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Statement

Theorem 1 (p. 117, quoted). "For $N>N_0$, there exists a sequence $\mathcal A\subset\{1, 2, \ldots, N\}$ such that (1) $|\mathcal A|>\frac{1}{248}\log N$ and $a+a'$ is squarefree for all $a\in\mathcal A$, $a'\in\mathcal A$."

The condition includes $a=a'$, so every $2a$ is squarefree as well. The threshold $N_0$ is not made explicit.

Proof pointer

Section 2, pp. 118--120. With $p_i$ the $i$-th prime, choose $K$ so that the product of $p_i^2$ over $i<K$ is below $N^{1/2}$ and over $i\le K$ is at least $N^{1/2}$ (display (3)), and let $P$ be the latter product, so $\log P\sim\frac12\log N$. The integers $n\equiv2\pmod 4$ that are divisible by no $p_i^2$ with $2\le i\le K$ fill $\prod_{i=2}^K(p_i^2-1)>P/5$ residue classes modulo $P$; intersected with $\{1,\ldots,N\}$ they give that many arithmetic progressions of difference $P$, each of about $N/P$ terms. A non-squarefree term of any of them is divisible by $p_i^2$ for some $K<i\le\pi(N^{1/2})$, and these number fewer than $3N/(P\log P)$ in all, so one progression has fewer than $15N/(P\log P)$ of them. Between consecutive non-squarefree terms of that progression, or between one of them and an end of $\{1,\ldots,N\}$, lies a run of $M>\frac1{124}\log N$ consecutive squarefree terms $2b,2b+P,\ldots,2b+(M-1)P$ (all terms are even), and the set $\{b,b+P,\ldots,b+[\frac{M-1}{2}]P\}$ has all its pairwise sums in that run and $[\frac{M+1}2]\ge\frac M2>\frac1{248}\log N$ elements (the print writes $[\frac{M+1}2]>\frac M2$, which fails for even $M$; the bound $\ge$ is enough).

Read depth

Claims checked: the statement was read clause by clause on the page image of the print (p. 117), and the proof on pp. 118--120 was followed. Nothing here is independently reviewed.

Dependencies

None in the corpus. External input: the prime number theorem.

Source. P. Erdős and A. Sárközy, On divisibility properties of integers of the form $a+a'$, Acta Math. Hungar. 50 (1987), no. 1--2, 117--122, doi:10.1007/BF01903370; the edition read is named on the source card.

Bears on

• Problem 1109: gives $f(N)>\frac1{248}\log N$ for $N>N_0$, the lower bound the problem's estimates start from; it answers neither of the problem's questions, which ask for upper bounds.