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Statement

With $w(A)=\sum_{1<a\in A}1/a$:

Theorem 6 (p. 8): "Consider any partition of the set of perfect squares greater than $1$ into two non-empty parts, $X$ and $Y$. Then there exist non-empty finite subsets $X'$ and $Y'$ of $X$ and $Y$ respectively such that $w(X')=w(Y')$."

Source. D. Larsen, Sufficiently abundant numbers are pseudoperfect, 9-page manuscript (GitHub Larsen-Daniel/Erdos-318, 318.pdf, commit 39139e2b of 1 February 2026); Theorem 6 on p. 8, proof on pp. 8--9; $w$ defined on p. 2. Read on the page image of p. 8 and in the text layer.

Read depth. Claims checked: the statement and the definition of $w$ were read clause by clause. The proof was read for structure only (below) and is not verified here.

Relation to Problem 318

Let $A=\{n^2:n\ge2\}$ and let $f:A\to\{-1,1\}$ be non-constant. Put $X=f^{-1}(1)$ and $Y=f^{-1}(-1)$; both are non-empty and they partition $A$. Theorem 6 gives non-empty finite $X'\subseteq X$, $Y'\subseteq Y$ with $w(X')=w(Y')$, and since $1\notin A$ here $w(B)=\sum_{b\in B}1/b$. Then $S=X'\cup Y'$ is finite and non-empty and $\sum_{n\in S}f(n)/n=w(X')-w(Y')=0$. Conversely, a finite non-empty $S$ with $\sum_{n\in S}f(n)/n=0$ meets both $X$ and $Y$ (a sum of one sign is not zero), and $S\cap X$, $S\cap Y$ have equal $w$. So Theorem 6 is exactly the affirmative answer to the third question of Problem 318. The deduction is written here for that page and is not taken from the source.

Proof pointer and sketch (pp. 8--9)

Assume $X$ is infinite. Take $N$ large in terms of $\min X$ and $\min Y$ and $z=N^24^N$; let $\mathcal B$ be the $N$-smooth squares below $z$ and $\mathcal Q$ the squares of the primes in $(N,\sqrt z)$, so that $\mathcal Q$ splits into dyadic blocks; let $Z=\mathcal B\cdot\mathrm{Div}(\mathcal Q)$ and $Y_{\mathrm{fin}}=Y\cap Z$. Replacing $Y'$ by its complement in $Y_{\mathrm{fin}}$ turns the goal into a subset $Z'\subseteq Z$ containing an element of $X$ with $w(Z')=w(Y_{\mathrm{fin}})$. A greedy pull-back over the $N$-smooth $d_i$ in $(1,N^2)$ (the recursion $a_i=a_{i-1}-1/d_i^2$, $b_i=b_{i-1}-1/d_i^2$ when $b_{i-1}\ge a_i(1+\varepsilon/8)^{-1}$) produces a target $b_l$ with $1+\varepsilon/8\le a_l/b_l=O(\min X+\min Y)$; the removed squares form a set $S$ with $w(S)=b_0-b_l$. Theorem 4 is then applied with $\beta=1/2$, $\epsilon=(\min X)^{-1}$, $L=N^2$ and $\ell/k=b_l$ to $\mathcal D=Z\setminus\{1,d_1^2,\ldots,d_l^2\}$ (Hypothesis 3 is checked from the primes $p$ with $p^2/4<h<p^2/2$), giving a second subset $\mathcal D'\ne Y_{\mathrm{fin}}\setminus S$ with $w(\mathcal D')\equiv b_l\pmod 1$, hence $=b_l$ since $w(\mathcal D)<1$; $Z'=S\cup\mathcal D'$ then has $w(Z')=w(Y_{\mathrm{fin}})$ and is not a subset of $Y_{\mathrm{fin}}$, so it contains an element of $X$.

Dependencies

Theorem 4 and Hypothesis 3 of the same paper (the circle-method theorem, stated on p. 5 with its proof on pp. 5--7, not checked here); the tail estimate $\sum_{j>d}1/j^2$ used in display (15); the count of primes $p$ with $p^2/4<h<p^2/2$ behind the bound $\sum_{d}h_d^2/d^2\gg\sqrt h/\log h$ that checks Hypothesis 3 (p. 9).

Standing

An unrefereed manuscript with a declared AI-assistance acknowledgment (proofreading), read statically; the site's Problem 318 page accepts the result (last edited 1 April 2026), and no independent review or journal record was found on 2026-09-18. Consumers state the theorem with this qualification.

Bears on. #318: the third question, answered yes by the equivalence above.