Goldbach Through the Ring

48/105 = phi(210) / 105

Every even number is a sum of two primes -- but can they share a residue class? In Z/210 = 2*3*5*7, the answer depends on null coordinates. Two or more nulls among {3, 5, 7} 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 residue class (mod 210) if and only if n has at most one null coordinate among {3, 5, 7}. Success rate: 48/105 = phi(210) / 105 = 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 mod-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 / 2 = 105

Even residues mod 210. The denominator of the success ratio.

Success rate

48/105 = phi(210) / 105

Exactly the ratio of units to even residues. Not approximate.

Z/210

2 * 3 * 5 * 7 = 210

The fourth ring in the chain. Where residue classes live. Four channels.

Null Coordinates

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

Nulls	Example	Same-class?	Why
0 nulls	n = 100 (mod 210)	YES	All channels open, many class pairs
1 null (mod 3)	n = 42 (mod 210)	YES	One channel closed, still enough room
2 nulls (3, 5)	n = 30 (mod 210)	NO	CRT forces p = 3 AND p = 5, impossible
3 nulls (3, 5, 7)	n = 210 (mod 210)	NO	All odd channels null, completely blocked

Contrast

Aspect	Classical Goldbach	Same-Class Goldbach
Statement	Every even n > 2 = p + q	n = p + q with p = q (mod 210)
Condition	None (always works, conjectured)	n has at most 1 null coordinate in {3, 5, 7}
Why it works	Heuristic (unproved since 1742)	CRT forces: 2+ nulls make same-class impossible. PROVED
Success rate	100% (conjectured)	48/105 = phi(210)/105 = 45.7% exactly
Deeper meaning	Primes are dense enough	Ring 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 (mod-3 null, still works), 30 (mod-3 + mod-5 null, impossible), 210 (all null, impossible), 9998 (large, see how many).

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