Suppose $\mathscr{A} \subset \mathbb{N}$ enjoys for all large enough cutoffs $X$ the following properties:

  • $|\mathscr{A} \cap [1,X]| > \sqrt{X}/10$; and
  • the discriminant $\prod_{\alpha \neq \beta}|\alpha-\beta|$ over $\alpha,\beta \in \mathscr{A} \cap [1,X]$ is composed only of primes smaller than $10\sqrt{X}$.

Is there a quadratic $an^2 + bn+c$, with $a > 0$, such that $\mathscr{A} \subset \{an^2+bn +c \mid n \in \mathbb{N} \}$?

Notes. The values of any such quadratic has both properties, upon replacing the number $10$ with an appropriate constant. The question is whether those are essentially all examples. It is motivated by a similar unsolved problem in the so-called "inverse large sieve," in which condition two is replaced by a sifting condition $|\mathscr{A} \mod{p}| \leq (p+1)/2$ for all primes $p$ (and the same conclusion is expected to hold up to finitely many exceptions); cf. Ben Green and Adam Harper's recent GAFA paper, Inverse questions for the large sieve. I am interested even in a proof that $\liminf |\mathscr{A} \cap [1,X]|/\sqrt{X} < \infty$.

  • 5
    $\begingroup$ This is a pretty question, and it could be hard. I would phrase it as follows: let $B \subset \{1,\dots, X\}$ be the set of $10\sqrt{X}$-smooth numbers, so $|B| \sim c X$ for a certain $c$ given by the Dickman de-Bruijn function. What is the largest $S$ for which $S - S \subset B$? In general it's very, very hard to show that $|S| < \sqrt{X}$ in problems of this type, because Fourier-based methods don't work. For a generic $B$, one expects the biggest $S$ to have size around $O(\log X)$ (in fact I proved this) so to have such a large $S$ indicates an unusual property of this particular $B$. $\endgroup$ – Ben Green Nov 17 '15 at 18:14
  • $\begingroup$ @BenGreen: Thank you, that's the better way to think of this problem. What is a reference for the proof of the logarithmic bound for generic $B$? $\endgroup$ – Vesselin Dimitrov Nov 18 '15 at 1:52
  • 1
    $\begingroup$ Here is the paper. arxiv.org/abs/math/0304183 $\endgroup$ – Ben Green Nov 18 '15 at 18:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.