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Claim. Rafik Zeraoulia's post of 27 April 2026 in the discussion thread of Problem 261 claims a positive answer to the weakened question of Erdős's 1988 paper, whether some rational $x$ has two representations $x=\sum_{a\in A}a/2^a$ over sets $A$ of distinct positive integers. From $4/2^4=5/2^5+6/2^6$ and $\sum_{m\ge1}m/2^m=2$, the rational $7/4$ is the sum over $\mathbb N\setminus\{5,6\}$ and over $\mathbb N\setminus\{4\}$. The post notes that the two representations share a large common tail and that a stronger non-degeneracy condition would make a different question.

Submission note. Posted to the erdosproblems.com forum by Rafik Zeraoulia on 27 April 2026:

I would like to point out a small observation about the weakened question mentioned in [Er88c], namely the existence of a rational number $x$ which has two representations.

Interpreting a representation as a choice of a set $A\subseteq \mathbb N$ of distinct positive integers, so that

$$x=\sum_{a\in A}\frac{a}{2^a},$$

one gets

such an example from the finite identity

$$> \frac{4}{2^4}=\frac{5}{2^5}+\frac{6}{2^6}.$$

Both sides are equal to $1/4$.

Now use the classical identity

$$\sum_{m=1}^{\infty}\frac{m}{2^m}=2.$$

Let

$$> A=\mathbb N\setminus\{5,6\}, \qquad B=\mathbb N\setminus\{4\}.$$

Then

$$> \sum_{a\in A}\frac{a}{2^a} = 2-\frac{5}{2^5}-\frac{6}{2^6} = 2-\frac14 = > \frac74,$$

whereas

$$\sum_{b\in B}\frac{b}{2^b} = 2-\frac{4}{2^4} = 2-\frac14 > = \frac74.$$

Therefore the rational number

$$x=\frac74$$

has two different

representations:

$$\frac74 = \sum_{a\in \mathbb > N\setminus\{5,6\}}\frac{a}{2^a} = \sum_{b\in \mathbb > N\setminus\{4\}}\frac{b}{2^b}.$$

Thus, under this literal interpretation, the

weakened question has a positive answer. Of course, this uses two representations with a large common tail; if the intended question requires some stronger non-cancellation or non-degeneracy condition, then that would be a different version.

Why it is rejected. It answers a variant, not the Statement. The third question of Problem 261 asks for a rational with at least $2^{\aleph_0}$ representations. Erdős's 1988 paper asks only for a rational with two, a weakening that the problem page records as a variant and that, as a reply in the thread reports, is generally believed to be a misprint for $2^{\aleph_0}$. Two representations of $7/4$ say nothing about whether any rational has $2^{\aleph_0}$ of them, and nothing about the first two questions, so its mathematics answers the variant and does not settle this problem; the value proved is its answer to that variant.

Standing. Rejected, as answering a variant. The identities are elementary and the thread disputes none of them; this corpus has not checked them. formal-conjectures marks its variant erdos_261.variants.two_representations research solved and cites a Lean proof that follows the post; this corpus has not built it.

Depends on. No page of this wiki.