---

Claim. Let $A$ be the set of squarefull positive integers, those in which every prime factor occurs to exponent at least two, $1$ included, and let $r(n)$ count the ordered pairs $(a,b)\in A^2$ with $a+b=n$; the write-up's Problem 1.1 restates the question of Problem 943 as whether $r(n)=n^{o(1)}$. Theorem 1.2 of the 24-page write-up Erdős Problem 943 on sums of two squarefull numbers, dated 18 March 2026 (the first link above), states that for every $\varepsilon>0$ and every $n\ge1$

$$r(n)\ll_\varepsilon n^{2/5+\varepsilon}.$$

The proof (Section 3) writes each summand uniquely as $x^2y^3$ with $y$ squarefree and sorts the solutions of $x_1^2y_1^3+x_2^2y_2^3=n$ into dyadic boxes $x_i\sim X_i$, $y_i\sim Y_i$. With $y_1,y_2$ fixed, the solutions inject into the representations of one integer $M\le n^{4/3}$ by a form $X^2+dY^2$ with $d$ squarefree, which number at most $6\tau(M)$ by an ideal count in $\mathbb{Q}(\sqrt{-d})$ (Lemma 2.2 and Proposition 3.1). With $x_1,x_2$ fixed, the pairs $(y_1,y_2)$ are primitive points of height at most $n^{1/3}$ on a diagonal cubic, whose number is $O_\varepsilon(n^\varepsilon)$ by Heath-Brown's uniform count (Proposition 3.2). So a box holds $\ll_\varepsilon n^{\varepsilon}\min(X_1X_2,Y_1Y_2)$ solutions, and $(X_1X_2)^2(Y_1Y_2)^3\le n^2$ makes that minimum at most $n^{2/5}$. Sections 4 to 10 study the coprime subcount, the representations with $(a,b)=1$: they give reformulations, bounds for a fixed squarefree cube factor of the first summand and for dyadic congruence boxes, and a second-moment barrier, and pose as Problem 10.1 a first-moment estimate that would make the coprime subcount $n^{o(1)}$. The write-up states that no reduction of $r(n)$ to this model is known, since dividing a representation by $\gcd(a,b)$ need not leave both summands squarefull (Remark 4.1). Section 11 surveys the literature of 2025 and 2026.

Submission note. Posted to the erdosproblems.com forum by Przemek Chojecki (account Przemek) on 18 March 2026:

I did an intensive back-and-forth with GPT-5.4 Pro on this problem and here is the write-up with:

a) an unconditional partial result

$r(n):=\#\{(a,b)\in A^2:\ a+b=n\} \ll_\epsilon n^{2/5+\epsilon}$

b) the strongest exact reformulations we currently have for a coprime model, and isolate where the square-sieve approach really breaks.

c) summarization of the 2025–2026 literature surrounding squarefull values, modular square roots, quadratic large sieves, and related averaging problems, and explain precisely which theorem is still missing. There is an on-going research work, so noting this will probably help finish the full problem in the near future (either by humans or new LLMs).

Covers. A partial result toward the answer yes: the bound $1_A\ast1_A(n)\ll_\varepsilon n^{2/5+\varepsilon}$ in place of the $n^{o(1)}$ the question asks for. Whether $1_A\ast1_A(n)=n^{o(1)}$ is not settled. The write-up proves no sharper bound on the full count, and its results on the coprime subcount do not bound it.

Claimant and assistance. The write-up prints Przemyslaw Chojecki as its only author and names no AI system. Chojecki posted it in the erdosproblems.com discussion thread on 18 March 2026, under the name Przemek Chojecki, saying it came out of an intensive back-and-forth with GPT-5.4 Pro. Chojecki's earlier comment in the thread, of 3 February 2026, proposed a reading of the problem through Campana points and claimed no result; Section 11 of the write-up refers to it.

Relation to Plasma AI's claim. Plasma AI's claim page records the same bound for the same count, proved in L24, an author-recorded result of later date; that page names this write-up as a pending claim with the same bound. L24 uses a norm count of the same kind for fixed cube factors. In place of the diagonal-cubic count it uses the counts of either summand's two parameters, which determine the representation, and combines three bounds in each dyadic box, so it takes no input from Heath-Brown's preprint.

Depends on. No page of this wiki. Beyond the divisor bound and the lemmas the write-up proves, its proof rests on one cited theorem, Theorem 1.3 of D. R. Heath-Brown's preprint Counting square-full solutions to $x+y=z$ (arXiv:2601.07817), the write-up's Lemma 2.3; that preprint is not filed in the library.

Standing. Claimed. This page rests on the write-up as read at the first link on 2026-10-18, whose title page is dated 18 March 2026. The erdosproblems.com curator does not credit it: the erdosproblems.com page (last edited 31 October 2025) does not mention the write-up, and its proof-claims tab lists no claim. A reply in the thread the same day, from Nat Sothanaphan, reads "Thanks. Standard check finds no issues.", its word "check" linking a shared ChatGPT conversation; that is not a named review of the proof. No arXiv version or journal record was found on 2026-10-18, there is no Lean proof, and nothing in this repository has verified the proof.