Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Statement
The paper gives its theorem no number. The abstract (printed p. 264) states it: "Let be a positive integer and let , be two sets of positive integers such that the product set consists of distinct numbers. Then, for a certain positive constant , , establishing a conjecture made by P. Erdös." The body (p. 264) states the problem with the hypothesis the abstract omits, "Let and be two sequences of integers so that the products are all distinct", and the conjecture as display (2),
"In this paper we wish to prove (2)." The closing line of the proof (p. 269) is the theorem in the paper's final form: "Thus, our result is that for we have , where ", reduced in three printed steps to . Here (Lemma 1, p. 265), is the constant of the Brun-sieve Lemma 2 (p. 266), and are the Mertens constants with for (p. 267), (p. 267) and (p. 268). The paper's footnote fixes "integers" to mean the natural numbers. In the notation of the problem page: for with the products , , , all distinct, for all .
Source. E. Szemerédi, On a Problem of P. Erdös, Journal of Number Theory 8 (1976), no. 3, 264--270; the abstract and display (2) on printed p. 264 (PDF p. 1 of the publisher scan), the closing statement with the constant on p. 269 (PDF p. 6), the proof on pp. 265--269 (PDF pp. 2--6), read on the page images. The edition is identified in the source digest.
Read depth. Claims checked: the abstract, the statement of the problem with (1) and (2), Lemmas 1--3, the two cases and the closing statement with were read clause by clause on the page images. The proof (pp. 265--269) was read in full on the page images for its structure, and no step was checked; the reduction of to its final form was not checked. Nothing here is independently reviewed.
Proof pointer
Pages 265--269. For a set and a prime , is the set of members of divisible by and the set of their quotients by . Lemma 1 (p. 265) passes to , with in which every prime that divides some member divides many: , likewise for ; the paper then assumes this of and . Lemma 2 (p. 266, Brun's method, cited to Halberstam and Roth): for , the integers up to coprime to a set of primes number at most , with absolute. Lemma 3 (p. 266): if for primes then , since and give the equal products ; this is the only use of the distinct-products hypothesis. With , Case I (p. 267), for every : Lemma 2 applied to with the primes not dividing any member of , and to , together with the Mertens bounds and , gives . Case II (pp. 267--269), for a largest such : the quotients in , , are at most , so Lemma 2 bounds and the same for ; if or, by symmetry, is below a threshold of order times a sieve product, the bound follows directly (two subcases on p. 268 according to whether the sieve product over is empty); otherwise Lemma 1 makes at least times the size of the union, so some with has a common element in all , , and then forces two primes with , against Lemma 3 (p. 269). The constant collects the three bounds and the factor of Lemma 1.
Dependencies
Brun's sieve in the form of Lemma 2, cited to Halberstam and Roth, Sequences I (1966), not held; the Mertens product bounds , quoted as well known; the convergence of and of . External premises are taken at statement level; none was checked here. The second proof, Theorem 1 of the 1976 Erdős--Szemerédi paper, has the same outline (its "associated" primes, dyadic blocks, distinct quotient pairs and Brun--Mertens count answer to Lemma 1, the sets , Lemma 3 and Lemma 2 here) and describes itself as "a simpler proof of (4), which nevertheless uses many of the ideas of the original proof" (p. 420).
Bears on
- Problem 490: the theorem is the problem's inequality for with all products distinct, in the paper the site names as its proof; the constant is explicit in terms of the Brun and Mertens constants but not evaluated. On the limit of the paper only records the hope to investigate whether the bound holds "for avery [sic] if " (p. 265), and its construction (1) (p. 264, the primes in against the integers up to not divisible by any of them) gives .