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Statement

Lemma 2 (p. 2). Let $d$ be a positive integer and $X$ a finite set of positive integers, and put

$$s=\sum_{x\in X}\frac1x,\qquad n=\sum_{x\in X}x^d$$

(the paper's (1)). Then

$$|X|\le s\sqrt[d+1]{\frac ns}$$

(the paper's (2)) and

$$\Bigl\lceil\frac1s\Bigr\rceil\le\min X\le\Bigl\lfloor\min\Bigl\{\sqrt[d+1]{\frac ns},\ \sqrt[d]{n}\Bigr\}\Bigr\rfloor$$

(the paper's (3)).

The lemma is stated for every finite set $X$; it is not restricted to representations, where $s=1$.

Source. Max A. Alekseyev, On partitions into squares of distinct integers whose reciprocals sum to 1, in The Mathematics of Various Entertaining Subjects, Volume 3 (2019), pp. 213--221, read in the arXiv version identified on the source card: the lemma and its proof on p. 2, in Section 1 (pp. 2--3).

Read depth. Claims checked: the statement and its hypotheses were read clause by clause on the page image; the short proof was read and its steps followed.

Proof pointer

P. 2. With $X=\{x_1<\cdots<x_k\}$, the harmonic mean $k/s$ is at most the $d$-th power mean $\sqrt[d]{n/k}$, which rearranges to (2). The lower bound in (3) comes from $1/x_1\le s$, and the two upper bounds from $s\le k/x_1$ combined with (2), and from $x_1^d\le n$.

Use in the paper

With $d=2$ the bounds (3), applied to the remaining reciprocal sum and the remaining sum of squares after each chosen element, give the range of the next element in a backtracking search (Algorithm 1, p. 3); the bound (2) makes the search terminate. Run on $m=8542$ with $s=1$, the search finds no representation, which is the paper's Lemma 3 (p. 3), the sharpness half of Theorem 1. The search was not rerun here.

Bears on

• Problem 283: no case of the problem on its own; the lemma bounds the computation behind the paper's Lemma 3, that $8542$ is not a sum of squares of distinct positive integers whose reciprocals sum to $1$, which makes the threshold of Theorem 1 for $p(x)=x^2$ exact.