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The Block Proof · An Attempt on Goldbach

Fold the number line. Punch out the composites. See what survives.

A visual retelling of the folded-strip approach to the Goldbach conjecture — where every even E becomes a row of columns, composites become holes, and a single surviving column is a proof for that number. It works beautifully… right up to the one inequality nobody can close.

00The strip, live

Pick an even number. Each column is a pair (x, E−x). If either side is composite, that side is a hole. Goldbach holds for E the moment one column has no hole on either side.

E = 122E = 100E = 7426E = 210E = 512E = 1000E = 48
prime block ■hole × (composite) — label = smallest factorsurvivor: both prime ⇒ E = x + (E−x)
E
122
first survivor g(E)
13
its partner
109
Goldbach pairs
4
√E (hole-makers ≤)
11.05
E^(1/4)
3.32
representation found122 = 13 + 109 — the first column with no hole on either side. Both 13 and 109 are prime.
x
E−x
■
7
3
■
3
5
■
5
7
3
■
9
■
3
11
■
■
13
3
■
15
■
3
17
■
■
19
3
■
21
■
3
23
5
■
25
3
5
27
■
3
29
■
7
31
3
■
33
5
3
35
■
5
37
3
■
39
■
3
41
■
■
43
3
7
45
■
3
47
7
■
49
3
■
51
■
3
53
5
■
55
3
5
57
■
3
59
■
■
61
Top row = left block x. Middle row = its reflection E−x. A number = the smallest prime factor punching that hole; ■ = prime (no hole). A column boxed in green is the first place both sides survive.

IISieve the odds, pair the survivors

A second, more tactile way to see it: draw every odd number below E as a pile of unit blocks, let each composite's smallest factor carve it into equal groups, and watch which piles survive as primes — then two survivors rebuild E. It lives on its own page.

▶ Open the unit-block sieve

01The fold

Take an even number E. Instead of scanning all the way to E, use only the left half of the folded strip:

3, 5, 7, … < E/2

Each block x is paired with its reflection E − x. A block is solid (■) when it is prime and a hole (×) when it is composite. Goldbach fails for E exactly when every single column carries at least one hole:

x E-x ■ | × × | ■ × | × ■ | × ... there can never be ■ | ■

One clean column — both sides prime — and you have written E = x + (E−x).

02Every hole has an owner

Label each hole by its smallest prime factor. That avoids all double-counting and gives each hole a single owner:

3 owns x = 3q, q ≥ 3 5 owns x = 5q, q ≥ 5, 3 ∤ q 7 owns x = 7q, q ≥ 7, 3 ∤ q, 5 ∤ q p owns x = pq, q ≥ p, q has no prime factor < p

An immediate consequence: a prime p cannot punch a genuinely new hole until p². Late primes arrive with almost no territory left.

the shrinking share
3 : ~ 1/3 5 : (1/5)(1 − 1/3) = 2/15 7 : (1/7)(1 − 1/3)(1 − 1/5) = 8/105 11: (1/11)(1 − 1/3)(1 − 1/5)(1 − 1/7) ...

03Why pure hole-counting can’t win

For failure, holes must cover every column: H_L + H_R ≥ L. So a counting proof would need fewer than half the blocks to be composite. For large E that is simply false — primes thin out to density 1 / log E, so there are far more than enough holes to go around.

route eliminatedPure hole counting cannot prove Goldbach. There are numerically enough holes to cover everything. The holes are just not freely placeable — and that is the whole game.

04Positioning: the two tracks

A column is killed by prime p exactly when x or E−x is divisible by p:

x ≡ 0 (mod p) or x ≡ E (mod p)

Two residue classes per prime — and if p | E they collapse into one. So among any p consecutive blocks, at least p − 2 escape p. The holes repeat on rigid, E-locked tracks; they cannot be slid around to plug gaps by hand.

rigorousOver a complete multiplication packet P = 3·5·7···p, the number of surviving positions is
C_p(E) = ∏(q−2) [q ∤ E] · ∏(q−1) [q | E] > 0
A full packet can never be entirely holed.

05The square-root layering

Any composite x ≤ L has a factor ≤ √L. So the first L blocks are decided by small primes only — and the structure nests:

E → E^(1/2) → E^(1/4) → E^(1/8) → …

Every genuinely new hole from a prime p > E^(1/4) lands above √E. The early strip is controlled entirely by tiny primes; the heavy factor-complexity is geometrically pushed to the right.

0 -------------- √E -------------- E/2 -------------- E ↑ large hole-makers CANNOT own anything here ↑ their new holes begin over here

06The whole thing reduces to one line

Define g(E) as the first x ≥ 3 where neither x nor E−x is punched. After sieving through √E, an unpunched number below E must be prime. So:

g(E) < E/2 for every even E > 4 ⟹ g(E) prime and E − g(E) prime ⟹ E = g(E) + (E − g(E)) ← Goldbach

Equivalently, the Finite Two-Track Covering Bound: the linked tracks x ≡ 0, E (mod p) for odd primes p ≤ √E cannot cover the whole prefix [3, E/2].

even sharper formA counterexample would force: for every prime q ≤ √E, the number E − q is composite. So the tracks q ≡ E (mod p) would have to cover every prime q ≤ √E. One uncovered prime finishes it. (But note: this stronger statement can fail even when Goldbach is true via a large prime — so it can’t just be assumed.)

07Where it honestly stops

proved
  • Every composite hole below E is generated by a prime ≤ √E.
  • A prime p makes no genuinely new hole before p².
  • New p-holes have the shape p·q with q ≥ p and q free of smaller factors.
  • Between consecutive prime squares, no new hole-maker appears.
  • In the folded strip, p-holes sit only on x ≡ 0, E (mod p).
  • Over a complete packet, the tracks cannot cover everything.
not provedThat the tracks cannot cover the particular finite prefix [3, E/2]. Complete cycles guarantee gaps; finite prefixes can behave differently. The inclusion–exclusion error terms from all those track intersections can drown the positive main term — which is exactly the parity barrier where sieve methods stall (Chen’s theorem gets prime + almost-prime, and no further by these means).

Empirically the openings appear absurdly early: sweeping every even E < 20,000, no strip was ever fully covered, and the latest first survivor found was x = 173 at E = 7426 — try it in the explorer above. But computation is not proof. The missing piece is a finite-gap theorem, not another fact about multiplication.