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Statement

Setting (pp. 1--2). The paper's equation (1) is

$$\frac{n}{2^n}=\sum_{i=1}^{k}\frac{a_i}{2^{a_i}},\qquad k>1,$$

in integers $n,k,a_1,\ldots,a_k$ with $a_i<a_{i+1}$ for $i=1,\ldots,k-1$ (abstract); the solutions are sought in positive integers.

Theorem 2.1 (p. 2).

1. For fixed $k$, if (1) has a solution then $n\le2^{k+1}-k-2$.
2. If (1) holds then $n+1\le a_1\le n+3$ and $2^{a_k-a_{k-1}}$ divides $a_k$. Moreover, if $n\ge2^{j+1}-j$ for some $1\le j<k$, then $a_i=n+i$ for $i=1,\ldots,j$.

Remark 2.2 (p. 3) notes that the bound in part 1 is attained: for fixed $k$ and $n=2^{k+1}-k-2$, the choice $a_i=n+i$ ($i=1,\ldots,k$) solves (1), an identity the paper attributes to Borwein and Loring.

Source. Sz. Tengely, M. Ulas and J. Zygadło, On a Diophantine equation of Erdős and Graham, J. Number Theory 217 (2020), 445--459, doi:10.1016/j.jnt.2020.05.006, read in arXiv:2008.01501v1 as identified on the source card; labels and pages are that preprint's. Theorem 2.1 on p. 2, its proof on pp. 2--3, Remark 2.2 on p. 3.

Read depth. Claims checked: the statement and Remark 2.2 were read clause by clause on the page images. The proof was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Pages 2--3. Since $x/2^x$ decreases for $x\ge1$, a solution has $a_i\ge n+i$, and comparing with $\sum_{i\le k}(n+i)/2^{n+i}$ gives part 1. If $a_1\ge n+4$ the same comparison forces $n<1$. Multiplying (1) by $2^{a_{k-1}}$ shows $a_k/2^{a_k-a_{k-1}}$ is an integer. The last claim is an induction on $j$, bounding the tail by $\sum_{i>n+j+1}i/2^i$ when $a_{j+1}\ge n+j+2$.

Dependencies

None.

Bears on

• Problem 261: the problem's finite sums with $t\ge2$ distinct terms are the solutions of (1) once the terms are ordered. The theorem restricts which numbers of terms and which first terms a given $n$ can use; by itself it neither produces a representation nor rules one out for any $n$, and it does not touch the question on rationals with $2^{\aleph_0}$ representations. Remark 2.2 records Borwein and Loring's identity, which gives a representation for $n=2^{k+1}-k-2$ with each $k>1$ and so infinitely many $n$ for the problem's first question; the result is Borwein and Loring's, not this paper's.