Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Statement
Theorem 3 (p. 3). "For any set of natural numbers , there are at least natural numbers in the set ."
The paper offers it as the symmetric version of its Unsolved problem, since is not symmetric in and , and calls it "a best possible result" (p. 3): taking gives equality, as the Remark on p. 3 notes. The statement places no finiteness hypothesis on ; its vector form, Theorem 4, is stated for finite sets.
In exponent vectors over the primes dividing members of , the number has exponent vector $d(\mathbf a,\mathbf b)=(|a_1-b_1|,\dots,|a_n-b_n|) =\delta(\mathbf a,\mathbf b)+\delta(\mathbf b,\mathbf a)$, and the paper states that Theorem 3 is equivalent to Theorem 4 (p. 3).
Equality. The Remark (p. 3) announces that equality holds only for the sets of integers , where are positive rationals with and for , the exponents satisfy for some bounds , the paper writes for a subgroup of (the exponents taken modulo 2), and makes all these numbers integers. It gives , with squarefree dividing , as a further example. The Remark defers the proof to section 3 (pp. 6--7), which states the characterization in vector form as Proposition 1 (p. 6) and only sketches its proof.
Source. A. Granville and F. Roesler, The set of differences of a given set, Amer. Math. Monthly 106 (1999), no. 4, 338--344; Theorem 3 and the Remark on p. 3 of the authors' eight-page preprint, read on the page image. The journal version was not compared.
Read depth. Claims checked: the statement, the Remark and the reduction to Theorem 4 were read clause by clause on the page image of p. 3. The equality characterization (Proposition 1, pp. 6--7) was read for its statement only.
Proof pointer
Equivalent to Theorem 4 through the exponent-vector translation above; the proof of Theorem 4 is in section 2 (pp. 4--5).
Dependencies
Bears on
None recorded. The quantity is the symmetric analogue of the ratio count of Problem 539; the paper derives no bound on that problem's from it.