The site writes $1_A\ast 1_A(n)$, which counts ordered pairs $(a,b)\in A\times A$ with $a+b=n$, $a=b$ included. Under that count the question fails trivially at $C=2$: two distinct elements $x<y$ of one part give the two ordered pairs $(x,y)$ and $(y,x)$ for $x+y$, so the bound $1_{A_i}\ast1_{A_i}(n)<2$ allows at most one element in each part, while an infinite Sidon set such as the powers of two has $1_A\ast1_A(n)\le2$ for every $n$ and cannot be split into finitely many parts of one element. The answer at $C=2$ then says nothing about the question Erdős and Newman asked; the case $C=1$ is answered no under every count, as in the poser's texts, and is not counted as a failure.

The change replaces $1_A\ast1_A(n)$ and $1_{A_i}\ast1_{A_i}(n)$ by $f_A(n)$ and $f_{A_i}(n)$ and inserts the definition of $f$; nothing else changes. The evidence is the poser's own statement of the question, which the site's wording renders. Erdős and Graham [ErGr80], printed p. 48 (Old and new problems and results in combinatorial number theory), write: "For the sequence $A = \{a_k\}$, let $f(n) = f_A(n)$ denote the number of solutions to $n = a_i + a_j$, $i < j$", and their question 7 (pp. 48--49), credited to Erdős and D. J. Newman, asks with that $f_A$ whether a set with $f_A(n)\le c$ for all $n$ can always be partitioned into $t=t(c)$ subsets $A_k$ with $f_{A_k}(n)<c$ for all $k$ and $n$. Erdős [Er80e], p. 43 (Some applications of Ramsey's theorem to additive number theory), also counts representations without regard to order: in the proof of his Theorem 1 a sum of four distinct terms $n_i+n_j+n_r+n_s$ has three representations, and $2n_i+n_r+n_s$ has one. That paper counts the sum $a+a$ once ($2n_i+2n_j$ has one representation), where [ErGr80] excludes it. The difference changes no recorded answer: every construction in [Er80e] reaches $C$ representations only through sums of two distinct elements, and [ErGr80], p. 49, reports that Nešetřil and Rödl answered the question as it states it in the negative for all $c$. No text of the poser counts ordered pairs, so the defect is the site's. The form follows the poser's statement of this question, not the results that settle it.

Results about the site's wording are credited here and count for nothing. AxiomMath published on 2026-06-18 a Lean 4 development generated by its prover AxiomProver (pinned file) that refutes the ordered-count statement at $C=2$ with the powers of two, with copies in the repository Jayyhk/erdos-lean (added 2026-06-22) and in Boris Alexeev's lean-proofs collection (added 2026-08-26). It answers the site's ordered count, not the corrected Statement's count of sums of two distinct elements, so it does not count toward the standing and its claim page (AxiomMath, 2026) is rejected.