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Erdős #1194 · additive combinatorics / additive basis

Let A⊂NA\subset\mathbb{N} be such that every integer n≥1n\geq 1 can be written uniquely as an−bna_n-b_n for some an,bn∈Aa_n,b_n\in A. How fast must an/na_n/n increase?

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Source reports OPENAs of 2026-09-08Source & evidence ↓

Problem statement

Let A⊂NA\subset\mathbb{N} be such that every integer n≥1n\geq 1 can be written uniquely as an−bna_n-b_n for some an,bn∈Aa_n,b_n\in A. How fast must an/na_n/n increase?

Lean 4

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Background

Such a set is called a perfect difference set; they can be constructed, for example, by a greedy approach (as detailed by Lev [Le04]). The greedy construction achieves an≪n3a_n\ll n^3.

Erdős [Er80] wrote it is 'easy to see' that

lim sup⁡n→∞ann=∞.\limsup_{n\to \infty}\frac{a_n}{n}=\infty.

Indeed, since AA must be a Sidon set, if an≤f(n)na_n \leq f(n)n for all nn, then all differences in [1,x/f(x)][1,x/f(x)] can be realised by differences from A∩[1,x]A\cap [1,x], and hence ∣A∩[1,x]∣2≫x/f(x)\lvert A\cap [1,x]\rvert^2 \gg x/f(x). Erdős (see for example [HaRo66]) proved that for infinitely many xx ∣A∩[1,x]∣≪(x/log⁡x)1/2\lvert A\cap [1,x]\rvert \ll (x/\log x)^{1/2}, which is a contradiction for a suitable choice of ff.

This argument proves that an≫nlog⁡na_n\gg n\log n for infinitely many nn.

Cilleruelo and Nathanson [CiNa08] describe a method for construction dense perfect difference sets from dense Sidon sets, allowing many of the results for the latter to transfer to the former.

GPT-5.4 Pro (see the comments) has given an argument which proves that an≫n2−o(1)a_n\gg n^{2-o(1)} infinitely often; in fact an≫n2/f(n)a_n \gg n^2/f(n) infinitely often for any function ff such that ∑1nf(n)\sum \frac{1}{nf(n)} diverges.

Progress & source assessment

The source labels this entry OPEN in the 2026-09-08 snapshot. Editorial progress notes are reproduced as source commentary, not independently verified results. Prize amounts and Lean badges at the source are not local awards or verification.

References
  1. [1]Thomas Bloom (ed.), Erdős Problems #1194 link

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