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Statement

For a set $\mathcal A$ of integers, $h^\wedge\mathcal A$ denotes "the set of integers which can be represented as a sum of $h$ distinct elements from $\mathcal A$", and $\mathcal A$ is admissible when $s^\wedge\mathcal A\cap t^\wedge\mathcal A=\emptyset$ for all $s\ne t$ (printed p. 33): an integer representable as a sum of $s$ distinct elements of $\mathcal A$ determines $s$.

Theorem 1 (printed p. 34). "There exists a constant $C$ such that any admissible set $\mathcal A$ included in $[1,N]$ satisfies $\operatorname{card}\mathcal A\le2N^{1/2}+CN^{5/12}$."

The abstract states the same as "the cardinality of such an admissible subset $\mathcal A$ is at most $(2+o(1))\sqrt N$. As shown by Straus, the constant 2 cannot be improved upon." The introduction (p. 34) records the earlier bounds it improves, Erdős's $O(N^{5/6})$ and Straus's $(4/\sqrt3+o(1))\sqrt N$ with the constant "recently reduced" by Erdős, Nicolas and Sárközy (Théorème 1), and Straus's example of an admissible $\mathcal A\subset[1,N]$ with $|\mathcal A|=\lfloor2\sqrt N-1\rfloor$, which shows the constant 2 is best possible. The proof (Section 6) proves the theorem with $C=10^6$ for all sufficiently large $N$: it assumes $\operatorname{card}\mathcal A>2N^{1/2}+10^6N^{5/12}$ and derives a contradiction; the theorem's constant $C$ absorbs the small $N$.

Source. J-M. Deshouillers and G. A. Freiman, On an additive problem of Erdős and Straus, 1, Israel J. Math. 92 (1995), 33--43, doi:10.1007/BF02762069; the definition on printed p. 33 (PDF p. 1), Theorem 1 on printed p. 34 (PDF p. 2), the proof on printed pp. 41--42 (PDF pp. 9--10) of the publisher's PDF, read on the page images (the OCR text layer garbles the mathematics). The artifact is identified in the source digest.

Read depth. Claims checked: the definition, the abstract, the account of the earlier bounds and Theorem 1 were read clause by clause on the page images on 2026-09-22. The proof (Section 6, pp. 41--42) was read in full on the page images and its reduction to Theorem 2 followed; the inequality $4dM+1\le d^2+(S-1)^2$ it ends with was checked here against its stated inputs $S\ge2\sqrt N+1$ and $dM\le N$. Theorem 2, which the proof uses, was checked at its statement only (its proof, Sections 1--5, read for structure). Nothing here is independently reviewed.

Proof pointer

Section 6 (pp. 41--42). For $N$ large and $\operatorname{card}\mathcal A>2N^{1/2}+10^6N^{5/12}$, Theorem 2 gives $\mathcal C\subset\mathcal A$, a difference $d$ and an integer $t$ such that $t^\wedge\mathcal C$ contains $u,u+d,\ldots,u+ld$ with $l>2N^{5/6}$, and $\mathcal A\setminus\mathcal C=\{a_1<a_2<\cdots\}$ lies in a progression of difference $d$ with at most $N^{7/12}$ terms. Choose $S>2N^{1/2}+1$ with $S\equiv d\pmod2$ and $\operatorname{card}(\mathcal A\setminus\mathcal C)>S$, and put $U=(S+d)/2$. The $U$-fold sums $a_1+\cdots+a_{U-1}+a_j$ ($U\le j\le S$), $a_1+\cdots+a_U+a_S$, ..., $a_{S-U+1}+\cdots+a_S$ are congruent modulo $d$ with consecutive gaps at most $dN^{7/12}$, so adding the progression in $t^\wedge\mathcal C$ shows that $t^\wedge\mathcal C+U^\wedge(\mathcal A\setminus\mathcal C)$ contains every integer congruent to $Ua_1+u$ modulo $d$ in $\mathcal J=[u+a_1+\cdots+a_U,\;u+a_{S-U+1}+\cdots+a_S]$. The integer $u+a_{U+1}+\cdots+a_S$ lies in $t^\wedge\mathcal C+(U-d)^\wedge(\mathcal A\setminus\mathcal C)$ and in the same residue class, and it lies in $\mathcal J$ as soon as $a_1+\cdots+a_U\le a_{U+1}+\cdots+a_S$ (the paper's $(*)$). With $Md$ a multiple of $d$ in $[a_U,a_{U+1})$, the left side is at most $(M-U+1)d+\cdots+Md$ and the right side at least $Md+\cdots+(M+S-U-1)d$, so $(*)$ follows from $4dM+1\le d^2+(S-1)^2$, which holds since $S\ge2\sqrt N+1$ and $dM\le a_{U+1}\le N$. The two sets $t^\wedge\mathcal C+ (U-d)^\wedge(\mathcal A\setminus\mathcal C)\subset(t+U-d)^\wedge\mathcal A$ and $t^\wedge\mathcal C+U^\wedge(\mathcal A\setminus\mathcal C)\subset (t+U)^\wedge\mathcal A$ then share an element, against admissibility.

Dependencies

Within the paper: Theorem 2 (p. 34), proved in Sections 1--5 (pp. 35--41) from Proposition 1 (a small $s^\wedge\mathcal A$), Proposition 2 (a subset $\mathcal B$ with $|4^\wedge\mathcal B|<5.8|\mathcal B|$), Freiman's inverse theorem in its easiest case (Proposition 3.1, cited to Freiman's 1973 monograph, Thm. 1.9, and his 1959 paper, neither held) and Proposition 4, the special case $\lambda=5.8$ of Theorem 3. Outside it: Straus's upper bound $(4/\sqrt3+o(1))\sqrt N$ (J. Math. Sci. 1 (1966), 77--80, not held), used as the standing assumption $\operatorname{card}\mathcal A\le2.31\sqrt N$ (pp. 34--35); the bound is reproved as Lemme 2 of the 1991 paper.

Bears on

• Problem 874: the problem's $k(N)$ is the largest admissible subset of $\{1,\ldots,N\}$, so Theorem 1 gives $k(N)\le2N^{1/2}+CN^{5/12}$, and with Straus's block $k(N)=2N^{1/2}+O(N^{5/12})$, hence $k(N)\sim2N^{1/2}$, the affirmative answer to the site's asymptotic question; this is the paper's "$(2+o(1))\sqrt N$", cited on the problem page. For large $N$ the bound is superseded by the exact Theorem 1 of the 1999 sequel, $k(N)\le2\sqrt{N+1/4}-1$, which the problem's status rests on.
• Problem 875: for an infinite admissible $A=\{a_1<a_2<\cdots\}$ the set $A\cap[1,x]$ is admissible, so $A(x)\le2x^{1/2}+Cx^{5/12}$ for all $x$; with $x=a_n$ this gives $a_n\ge(1+o(1))n^2/4$, and a gap bound $a_{n+1}-a_n\le n^c$ for all large $n$ forces $c\ge1$, the deductions the problem page makes from the sharper 1999 bound. Deduction made here.