# Erdős conjecture on arithmetic progressions

https://en.wikipedia.org/wiki/Erdős_conjecture_on_arithmetic_progressions

Erdős' conjecture on arithmetic progressions, often referred to as the Erdős–Turán conjecture, is a conjecture in arithmetic combinatorics (not to be confused with the Erdős–Turán conjecture on additive bases). It states that if the sum of the reciprocals of the members of a set A of positive integers diverges, then A contains arbitrarily long arithmetic progressions.

Formally, the conjecture states that if A is a large set in the sense that

${\displaystyle \sum _{n\in A}{\frac {1}{n}}\ =\ \infty ,}$

then A contains arithmetic progressions of any given length, meaning subsets of the form ${\displaystyle \{a,a{+}c,a{+}2c,\ldots ,a{+}kc\}\subset A}$ for arbitrarily large k.

## History

In 1936, Erdős and Turán made the weaker conjecture that any set of integers with positive natural density contains infinitely many 3 term arithmetic progressions. [1] This was proven by Klaus Roth in 1952, and generalized to arbitrarily long arithmetic progressions by Szemerédi in 1975 in what is now known as Szemerédi's theorem.

In a 1976 talk titled "To the memory of my lifelong friend and collaborator Paul Turán," Paul Erdős offered a prize of US$3000 for a proof of this conjecture. [2] As of 2008 the problem is worth US$5000. [3]

## Progress and related results

Unsolved problem in mathematics:

Does every large set of natural numbers contain arbitrarily long arithmetic progressions?

Erdős' conjecture on arithmetic progressions can be viewed as a stronger version of Szemerédi's theorem. Because the sum of the reciprocals of the primes diverges, the Green–Tao theorem on arithmetic progressions is a special case of the conjecture.

The weaker claim that A must contain infinitely many arithmetic progressions of length 3 is a consequence of an improved bound in Roth's theorem, which appears as the main result in a 2020 preprint by Bloom and Sisask. [4] The former strongest bound in Roth's theorem is due to Bloom. [5]

The converse of the conjecture is not true. For example, the set {1, 10, 11, 100, 101, 102, 1000, 1001, 1002, 1003, 10000, ...} contains arithmetic progressions of every finite length, but the sum of the reciprocals of its elements converges.

4. ^ Bloom, Thomas F.; Sisask, Olof (2020). "Breaking the logarithmic barrier in Roth's theorem on arithmetic progressions". arXiv:. Cite journal requires |journal= ( help)