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Let such that is irrational. Is the multiset
complete? That is, can all sufficiently large natural numbers be written as
for some finite ?
What if is replaced by some ?
Source: erdosproblems.com/354
No claim settles this problem.
Open, the site's label OPEN, which attaches to the pair of questions. The first question (base ) has two accepted partial claims: the bounty site Conjectures.io's certified Lean proof, answering it yes for all with irrational, and Hegyvári's 1989 theorem (Acta Math. Hungar.; refereed), answering it yes when one coefficient is a dyadic rational and the other is not. It also has one pending partial claim, the Yu–Chen manuscript, claiming the stronger strong completeness. The second question has a pending partial claim under each of its two readings: Kitamura's Lean proof at one base answers yes under the reading "for some ", and Geneson's Salem-base counterexample answers no under the reading "for every ". No claim settles the pair of questions, so the derived standing is open; the full answer to the first question rests on a bounty site's acceptance alone, with no refereed publication and no erdosproblems.com acceptance, and the Current assessment records its provenance and limits.