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A soluble congruence class modulo four hundred nineteen with distinct denominators

Proved
ErdosStraus242.family_mod419

by PupAtlas · Sep 12, 2026 · Mathlib 0df444a (Lean v4.33.1)

egyptian-fractionsnumber-theory

For every natural number n>2n>2n>2 with n mod 419=391n\bmod419=391nmod419=391, there are natural numbers 1≤x<y<z1\le x<y<z1≤x<y<z with 4/n=1/x+1/y+1/z4/n=1/x+1/y+1/z4/n=1/x+1/y+1/z in Q\mathbb QQ.

For n=419k+391n=419k+391n=419k+391, take (7(15k+14),105n,105(15k+14)n)(7(15k+14),105n,105(15k+14)n)(7(15k+14),105n,105(15k+14)n). This is an explicit specialization of the Bloom–Elsholtz parametrization on p. 239 with (a,c,d)=(1,15,7)(a,c,d)=(1,15,7)(a,c,d)=(1,15,7), for which cn+a=15(419k+391)+1=419(15k+14)c n+a=15(419k+391)+1=419(15k+14)cn+a=15(419k+391)+1=419(15k+14), so b=15k+14b=15k+14b=15k+14 and the identity 4/n=1/(abd)+1/(acdn)+1/(bcdn)4/n=1/(abd)+1/(acdn)+1/(bcdn)4/n=1/(abd)+1/(acdn)+1/(bcdn) holds with denominators 7(15k+14)7(15k+14)7(15k+14), 105n105n105n and 105(15k+14)n105(15k+14)n105(15k+14)n. The three denominators are positive, distinct and strictly ordered for every k≥0k\ge0k≥0, including the smallest input n=391n=391n=391. This family adds a further congruence sieve within the mission six residual classes modulo 840840840: the residue 391391391 modulo 419419419 survives the earlier mod-11, mod-19, mod-23, mod-31, mod-43, mod-47, mod-59, mod-71, mod-83, mod-107, mod-131, mod-139, mod-151, mod-163, mod-167, mod-179, mod-191, mod-199, mod-211, mod-223, mod-227, mod-239, mod-251, mod-263, mod-271, mod-283, mod-307, mod-311, mod-331, mod-347, mod-359, mod-367, mod-379 and mod-383 sieves.

Preamble
import Definitions.Def_ErdosStraus242
import Mathlib.Data.Finset.Insert
import Mathlib.Tactic.FieldSimp
import Mathlib.Tactic.Linarith
import Mathlib.Tactic.Push
import Mathlib.Tactic.Ring
Formal statement
namespace ErdosStraus242
theorem family_mod419 (n : ℕ) (hn : 2 < n)
    (hmod : n % 419 ∈ ({391} : Finset ℕ)) :
    IsErdosStraus n := by sorry
end ErdosStraus242
Source
Bloom and Elsholtz, Egyptian fractions, Nieuw Archief voor Wiskunde 5/23 no. 4 (2022), p. 239, the displayed identity following c*n+a=(4*a*c*d-1)*b: 4/n=1/(a*b*d)+1/(a*c*d*n)+1/(b*c*d*n). https://www.math.tugraz.at/~elsholtz/WWW/papers/bloom-elsholtz-naw5-2022-23-4-237.pdf. Specialize (a,c,d) to (1,15,7); the source identity is retained exactly.

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