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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. P. Erdős and M. Rosenfeld, The factor-difference set of integers, Acta Arith. 79 (1997), no. 4, 353--359 (card), Proposition 3.2: for every positive integer kk there are integers N1<⋯<NkN_1<\cdots<N_k with ∣⋂i=1kD(Ni)∣≥2\lvert\bigcap_{i=1}^k D(N_i)\rvert\ge2. The proof takes distinct odd primes p1,…,pnp_1,\ldots,p_n, sets $2\alpha=p_1\cdots p_k+p_{k+1}\cdots p_n$ and 2β=p1⋯pk−pk+1⋯pn2\beta=p_1\cdots p_k-p_{k+1}\cdots p_n, and turns each factorization of p1⋯pnp_1\cdots p_n into an integer x2−α2=y2−β2x^2-\alpha^2=y^2-\beta^2 whose factor difference set contains 2α2\alpha and 2β2\beta. With k=2k=2 this answers Problem 885 yes for k=2k=2. The paper also prints two triples, found by Barry Guiduli as the paper credits, whose sets share four values:

{420,3780,14940,76860}⊂D(6925500)∩D(37901500)∩D(108448956),\{420,3780,14940,76860\}\subset D(6925500)\cap D(37901500)\cap D(108448956), {420,3780,61695,154332}⊂D(2778300)∩D(862552800)∩D(5400442044).\{420,3780,61695,154332\}\subset D(2778300)\cap D(862552800)\cap D(5400442044).

All twenty-four memberships hold, since for each listed dd and NN the number d2+4Nd^2+4N is a square of the parity of dd. Either triple answers the problem yes for k=3k=3. The year is the only date the journal record gives, so this page carries the first of January.

Covers. The instances k=2k=2 and k=3k=3.

Depends on. No page of this wiki.

Acceptance. Refereed: Acta Arithmetica (the paper thanks its referee). The site labels the problem OPEN, so its commentary crediting the paper is not acceptance.