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Statement

Setting (Section 2, p. 47). Let a1,…,a2na_1,\ldots,a_{2n} be distinct integers with 1≤ai≤4n1\le a_i\le4n, and let b1,…,b2nb_1,\ldots,b_{2n} be the remaining integers of the interval (1,4n)(1,4n) (the complement of the aia_i among the integers up to 4n4n).

Question (p. 47). Is there always an integer tt for which the number of solutions of ai+t=bja_i+t=b_j is at least nn? The paper says it could not decide this, and notes that the choice ai=n+ia_i=n+i, 1≤i≤2n1\le i\le2n, shows that nn would be best possible.

Theorem (p. 47). There is always an integer tt for which the number of solutions of ai+t=bja_i+t=b_j is at least n/2n/2.

Reported result (p. 47). Scherk (the paper's reference [3]) proved that for a suitable tt the number of solutions is at least (2−2)n(2-\sqrt2)n.

The English summary (p. 48) states the theorem as: "I show that there exists an integer xx so that there are at least n/2n/2 bb's among the integers ai+xa_i+x. Scherk improved this to (2−2)n(2-\sqrt{2})n. It is not known whether this can further be improved to nn."

Proof pointer

Averaging (p. 47): over the shifts −4n<x<4n-4n<x<4n the equation ai+x=bja_i+x=b_j has 4n24n^2 solutions in all, one for each pair (ai,bj)(a_i,b_j), and there are fewer than 8n8n shifts, so some shift carries at least n/2n/2 of them.

Read depth. Claims checked: the setting, the question, the theorem, the example ai=n+ia_i=n+i and Scherk's bound were read clause by clause on the page images of pp. 47 and 48; the averaging was re-derived here.

Source. P. Erdős, Some remarks on number theory (in Hebrew), Riveon Lematematika 9 (1955), 45--48; the edition read is named on the source card.

Dependencies

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Bears on

  • Problem 36: with N=2nN=2n, the aia_i and bjb_j form a partition of {1,…,2N}\{1,\ldots,2N\} into two halves of size NN, and the count of solutions of ai+t=bja_i+t=b_j is the problem's count of differences b−a=tb-a=t (the problem counts a−b=xa-b=x, which is the same quantity with the halves' names exchanged). In the problem's normalization the theorem is c≥1/4c\ge1/4, Scherk's bound is c≥1−1/2c\ge1-1/\sqrt2, and the paper's question asks whether c=1/2c=1/2 is admissible, which the example ai=n+ia_i=n+i shows would be optimal; the paper poses this as a question it could not decide, not as a conjecture. The paper treats only totals 4n4n, that is even NN, and proves nothing beyond the bound 1/41/4.