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Statement

Notation (p. 204): $\Gamma(N)$ is the set of the subsets of $\{1,\ldots,N\}$, $A(n)$ and $B(n)$ count elements up to $n$, and $[x]$ is the integer part.

Theorem 6 (p. 213, quoted with its displays). "Let"

$$0<\varepsilon<\frac14.\qquad(31)$$

"If $N>N_0(\varepsilon)$, $B\subset\Gamma(N)$ and"

$$B(N)<\frac{1}{2\log1/\varepsilon}\log N\qquad(32)$$

"then there exists a sequence $A\subset\Gamma([N/2])$ such that"

$$A([N/2])>\Bigl(\frac12-\varepsilon\Bigr)\Bigl[\frac N2\Bigr]\qquad(33)$$

"holds and"

$$a_x+a_y=b_z\qquad(34)$$

"is not solvable."

Here $B\subset\Gamma(N)$ and $A\subset\Gamma([N/2])$ mean sets of integers in $\{1,\ldots,N\}$ and $\{1,\ldots,[N/2]\}$. The sum in (34) allows $x=y$.

Role in the paper (p. 212). Section 4 states that for sum intersector sets $B(N)\to+\infty$ must hold, in contrast with Theorem 5 for differences; Theorem 6 is the quantitative form: a $B$ with fewer than $\log N/(2\log1/\varepsilon)$ elements misses the sums of a set $A$ as large as (33).

Sharpness left open (pp. 222--223). The paper does not know whether Theorem 6 is best possible and asks, as question (i): is it true that if $\lim_{N\to+\infty}f(N)=+\infty$, then for $\varepsilon>0$ and $N>N_0(\varepsilon)$ there is a $B\subset\Gamma(N)$ with $B(N)<f(N)\log N$ such that $A\subset\Gamma([N/2])$ and $A([N/2])>\varepsilon[N/2]$ imply the solvability of (34)?

Source. P. Erdős and A. Sárközy, On differences and sums of integers, II, Bull. Soc. Math. Grèce (N.S.) 18 (1977), no. 2, 204--223: the statement on p. 213, the proof on pp. 213--216, question (i) on p. 223. The edition read is identified on the source card.

Read depth. Claims checked: the statement and question (i) were read clause by clause on the printed pages. The proof was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Pp. 213--216. Apply Dirichlet's simultaneous approximation theorem (35) to the $k=B(N)$ numbers $b_i/[N/2]$ with $Q=\frac\varepsilon4[N/2]$, giving $q\le Q$ with every $\lVert qb_i/[N/2]\rVert<Q^{-1/k}$ (36). Let $A$ be the integers $a\le[N/2]$ whose fractional part $\{qa/[N/2]\}$ lies strictly between $\frac1{2Q^{1/k}}$ and $\frac12-\frac1{2Q^{1/k}}$ (40). Every sum of two elements then has $\lVert q(a_x+a_y)/[N/2]\rVert>Q^{-1/k}$, so no $b_z$ is such a sum. The fractional parts are equidistributed over the residues modulo $s$, where $r/s=q/[N/2]$ in lowest terms (37)--(39), which gives $A([N/2])\ge[N/2](\frac12-Q^{-1/k}-\frac\varepsilon2)$ (41), and (32) gives $Q^{-1/k}<2\varepsilon^2\le\varepsilon/2$ for large $N$ (42).