---

Statement

The sets are sets of positive integers $1\le a_1<\cdots<a_k\le n$, and the sums $a_i+a_j$ run over $i\le j$, equal summands allowed (the count $\binom{k+1}2$ of all formal sums on pp. 196 and 205). Page 204 introduces the proposition: "If we omit the restriction on $k$ (the number of elements in the set), then obviously nearly all numbers up to $n$ can have a unique representation as $a_i+a_j$".

Proposition 2. "We can construct a set of positive integers $1\le a_1< \cdots<a_k\le n$ so that at least $n-2^{3/2}n^{1/2}$ numbers up to $n$ have a unique representation as $a_i+a_j$."

Remark. "We think that the term $2^{3/2}n^{1/2}$ cannot be replaced by $o(n^{1/2})$ in Proposition 2, but we cannot prove this even if we take a much larger set of $Cn^{1/2}$ elements in the interval $[1,n]$."

Both as printed on p. 204. In the notation of Problem 14, with $B$ the set of integers representable in exactly one way as a sum of two elements of $A$, the proposition gives an $A$ with $|{1,\ldots,N}\setminus B|\le 2^{3/2}N^{1/2}$, and the Remark is the problem's second question, whether $o(N^{1/2})$ is possible, stated as the authors' expectation that it is not.

Source. P. Erdős and R. Freud, On Sums of a Sidon-Sequence, J. Number Theory 38 (1991), 196--205; Proposition 2 with its proof and the Remark on printed p. 204 (PDF p. 9 of the publisher's open-archive scan), read on the page image. The artifact is identified in the source digest.

Read depth. Claims checked: the statement, the introducing sentence and the Remark were read clause by clause on the page image. The proof (two sentences) was read in full and followed, with the count spelled out below. Nothing here is independently reviewed.

Proof pointer

Page 204. "Take the set $1,2,\ldots,w,2w,\ldots$ with $w=\lceil(n/2)^{1/2} \rceil$ having about $3(n/2)^{1/2}$ elements. Then all numbers below $n$ which are greater than $2w$ and are not multiples of $w$ have a unique representation as $a_i+a_j$." Spelled out here, not a review verdict: a number $m=qw+r$ with $q\ge2$ and $1\le r\le w-1$ is $qw+r$ with $qw$ and $r$ in the set; any other representation $a+b$ with $a\ge b$ has $a$ a multiple $jw$ (since $a\le w$ forces $m\le2w$) and $b=m-jw\in[1,w]$, which forces $j=q$. The excluded numbers are the $2w$ numbers up to $2w$ and the about $n/w$ multiples of $w$, together $2w+n/w\sim2^{3/2}n^{1/2}$ at $w\sim(n/2)^{1/2}$. Equal summands change nothing: $2a\le2w$ for $a\le w$, and $2jw$ is a multiple of $w$, so no double lands on a counted number.

Dependencies

None; the construction is explicit and elementary.

Bears on

• Problem 14: the Remark is the problem's second question, and the proposition is the construction showing the exceptional set can be as small as $2^{3/2}N^{1/2}$; the paper's convention (positive integers, $i\le j$) is the one recorded on the problem page.