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Problem 456

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claims/: The 2 claim pages of Problem 456, one per claimant's result; the problem's standing derives from them.


Statement. Let pnp_n be the smallest prime ≡1(modn)\equiv 1\pmod{n} and let mnm_n be the smallest integer such that n∣ϕ(mn)n\mid \phi(m_n).

Is it true that mn<pnm_n<p_n for almost all nn? Does pn/mn→∞p_n/m_n\to \infty for almost all nn? Are there infinitely many primes pp such that p−1p-1 is the only nn for which mn=pm_n=p?

Status. Open. The site's proof-claims tab carries one full proof claim, by David Turturean (naming GPT-6-Astra Pro as the system used, with earlier work by ChatGPT-5.5-Pro and Claude), submitted 2026-09-23 with an Overleaf write-up and a Lean repository: it answers the three questions no, no and yes, unconditionally, through a positive lower density for the set of nn with mn=pnm_n=p_n and a lower bound of order X/(log⁡X)55X/(\log X)^{55} for the uniqueness primes up to XX; an earlier manuscript of the same author, posted on 4 May 2026, answered the third question only under Dickson's conjecture. The site labels the problem OPEN (page last edited 07 October 2025), no referee or outside reviewer has accepted the argument, and the claim is recorded as pending on its claim page; the derived standing is claimed.

Source. erdosproblems.com/456, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #456, https://www.erdosproblems.com/456.

References.

Formalization. Statement in the file ErdosProblems/456.lean of formal-conjectures as of its last change, 18 September 2026: erdos_456.parts.i, parts.ii and parts.iii, the three questions, each an answer(sorry) equivalence marked research open with no formal_proof attribute.

Current assessment

The question (site formulation). With pnp_n the least prime ≡1(modn)\equiv1\pmod n and mnm_n the least integer with n∣ϕ(mn)n\mid\phi(m_n): whether mn<pnm_n<p_n for almost all nn, whether pn/mn→∞p_n/m_n\to\infty for almost all nn, and whether infinitely many primes pp have p−1p-1 as the only nn with mn=pm_n=p. OPEN.

Standing. One pending full claim, by David Turturean (2026-09-23; claim page): the answers no, no and yes, through a positive lower density for {n:mn=pn}\{n:m_n=p_n\} and a lower bound of order X/(log⁡X)55X/(\log X)^{55} for the uniqueness primes up to XX. His earlier manuscript of 4 May 2026, which answers the first two questions no (claim page), is a pending partial claim. No referee, outside reviewer or the site's curator has accepted either argument, so the derived standing is claimed and the mathematical standing is unsettled.

Search scope. The site's page, its comments and its proof-claims thread (one claim, no comments), the formal-conjectures statement file, the community database (which lists the problem as open with its statement formalized, as of its last update on 2026-06-07) and the claimant's two manuscripts, the earlier one through its library card; no other literature search was made.

Known Results

  • Trivially mn≤pnm_n\le p_n for every nn, since n∣pn−1=ϕ(pn)n\mid p_n-1=\phi(p_n), and Linnik's theorem gives pn≤nO(1)p_n\le n^{O(1)} (site commentary).
  • If n≥2n\ge2 and n+1n+1 is prime then mn=pn=n+1m_n=p_n=n+1, since ϕ(m)<n\phi(m)<n for every m≤nm\le n (site commentary).
  • Erdős [Er79e] states, as easy to show, that mn<pnm_n<p_n for infinitely many nn and that mn/n→∞m_n/n\to\infty for almost all nn (site commentary).
  • van Doorn observes in the site's comments that n=22k+1n=2^{2k+1} with k≥1k\ge1 gives mn≤2nm_n\le2n and pn≥2n+1p_n\ge2n+1, so mn<pnm_n<p_n along that sequence.
  • Pending: Turturean's claim above. His earlier manuscript of 4 May 2026 (claim page; card) proves the positive-density theorem and the first two answers (its Theorem 1.1 and Corollary 1.2) and answers the third question only under Dickson's conjecture for the triple t,2t+1,8t+1t,2t+1,8t+1 (its Theorem 1.4); the claim of 2026-09-23 removes that hypothesis.

Linked library material

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