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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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Statement

Problem 7 (p. 346). The authors ask whether there is a Sidon set A⊂{1,2,…,n}\mathcal A\subset\{1,2,\ldots,n\} with ∣A∣≪n1/3|\mathcal A|\ll n^{1/3} that is "maximal" (their quotation marks) in the sense that no b∈{1,2,…,n}b\in\{1,2,\ldots,n\} with b∉Ab\notin\mathcal A leaves A∪{b}\mathcal A\cup\{b\} a Sidon set. In parentheses they add that the answer would throw more light on the role of the greedy algorithm in this field. The Sidon property is the paper's: the sums a+a′a+a' with a≤a′a\le a' in A\mathcal A are distinct (p. 330). The print leaves the quantifier on nn implicit; read with the implied constant independent of nn, the question asks for such a set for every large nn. The paper gives no result on it.

Source. P. Erdős, A. Sárközy, V. T. Sós, On Sum Sets of Sidon Sets, I, J. Number Theory 47 (1994), 329--347, doi:10.1006/jnth.1994.1040; §12, p. 346. The edition read is identified on the source card.

Read depth. Claims checked: the problem was read clause by clause on the page images of the journal print. A question has no proof to check; the note below is the corpus's own.

Note

The exponent 1/31/3 is the least possible. Adding b∉Ab\notin\mathcal A to a Sidon set creates a repeated sum exactly when 2b∈SA2b\in\mathcal S_{\mathcal A} or b+a∈SAb+a\in\mathcal S_{\mathcal A} for some a∈Aa\in\mathcal A. So in a maximal A\mathcal A every such b≤nb\le n is s/2s/2 or s−as-a with s∈SAs\in\mathcal S_{\mathcal A}, a∈Aa\in\mathcal A, and by (2.1) (p. 329) n≤∣A∣+(∣A∣+1)∣SA∣≤∣A∣+∣A∣(∣A∣+1)2/2n\le|\mathcal A|+(|\mathcal A|+1)|\mathcal S_{\mathcal A}|\le|\mathcal A|+|\mathcal A|(|\mathcal A|+1)^2/2, which gives ∣A∣≫n1/3|\mathcal A|\gg n^{1/3}. This deduction is not in the paper.

Dependencies

None.

Bears on

  • Problem 156: Problem 7 is this problem, with nn for NN and ∣A∣≪n1/3|\mathcal A|\ll n^{1/3} for O(N1/3)O(N^{1/3}). The paper poses it and does not resolve it.