A soluble congruence class modulo one hundred sixty-three with distinct denominators

Proved

ErdosStraus242.family_mod163

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

egyptian-fractionsnumber-theory

For every natural number $n>2$ with $n\bmod163=159$ , there are natural numbers $1\le x<y<z$ with $4/n=1/x+1/y+1/z$ in $\mathbb Q$ .

For $n=163k+159$ , take $(41k+40,41n,41(41k+40)n)$ . This is an explicit specialization of the Bloom–Elsholtz parametrization on p. 239 with $(a,c,d)=(1,41,1)$ , for which $c n+a=41(163k+159)+1=163(41k+40)$ , so $b=41k+40$ and the identity $4/n=1/(abd)+1/(acdn)+1/(bcdn)$ holds with denominators $41k+40$ , $41n$ and $41(41k+40)n$ . The three denominators are positive, distinct and strictly ordered for every $k\ge0$ , including the smallest input $n=159$ . This adds a further congruence sieve within the mission six residual classes modulo $840$ .

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_mod163 (n : ℕ) (hn : 2 < n)
    (hmod : n % 163 ∈ ({159} : 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 (1,41,1); the source identity is retained exactly.

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