Goldbach Through the Ring

48/105 = SEES / HYDOR

Every even number is a sum of two primes -- but can they share a signature class? In Z/210 = D*K*E*b, the answer depends on null coordinates. Two or more nulls among {K, E, b} make same-class sums impossible. The ring gates Goldbach.

The Same-Class Goldbach Theorem

Same-Class Goldbach (PROVED)
An even number n can be written as p + q where p, q are primes in the same signature class (mod 210) if and only if n has at most one null coordinate among {3, 5, 7}. Success rate: 48/105 = phi(210) / HYDOR = 45.7%.

Proof sketch: If n is divisible by two odd primes k, e from {3, 5, 7}, then same-class means p = q (mod k) AND p = q (mod e). But p + q = 0 (mod k) and p + q = 0 (mod e). Together: 2p = 0 (mod k), forcing p = 0 (mod k), so p = k. Similarly p = e. But p cannot equal both. Contradiction.

With at most one null, enough room remains in the class structure for same-class pairs to exist. The D=2 channel is always null for even numbers (trivially), so it does not constrain.

The Landscape of 210

48 unit classes
phi(210) = 48
The coprime residues mod 210. Primes (except 2,3,5,7) live here.
105 even residues
210 / D = HYDOR
Even residues mod 210. The denominator of the success ratio.
Success rate
48/105 = SEES/HYDOR
Exactly the ratio of units to even residues. Not approximate.
DATA ring
Z/210 = D*K*E*b
The fourth ring in the chain. Where signature classes live. Four channels.

Null Coordinates

A null coordinate at prime p means n = 0 (mod p). Each null restricts same-class possibilities by forcing prime factors. Two nulls from {3,5,7} leave no room.

NullsExampleSame-class?Why
0 nullsn = 100 (mod 210)YESAll channels open, many class pairs
1 null (K)n = 42 (mod 210)YESOne channel closed, still enough room
2 nulls (K,E)n = 30 (mod 210)NOCRT forces p = 3 AND p = 5, impossible
3 nulls (K,E,b)n = 210 (mod 210)NOAll odd channels null, completely blocked

Contrast

AspectClassical GoldbachSame-Class Goldbach
StatementEvery even n > 2 = p + qn = p + q with p = q (mod 210)
ConditionNone (always works, conjectured)n has at most 1 null coordinate in {K,E,b}
Why it worksHeuristic (unproved since 1742)CRT forces: 2+ nulls make same-class impossible. PROVED
Success rate100% (conjectured)48/105 = phi(210)/HYDOR = 45.7% exactly
Deeper meaningPrimes are dense enoughRing structure controls which sums are ALLOWED

Explore: Same-Class Pairs

Enter an even number. The widget finds its null coordinates, predicts whether same-class pairs exist, then searches for them.

Even number:

Try: 100 (0 nulls, many pairs), 42 (K null, still works), 30 (K+E null, impossible), 210 (all null, impossible), 9998 (large, see how many).

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