the smallest element and then dilate A until it intersects at least 2 residue classes modulo 3. Let This union is disjoint and moreover the sets A j + 3 • A are disjoint.Thus from the obvious fact (2) it follows that We say that A is fully distributed modulo 3 (FD mod 3) if A intersects all 3 residue classes modulo 3.By (3), if A is FD mod 3 then |A + q • A| ≥ 4|A| − 3 > 4|A| − 4 and the theorem follows.
Thus we may assume that A intersects exactly two residue classes modulo 3, so where the union is disjoint and

Suppose now that
After a translation by −a 1 , dilation by 1 q , and another translation by −a 1 , we obtain Therefore for any x ∈ B 1 , there is a y ∈ B 2 such that a 2 − a 1 + x + 3y ∈ B 1 + 3 • B 1 and so there is an x ′ ∈ B 1 such that a 2 − a 1 + x ≡ x ′ (mod 3).We may repeat this for x ′ in place of x so there is an element x ′′ ∈ B 1 such that a 2 − a 1 + x ′ ≡ x ′′ (mod 3).Since (a 2 − a 1 , 3) = 1, we have that x, x ′ , x ′′ are incongruent mod 3. Thus B 1 is FD mod 3 and it follows from (3) that Case 1. Suppose that B 1 and B 2 are both FD mod 3.
We can always find two more elements in the (disjoint) union of ( Consider the maximum element of A and let it be in A 2 and call it M. Let u ∈ A 1 be maximally chosen.Then it is clear that u Then by ( 3) Case 2. Suppose that B 1 and B 2 are both not FD mod 3.
Case 3. Suppose one of B 1 and B 2 is FD mod 3 and the other is not.
We may assume, without loss of generality, that B 1 is FD mod 3 while B 2 is not FD mod 3.This is the only case when we use the induction hypothesis.Since

By induction and (3), we have
In all cases we obtain |A + 3 • A| ≥ 4|A| − 4, thus completing the induction.

Utilizing (2), we obtain
We will say that A is fully distributed modulo p (FD mod p) if A intersects every residue class modulo p and A is fully distributed modulo q (FD mod q) if A intersects every residue class modulo q.Thus A is FD mod p and/or FD mod q if and only if r = p and/or s = q.Lemma 3.1.For any fixed 1 Similarly for any fixed 1 ≤ i ≤ r, either Proof.We only prove the first statement and the second statement is a symmetric argument.
Then for any m = j, we have It follows that for every x ∈ p•Q ′ j there is a y ∈ Q ′ m such that q m −q j +x+qy ∈ p•Q ′ j +q •Q ′ j , and so there is an x ′ ∈ p • Q ′ j such that q m − q j + x ≡ x ′ (mod q).We may repeat this argument with x ′ in place of x, and so on, and we may repeat for all m so that eventually we get for any x ∈ p • Q ′ j and any z = u 1 (q 1 − q j ) + • • • + u s (q s − q j ), where u 1 , . . ., u s are arbitrary integers, that there is an x ′ ∈ p • Q ′ j with z + x ≡ x ′ (q).Since A is reduced, the set of z describes all residues mod q by (5), it follows that p • Q ′ j is FD mod q and since (p, q) = 1, Q ′ j is FD mod q.
The previous lemma will be useful in finding new elements if any of the P ′ i are not FD mod p or if any of the Q ′ j are not FD mod q.For convenience, set A ij := P i ∩ Q j , where some of these sets may be empty.Then The fact that we may write A ij = a ij + pq • A ′ ij is precisely the Chinese remainder theorem.We have that where some of the summands are possibly zero.
Lemma 3.2.Fix 1 ≤ i ≤ r and 1 ≤ j ≤ s.Suppose P ′ i is FD mod p. Then either Similarly, suppose Q ′ j is FD mod q.Then either Proof.We prove the first statement, and the second statement follows by a symmetric argument.The result is trivial for An easy calculation reveals that Note that This means, for every x ∈ P ′ i there is an We are now ready to prove Theorem 1.1.Our strategy is simple: we start from Lemma 2.1 and gradually improve it in an iterative way.Proposition 3.3.Let q > p ≥ 1 are relatively prime integers.For every integer 3(p + q) ≤ m ≤ (p + q) 2 and for all finite sets of integers A, we have Proof.Observe that m = (p + q) 2 is precisely Theorem 1.1.We prove by induction on m.For m = 3(p + q), we claim |p • A + q • A| ≥ 3|A| − pq, which is even true with 3|A| − 2 by Lemma 2.1.Suppose now that Proposition 3.3 is true for a fixed 3(q + p) ≤ m < (p + q) 2 .For simplicity, we write C m = (pq) m+1−3(p+q) .
First, assume there is an 1 ≤ i ≤ r such that |P i | ≤ 1 p+q |A|.Then by the induction hypothesis, (2), and m ≤ (p + q) 2 , we obtain Thus we may assume that every P i and Q j has more than 1 p+q |A| elements.If there is a 1 ≤ i ≤ r such that P ′ i is not FD mod p then by Lemma 3.1 and the induction hypothesis we have since C m+1 ≥ 2C m .A symmetric argument works if there is a 1 ≤ j ≤ s such that Q ′ j is not FD mod q.Thus we may assume that every P ′ i is FD mod p and every Q ′ j is FD mod q.Fix 1 ≤ i ≤ r and 1 ≤ j ≤ s.Then using both parts of Lemma 3.2 we obtain either by the induction hypothesis or A ′ ij is FD mod p and FD mod q.In the latter case, by ( 6) since C m ≥ pq and (p + q) ≥ m+1 p+q .Note that this is the point which blocks us from gradually improving the lower bound beyond (p + q)|A|.
In either case |p • Q j + q • P i | ≥ m+1 p+q |A ij | − C m .By (7), it follows that

Theorem 2 . 2 .
Let A be a finite subset of the integers.Then |A + 3 • A| ≥ 4|A| − 4. Proof.If |A| = 1 the result is trivial.If |A| = 2 the result follows from Lemma 2.1.We use induction on the size of |A|.Assume |A| > 2, and for any proper subsets X of A, we have |X + 3 • X| ≥ 4|X| − 4.