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Source. The opening paragraph, p. 1, and §2 "Prime-power residue families", p. 3, of A Two-Copy Proof of Erdős Problem 126 (2026), a three-page preliminary exposition with no printed author, posted at https://www.erdosproblems.com/static/126-proof.pdf; the edition read is identified on the source card. The result is unnumbered in the print; this page names it the main theorem.

Statement

Main theorem (p. 1, quoted). "Let $A\subset\mathbb N$ be finite, and let $r$ be the number of primes dividing at least one sum $a+b$ with distinct $a,b\in A$. We prove $|A|\ll r^2+1$, where the implied constant is absolute."

The print lets $\mathbb N$ contain $0$: §2 (p. 3) removes $0$ from $A$ and restores it at the end. It concludes, also on p. 1, that the extremal function $f(n)$ of Problem 126 satisfies $f(n)\gg\sqrt n$ and hence $f(n)/\log n\to\infty$; the print does not define $f(n)$ itself.

Explicit constants (derived on this page, not printed). Write $S(A)$ for the set of primes in the statement, so $r=|S(A)|$. Tracking the constants in the proof of Proposition 1 gives $n\leq3r^2$ for a set of $n\geq2$ positive integers, and for every finite $A\subseteq\{0,1,2,\ldots\}$

$$|A|\leq3|S(A)|^2+2.$$

The additive $2$ is needed: $A=\{0,1\}$ has empty $S(A)$. With $f(n)$ the least value of $|S(A)|$ over $n$-element sets, as on the problem page, this gives $f(n)\geq\sqrt{(n-2)/3}$ for $n\geq2$. The same constants appear as card_le_three_sq and quadraticBound in the pinned formal module named on the source card.

Read depth. Claims checked: the statement and the argument of §2 were read clause by clause on the printed pages, and the explicit constants above were derived here from that argument. Nothing here is independently reviewed.

Proof sketch

P. 3. Remove $0$, index the remaining elements $a_i$, and let $\mathcal P$ be the primes dividing some off-diagonal sum. For each $p\in\mathcal P$ and each level $k\geq0$, the negation orbits $\{x,-x\}$ modulo $p^{k+1}$ with $x\ne-x$ and both classes occupied are retained as labelled supports of weight $\log p$; for a fixed $p$ they form a laminar family. Each element gets the sign of its $p$-free part, read modulo $p$ for odd $p$ and modulo $4$ for $p=2$, from a sign choice that is opposite on each pair $\{u,-u\}$; this separates the two classes of each retained orbit. The self-opposite classes are collected in a positive semidefinite kernel $B$.

Off the diagonal, unique factorization of $a_i+a_j$ gives $C+B=\log(a_i+a_j)$, displayed as (7), and of $|a_i-a_j|$ gives $R+B\leq\log|a_i-a_j|$, displayed as (8); since $|a_i-a_j|<a_i+a_j$, $R<C$. On the diagonal $C=0$ and $B(i,i)\leq\log(2a_i)$, so the [[arithmetic_functions/adamczewski_2026_erdos126/logarithmic_kernel|logarithmic kernel]] $L(i,j)=\log(a_i+a_j)$ exceeds $C+B$ only by a nonnegative diagonal. Hence $C$ is conditionally negative semidefinite, and Proposition 1 bounds $|A\setminus\{0\}|$. Restoring $0$ adds at most one element and does not enlarge the set of primes.

The constants above come from Proposition 1's bound $n\leq3r^2$ when $|A\setminus\{0\}|\geq2$; when $|A\setminus\{0\}|\leq1$, $|A|\leq2$.

Dependencies

Proposition 1 and the [[arithmetic_functions/adamczewski_2026_erdos126/logarithmic_kernel|logarithmic kernel identity]] of the same exposition; elementary prime factorization.

Bears on

• Problem 126: the problem asks whether $f(n)/\log n\to\infty$, where $f(n)$ is the largest number such that every $n$-element set of natural numbers has at least $f(n)$ distinct prime factors in its product of off-diagonal pair sums. That number is the least $|S(A)|$ over such sets, so the theorem gives $f(n)\gg\sqrt n$ and answers the question affirmatively. The exposition is not refereed; the problem's standing is recorded on its claim pages.