Extremal problems for GCDs

We prove that if $A \subseteq [X, 2X]$ and $B \subseteq [Y, 2Y]$ are sets of integers such that $\gcd(a,b) \geq D$ for at least $\delta |A||B|$ pairs $(a,b) \in A \times B$ then $|A||B| \ll_{\varepsilon} \delta^{-2 - \varepsilon} XY/D^2$. This is a new result even when $\delta = 1$. The proof uses ideas of Koukoulopoulos and Maynard and some additional combinatorial arguments.


Introduction and proof strategy
Fix ε ∈ (0, 1) throughout the paper; implied constants and thresholds may depend on ε but are otherwise absolute. Let p 0 be a threshold which (at the start of Section 3) will be taken to be sufficiently large. Given a finite set S ⊂ N, write P(S) for the set of primes dividing some element of S, and P small (S) for the set of primes p p 0 dividing some element of S.
Our main result is the following. Let us make some remarks on this theorem. 1. This obviously implies the cruder bound |A||B| ≪ δ −2−ε XY /D 2 , mentioned in the abstract. The more precise form we have stated seems of little additional interest in its own right, but is critical for the proof. Perhaps the most natural case is when A = B and X = Y , when the result says the following: if, for a proportion δ of all pairs (a, a ′ ) ∈ A × A we have gcd(a, a ′ ) D, then |A| ≪ δ −1−ε X/D.
2. We believe that the result is new even when δ = 1, that is to say when gcd(a, b) D for all a ∈ A and b ∈ B. In this case, the result is clearly sharp up to a multiplicative constant. Indeed, assuming that D is an integer, we may take tight examples for δ = 1 (see [1,Section 15]) in which there is no single d ≫ D that divides a positive proportion of A and B.
3. For δ ∈ (0, 1) the result is also sharp for a wide range of parameters, apart from the factor of δ −ε . To see this, let D δ −1 be given, set D 0 := ⌊δD⌋, and consider the sets The proportion of pairs of integers with gcd k is 1/k 2 ζ(2), and so the proportion of pairs of integers with gcd k is ≫ 1/k. It follows (at least if X/D 0 is big enough compared to D/D 0 ) that the number of such pairs (x, x ′ ) is ≫ δ|A||B|.
4. When A = B, X = Y and δ = 1, the result says the following: if gcd(a, a ′ ) D for all a, a ′ ∈ A, then |A| ≪ X/D. However (we are rather embarrassed to admit) Zachary Chase pointed out to the authors that this particular result is trivial, because the assumption implies that |a − a ′ | D whenever a = a ′ . This argument does not, however, appear to extend to the other cases.

5.
A straightforward dyadic decomposition argument would allow one to establish similar results under the assumption that A ⊂ [X] and B ⊂ [Y ]. We leave the details to the reader.
Notation. Our notation is standard. If p is a prime and a ∈ Z, we write v p (a) for the largest k such that p k |a. We extend this to rationals by v p (a/b) = v p (a)−v p (b). Implied constants in the O(), ≪ and ≫ notations are absolute (though they may depend on ε, which is fixed throughout the paper).
Strategy. Our strategy for proving Theorem 1.1 is essentially to proceed by induction on #P(A ∪ B), but we will phrase the argument in terms of a hypothetical counterexample with minimal #P(A ∪ B). The first main business is to show that such a minimal counterexample has a very specific structure.
Though such a statement does not appear explicitly in their work, this proposition should be considered essentially due to Koukoulopoulos and Maynard [1]. We will give a fairly short, self-contained proof. On some level this is equivalent to the argument of [1], but we phrase things rather differently.
To complete the proof of Theorem 1.1, we prove the following counterpart to Proposition 1.2. Evidently, this means that A, B do not in fact give a counterexample to Theorem 1.1. Combining Propositions 1.2 and 1.3 shows that no minimal counterexample to Theorem 1.1 exists, so Theorem 1.1 is true.
The proof of Proposition 1.3 uses some combinatorial arguments and is not found in [1].

Concentrated measures on Z 2
In this section we prove a result about concentration of probability measures on Z 2 . It is the key technical ingredient in the proof of Proposition 1.2, where it is used to concentrate the pair of valuation functions (v p (a), v p (b)) around a diagonal pair (k, k).
Here, as in the rest of the paper, we write q = 2 + ε and write q ′ for the conjugate index to q (i.e. 1 q + 1 q ′ = 1).
Then c 1 9 , and µ is highly concentrated near some point (k, k): Proof. We first prove the lower bound on c. Using (2.1), Z 2 µ(i, j) = 1, m∈Z λ |m| 9 and q ′ < 2, we have where the last step follows from the Cauchy-Schwarz inequality and the assumption The lower bound on c follows.
Bound for Σ 5 . A trivial modification to the argument used for Σ 1 (allowing i = j, which gives just a term with m = 0) shows that Σ 5 ≪ γ 2 q ′ .
Bound for Σ 5 + Σ 6 . By (2.1) and the fact that sup i x i y i = 1 − γ, where we used Cauchy-Schwarz yet again.

Properties of a minimal counterexample
We turn now to the proof of Proposition 1.2. We first reduce matters to the following "local" statement at a single prime p.  P(A, B) be a prime. Then p > p 0 (ε), and there is k p ∈ Z 0 and Ω p ⊂ Ω such that for all (a, b) ∈ Ω p we have |v p (a) −k p | + |v p (b) −k p | 1, and such that |Ω \ Ω p | ≪ p −1−ε/3 |Ω| . Proposition 1.2 follows quickly from this by taking N = p p kp and Ω ′ := p Ω p . We have if p 0 is big enough (this is the point at which p 0 is constrained). It remains, then, to establish Proposition 3.1. Fix, for the rest of this section, the prime p. For i, j ∈ Z 0 , we define A i := {a ∈ A : v p (a) = i}, B j := {b ∈ B : v p (b) = j}, and write α i := |A i | |A| and β j := |B j | |B| for the relative densities of these sets. Write , thus µ is a finitely-supported probability measure on Z 2 0 . For any i, j, consider the setsĀ i := p −i · A i andB j := p −j · B j . These are sets of integers, coprime to p, By the minimality assumption, these sets cannot be a counterexample to Theorem 1.1, and therefore we have the inequality (3.1) On the other hand, Note also that P(Ā i ,B j ) ⊂ P(A, B) \ {p}, and so Comparing (3.1), (3.2) and (3.3) gives, for all i and j, since ε < 1. This puts us in the situation covered by Lemma 2.1, with (in that lemma) The hypotheses of the lemma are satisfied, since p − 1 q 2 − 1 3 4/5. The lemma implies, first of all, that c > 1 10 ; this immediately tells us that p > p 0 . We conclude that there is some k such that This is precisely what is needed in Proposition 3.1, taking

Finishing the argument
In this section we complete the proof of Theorem 1.1 by establishing Proposition 1.3. That is, our task is as follows. Suppose that A ⊂ [X, 2X], B ⊂ [Y, 2Y ], that Ω ⊂ A × B has size δ 2 |A||B|, and that gcd(a, b) D whenever (a, b) ∈ Ω. Suppose that there is some positive integer N such that for all primes p and for all (a, b) ∈ Ω. We are to show that, under these assumptions, we have the bound Let us begin the proof. In the course of the argument it will be convenient to use a little of the language of graph theory. Thus if a ∈ A then we write deg(a) := #{b ∈ B : (a, b) ∈ Ω}, and analogously for b ∈ B. Write A ′ := {a ∈ A : deg(a) > 0} and If a ∈ A ′ then, by (4.1), v p (a/N) ∈ {−1, 0, 1} for all primes p. We define the defect a * to be the product of all primes for which v p (a/N) = 0. Now we make the crucial observation that if (a, b) ∈ Ω then (4.3) To prove this, we take p-adic valuations. It is easily seen that whenever (a, b) ∈ Ω. This follows immediately from (4.1), noting that v p (a * ) = 1 if v p (a/N) = ±1 and v p (a * ) = 0 otherwise, and similarly for v p (b * ). As a consequence of (4.3) and our assumptions, we see that whenever (a, b) ∈ Ω. This would allow us to conclude very quickly, were it not for the fact that the map a → a * need not be injective (see Section 5 for some further remarks on this point). Fortunately, we have the following substitute for injectivity.
Proof. If a ∈ A ′ , write a + for the product of all primes with v p (a/N) = 1, and a − for the product of all primes with v p (a/N) = −1. Thus (4.6) (4.7) Since A ⊂ [X, 2X], it follows by multiplying (4.6) and (4.7) that if a * T then Similarly, dividing (4.6) by (4.7), we see that if a * T then It follows from (4.8), (4.9) that the number of choices for the pair (a + , a − ) is at most 2T . However, if we know a + , a − and N then we can recover a uniquely, so the map a → (a + , a − ) is injective. The proof for B ′ is the same.  (a, b) ∈ Ω, so we have the upper bound (4.5). Comparing (4.5), (4.10), (4.11) immediately yields (4.2).

Further results and remarks
Suppose that A and B are finite sets of square-free positive integers. In this instance, we may assume that the positive integer N that satisfies (4.1) is also square-free, and thus the map a → a * is injective, since knowing N and a * determines a. This enables us to circumvent Lemma 4.1, and prove the following theorem.
Proof. The proof of Proposition 1.2 holds mutatis mutandis. Analysing the minimal counterexample as before, we conclude that a * b * Q (by analogy with (4.5)). Using graph theoretic language as before, there exists a setÃ ⊂ A with |Ã| δ|A|/4 such that deg(a) δ|B|/4 for all a ∈Ã. Since a → a * is injective, there is some a ∈Ã for which a * δ|A|/4. LettingB := {b ∈ B : (a, b) ∈ Ω}, we have |B| δ|B|/4 and for all b ∈B we have b * Q/a * . Therefore, since b → b * is injective, we have δ|A| .
This rearranges to |A||B| 16δ −2 Q, which shows that the minimal counterexample is not in fact a counterexample, thus settling the theorem.
One might wonder whether the bound |A||B| ≪ δ −2−ε Q holds for general finite sets of integers A and B (not just for square-frees). However there is a counterexample to this assertion, even with δ = 1, given by A = B = p X p m n : mn X, µ 2 (m) = µ 2 (n) = 1, gcd(m, n) = 1 .
One may establish that for all (a 1 , a 2 ) ∈ A × A one has the bound a 1 a 2 / gcd(a 1 , a 2 ) 2 X 2 . Yet |A| ≫ X log X. By this, one notes that the use of dyadic ranges in the proof of Lemma 4.1 was critical.