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Problem 1056

../

claims/: The 4 claim pages of Problem 1056, one per claimant's result; the problem's standing derives from them.


Statement. Let k≥2k\geq 2. Does there exist a prime pp and consecutive intervals I1,…,IkI_1,\ldots,I_k such that

∏n∈Iin≡1(modp)\prod_{n\in I_i}n \equiv 1\pmod{p}

for all 1≤i≤k1\leq i\leq k?

Formulation. The site's wording, like the formal-conjectures statement, puts no lower bound on an interval's length and admits the one-element interval {1}\{1\}, whose product is 11. Under it the case k=2k=2 holds at p=5p=5 with {1}\{1\} and {2,3}\{2,3\}, and, by Wilson's theorem, at every prime p≥5p\ge5 with {1}\{1\} and [2,p−2][2,p-2]. The sources exclude that interval: Erdős's example modulo 1111, Guy's A15, Prime Puzzles problem 27 (which counts the string {1}\{1\} only by a stated choice) and OEIS A060427 (least primes 1111, 1717 and 2323 for two, three and four products). A witness under the sources' reading is also one under the site's wording, and a witness under the site's wording for kk intervals gives a witness under the sources' reading for k−1k-1, so the question for every kk has the same answer under both readings. The page's standing targets the site's wording; the claim pages state their instances under the sources' reading. A witness whose common residue is 11, such as Mąkowski's or the tetrads below, gains one interval under the site's wording.

Status. Open.

Source. erdosproblems.com/1056, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #1056, https://www.erdosproblems.com/1056.

References.

  • [Gu04] Guy, Richard K., Unsolved problems in number theory. 3rd ed., Problem Books in Mathematics, Springer, New York (2004), xviii+437 pp. Section A15 "Congruent products of consecutive numbers", printed p. 54, which records Erdős's example 3⋅4≡5⋅6⋅7≡1 mod 113\cdot4\equiv5\cdot6\cdot7\equiv1\bmod11, asks for the least prime with three congruent products, and reports the least primes found for several numbers of products. Library home: guy_2004_unsolved_problems_number_theory.
  • [Ma83] Mąkowski, Andrzej, On a number-theoretical problem of Erdős. Elem. Math. (1983), 101-102.

Formalization. Statement in formal-conjectures.

Current assessment

The question for every kk is open; the instances settled so far are finite witnesses, and no construction gives kk intervals for every kk. The site labels the problem OPEN, and the derived standing stays open.

The case k=2k=2 is Erdős's example 3⋅4≡5⋅6⋅7≡1(mod11)3\cdot4\equiv5\cdot6\cdot7\equiv1\pmod{11}, from a letter of 31 October 1979 that Guy reports in A15 ([[problems/diophantine_problems/E1056/claims/1981_01_01_erdos|Erdős's claim page]]). The case k=3k=3 is Mąkowski's example modulo 1717 in Elemente der Mathematik [Ma83] ([[problems/diophantine_problems/E1056/claims/1983_01_01_makowski|Mąkowski's claim page]], accepted on that publication). Guy also reports Mąkowski's example modulo 2323, four intervals (row 66 of the Noll--Simmons table in A15), and examples sent by W. Narkiewicz, rows 77 to 99 of that table, which give up to eight intervals modulo 599599; Narkiewicz's examples are reported from correspondence, with no publication to page. Landon Noll and Chuck Simmons asked more generally for nn equal factorials q1!≡⋯≡qn!(modp)q_1!\equiv\cdots\equiv q_n!\pmod p, which give n−1n-1 adjacent intervals of product 11 when the common residue is nonzero, and Guy prints their table of least primes for n≤11n\le11, the last being p=3011p=3011 with nine intervals; it is a computed table reported by Guy, and the cases it settles are covered by Andersen's claim page.

J. K. Andersen extended the least primes to k=14k=14 on Prime Puzzles problem 27 in 2007, with explicit intervals; his k=14k=14 witness modulo 1042800710428007 settles every kk from 22 to 1414 ([[problems/diophantine_problems/E1056/claims/2007_05_04_andersen|Andersen's claim page]]). OEIS A060427 lists these least primes. Kenta Kitamura's forum post of 21 June 2026 restates that witness and is recorded on Andersen's page. Agustín-Aquino and Hernández Santiago prove that infinitely many primes have four distinct nn with n!≡1(modp)n!\equiv1\pmod p, which gives the case k=3k=3 for infinitely many primes ([[problems/diophantine_problems/E1056/claims/2026_06_03_agustin_aquino_hernandez_santiago|their claim page]]).

A forum post of 3 January 2026 by Lorenzo Luccioli, made with the help of Aristotle, gives a Lean derivation of the Noll--Simmons formulation from the original question; it relates two formulations, settles no instance and has no page. The other thread comments point to OEIS A060427 and to a related construction of Hardy and Subbarao, and claim no result.

Linked library material

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