Problem 51

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claims/: The 1 claim page of Problem 51, one per claimant's result; the problem's standing derives from them.

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Statement. Is there an infinite set $A\subset \mathbb{N}$ such that for every $a\in A$ there is an integer $n$ such that $\phi(n)=a$, and yet if $n_a$ is the smallest such integer then $n_a/a\to \infty$ as $a\to\infty$?

Status. Open. The site labels the problem OPEN (page last edited 2025-09-30), and its proof-claims tab carries no entry.

Source. erdosproblems.com/51, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #51, https://www.erdosproblems.com/51.

References.

• [Gu04] Guy, Richard K., Unsolved problems in number theory, third edition, Problem Books in Mathematics, Springer (2004), xviii+437 pp.; B36 "Euler's totient function", printed p. 139: Erdős's question whether for every $\epsilon$ there is an $n$ with $\phi(n)=m$, $m<\epsilon n$, and $\phi(t)\ne m$ for all $t<n$, "perhaps there are many such $n$". Library home: guy_2004_unsolved_problems_number_theory.

Formalization. Statement in formal-conjectures.

Current assessment

The site's label is OPEN (site record accessed 2026-09-04, problem page last edited 2025-09-30). No claim settles the question. One claim page is recorded, and it is rejected: a one-page proof posted to the site's discussion thread on 2026-01-11, produced by ChatGPT (free version) for the user who posted it, asserted a yes answer through the products $\prod_{i\le k}(p_i-1)$ and failed at its minimality step the same day, with the curator's counterexamples $\phi(3)=\phi(6)$ and $\phi(3\cdot5\cdot7)=\phi(5\cdot13)$ (the claim page). The one outside result bearing on the question, the OpenAI release's dichotomy for the counts of totients with least preimage between $kx$ and $(k+1)x$ (manuscript of 2026-09-25), gets no claim page: it proves a reformulation of the question, a dichotomy that decides neither answer and settles no instance beyond the known cases $k=1,2$, so Known Results records the theorem and its Lean declarations instead. No further literature search is recorded.

Known Results

The OpenAI release's manuscript An asymptotic formula for the number of totients (2026-09-25; the preprint at the pinned revision, digested on the intake card openai_2026_asymptotic_formula_number_totients) proves, as a companion to its asymptotic formula for the number of totients (whose fixed-scale limit answers the doubling question of Problem 416, and which is a pending claim on that problem's second question), a dichotomy for the totients by least preimage. Write $\ell(v)$ for the least $n$ with $\phi(n)=v$ (the $n_a$ of the Statement), $V(x)$ for the number of totients up to $x$ and, for an integer $k\ge1$, $N_k(x)=\#\{v\le x\ \text{totient}: kx<\ell(v)\le(k+1)x\}$. Its Theorem 2.2 states that $N_k(x)$ has an asymptotic formula on the counting scale of the main theorem, with a coefficient built from finite arithmetic data, and that for each $k$ exactly one of two alternatives holds: if some totient $d$ has $\ell(d)>kd$, then $N_k(x)\asymp_k V(x)$, so a positive proportion of all totients up to $x$ have least preimage in $(kx,(k+1)x]$; if no such $d$ exists, then $N_k(x)=0$ for every $x$. The first alternative holds for $k=1$ and $k=2$. The manuscript's own remark reads the theorem as this page does: a positive answer to the question is equivalent to the first alternative holding for every $k$, and neither the unboundedness of $\ell(d)/d$ over totients nor the classification of the $k$ is established there. The result is therefore a reformulation of the question, not progress on it, and no claim page records it. The release's Lean tree proves the theorem as OAI.TotientAsymptotic.weighted_totient_asymptotic (with weighted_totient_one_two for the $k=1,2$ seeds) and its zero alternative as OAI.TotientAsymptotic.companion_zero_case, in the folder lean/OAI/NumberTheory/TotientAsymptotic at the pinned revision, pinned by the comparator challenges TotientAsymptotic.lean and TotientCompanionZero.lean. These declarations state the manuscript's theorems faithfully, and no declaration of the family asserts or refutes the seed condition for all $k$; this corpus has not built them. The $k=1$ case of the same theorem bears on Problem 417, whose page records the item and why it is not a claim there.

Linked library material

These entries are derived from explicit links on library pages. They are navigation only and do not by themselves record mathematical progress.

• openai_2026_asymptotic_formula_number_totients
• openai_2026_asymptotic_formula_number_totients / theorem_2_2
• guy_2004_unsolved_problems_number_theory