Recognition Physics φ · Erdős–Straus Lab

A 78-year-old conjecture, searched by cost. 62,499 for 62,499.

The Erdős–Straus conjecture (1948) says that for every integer n ≥ 2 the fraction 4/n splits into just three unit fractions: 4/n = 1/x + 1/y + 1/z. Nobody has proved it. Most residue classes of n were settled decades ago by explicit formulas; the stubborn survivors live in n ≡ 1 (mod 4), the hard class, where no single formula is known to work. This script attacks exactly that class, guided by the recognition-cost reduction: instead of searching x, y, z blind, it searches the small parameter c = 4x − n (the cost class), where a Lean-certified identity turns each candidate into a factoring problem with an instant yes/no. Every hit is re-verified against the exact integer identity before it counts.

Distinction-mathematics demo · reduction certified in IndisputableMonolith.NumberTheory.ErdosStrausRCL · run verified 2026-07-08

62,499 / 62,499hard-class n solved
14max offset ever needed
95.8%solved at c = 3
stdlibno deps · ~4s
The reduction

From three unknowns to one small number

c = 4x − n, N = n·x (c·y − N)(c·z − N) = N²

Fix n, walk x up from ⌊n/4⌋+1. Each step fixes c and N; a solution exists iff N² has a factor pair with both factors ≡ −N (mod c). Factor N once, enumerate divisors, check the congruence. The three-unknown search collapses to a one-parameter walk plus a divisor scan.

The measured shape

The cost histogram is steep

Of 62,499 hard-class n up to 250,000: 59,867 solve at the very first cost class c = 3, another 2,265 at c = 7, and the tail dies fast; the rarest classes (c = 39, c = 59) appear exactly once each. No n ever needed more than 14 steps of the walk. Cost-ordering the search is the whole point: the cheapest recognition class almost always suffices.

Built-in verifier

No solution counts unchecked

4xyz = n(xy + xz + yz)

Every candidate (x, y, z) is re-verified against this exact integer identity, which is algebraically equivalent to 4/n = 1/x + 1/y + 1/z. A single failure would print as a miss. The run reports zero misses; that line is the falsifiable claim of this page.

Where the hard class lands, from the run

Number of hard-class n (up to 250,000) whose first solution appears at each cost class c, log scale. Three orders of magnitude separate c = 3 from the tail; the distribution is the measured fingerprint of the cost-guided walk.

c histogram from output.txt: 59867, 2265, 241, 42, 37, 32, 2, 11, 1, 1 across c = 3, 7, 11, 15, 19, 23, 27, 31, 39, 59. Counts on a log axis.

Run it yourself

One file, Python standard library only, about four seconds for the quarter-million sweep. Change the limit in the last line to push further.

$ python3 erdos_straus_search.py
checked hard class n = 1 mod 4, n <= 250000
misses: 0
max offset used: 14
c histogram:
  c=3: 59867
  c=7: 2265
  c=11: 241
  ...
  c=59: 1
↓ Search script Reference output
Honest scope

MEASURED, plus a THEOREM-grade skeleton, and explicitly NOT a proof of the conjecture. The reduction identities and several residue-class theorems (5 mod 12, 9 mod 12, and the divisor-pair certificates) are proved in Lean in IndisputableMonolith.NumberTheory.ErdosStrausRCL; the sweep result (62,499/62,499, max offset 14, the c histogram) is a finite computation, verified per-solution by the exact identity. Erdős–Straus itself remains OPEN: a finite search settles nothing asymptotically, and the interesting open question this instrument raises is whether the offset bound (≤ 14 here) and the c-histogram tail stay bounded as the limit grows. A hard-class n with no solution at any offset would falsify the conjecture itself; a runaway offset would falsify the cost-guided search's implicit bet that cheap classes suffice.

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