> Full-text translation from Chinese. The original remains authoritative for its research claims; this translation is not independent mathematical verification.

# Legendre Base-Space Program Phase 12
## Square $P_2$ Boundary / Bilinear Semiprime Capacity
### Cube-Root RootRough, Least-Factor Rows, Cofactor Switching and the Buchstab $\log 2$ Margin

**Version:** v0.1  
**Date:** 2026-10-02  
**Author:** Neo.K × AI collaboration  
**Institution:** EveMissLab / 一言諾科技有限公司  
**Research line:** Legendre Base-Space Prime-Fitting Program  
**Prerequisites:** Phases 1–11; incorporates the semiprime-boundary / cofactor-switching structure of SRBRD-13–20  
**Status:** Phase 12 exact structural reduction + finite audit + current-frontier comparison; Legendre's conjecture remains unproved

---

# 0. The Convergent Conclusion of This Phase

Phase 11 located the hardest residual obstruction after deep pre-sieving as:

$$
\boxed{
\text{prime}
\quad\text{vs}\quad
\text{single-pair semiprime}.
}
$$

Phase 12 does not continue increasing the number of factor-shell levels, but instead uses a more canonical cutoff:

$$
\boxed{
z^3
\ge
(n+1)^2-1.
}
$$

At this cube-root boundary, a RootRough survivor in the square cell **can only be prime or semiprime**.

Legendre is therefore reformulated exactly as:

$$
\boxed{
\#\text{prime}
=
\#\text{cube-root RootRough survivors}
-
\#\text{semiprime least-factor rows}.
}
$$

This is the simplest exact $P_2$ boundary identity in the series so far.

---

# 1. First Correct the Research Scale

Phases 10–11 retained a sufficient but stronger route:

$$
r_*(n)=o(n).
$$

It would imply that a prime appears before the upper endpoint:

$$
n^2+o(n)
$$

is reached.

This is much stronger than the original Legendre proposition:

$$
n^2<p<(n+1)^2
=
n^2+2n+1
$$

as stated above.

Phase 12 therefore demotes:

$$
\boxed{
\text{sublinear fit-depth}
}
$$

to an ‘optional strong route’, no longer a necessary target of the main line.

The main line returns to the full square cell:

$$
\boxed{
(n^2,(n+1)^2).
}
$$

---

# 2. Square-Cell Scale

Let:

$$
\boxed{
X=n^2,
}
$$

and:

$$
\boxed{
H=2n.
}
$$

Then the integers in the square cell are:

$$
X+1,\dots,X+H.
$$

Its right endpoint is:

$$
X+H
=
(n+1)^2-1.
$$

---

# 3. Canonical Cube-Root Cutoff

Define:

$$
\boxed{
z_3(n)
=
\min
\left\{
z\in\mathbb N:
z^3\ge (n+1)^2-1
\right\}.
}
$$

Asymptotically:

$$
\boxed{
z_3(n)
=
n^{2/3}(1+o(1)).
}
$$

In terms of $X=n^2$:

$$
\boxed{
z_3
=
X^{1/3}(1+o(1)).
}
$$

---

# 4. RootRough Predicate

Use the canonical definition from SRBRD.

For:

$$
q\ge2,
$$

define:

$$
\boxed{
\operatorname{RootRough}_z(q)
:
\Longleftrightarrow
\neg\exists p\in\mathbb P
\quad
p\le z,
\quad
p^2\le q,
\quad
p\mid q.
}
$$

In the large-scale region of this square cell,

$$
z_3<\sqrt q,
$$

so it agrees with ordinary roughness:

$$
P^-(q)>z_3
$$

in this range.

---

# 5. Cube-Root $P_2$ Boundary Theorem

## Theorem 5.1

If:

$$
n^2<q<(n+1)^2
$$

and:

$$
\operatorname{RootRough}_{z_3(n)}(q),
$$

then:

$$
\boxed{
q\text{ is prime or semiprime}.
}
$$

## Proof

If $q$ is composite, let:

$$
p=P^-(q).
$$

RootRough gives:

$$
p>z_3.
$$

If:

$$
\Omega(q)\ge3,
$$

then every prime factor is:

$$
>z_3,
$$

so:

$$
q>z_3^3.
$$

But:

$$
z_3^3
\ge
(n+1)^2-1
\ge q,
$$

a contradiction.

Therefore:

$$
\Omega(q)=2.
$$

so it is semiprime.

This completes the proof.

---

# 6. Square-Cell RootRough Survivor Set

Define:

$$
\boxed{
\mathcal S_3(n)
=
\left\{
q\in\mathbb Z:
n^2<q<(n+1)^2,
\quad
\operatorname{RootRough}_{z_3}(q)
\right\}.
}
$$

By Theorem 5.1:

$$
\boxed{
\mathcal S_3(n)
=
\mathcal P_\square(n)
\sqcup
\mathcal E_2(n),
}
$$

where:

$$
\mathcal P_\square(n)
=
\mathbb P\cap(n^2,(n+1)^2),
$$

and:

$$
\mathcal E_2(n)
$$

is the set of cube-root-rough semiprimes in the cell.

---

# 7. The First Exact Identity for the Prime Count

Thus:

$$
\boxed{
|\mathcal P_\square(n)|
=
|\mathcal S_3(n)|
-
|\mathcal E_2(n)|.
}
$$

Legendre is therefore equivalent to:

$$
\boxed{
|\mathcal S_3(n)|
>
|\mathcal E_2(n)|
\quad
\forall n.
}
$$

There is no longer any need to handle $P_3,P_4,\dots$:

the cube-root boundary automatically excludes them.

---

# 8. Least-Factor Semiprime Rows

For a prime:

$$
p>z_3
$$

define:

$$
\boxed{
\mathcal R_p(n)
=
\left\{
q\in\mathcal S_3(n):
p\mid q,
\quad
p^2\le q
\right\}.
}
$$

This definition does not require:

$$
q/p\in\mathbb P.
$$

---

# 9. Row Theorem

## Theorem 9.1

For:

$$
q\in\mathcal S_3(n),
$$

and a prime:

$$
p>z_3,
$$

we have:

$$
q\in\mathcal R_p(n)
$$

if and only if:

$$
\boxed{
q\text{ semiprime}
\quad\text{and}\quad
P^-(q)=p.
}
$$

## Proof

If:

$$
p\mid q,
\qquad
p^2\le q,
$$

since $q$ is prime or semiprime and:

$$
q>p,
$$

$q$ cannot be prime.

Thus:

$$
q=ab
$$

is semiprime, with $a\le b$.

If $p=b>a$, then:

$$
q=ab<p^2,
$$

a contradiction.

Hence:

$$
p=a=P^-(q).
$$

The converse is immediate.

This completes the proof.

---

# 10. Disjoint Least-Factor Partition

Since the least prime factor is unique:

$$
\boxed{
\mathcal R_p(n)
\cap
\mathcal R_r(n)
=
\varnothing
\quad
(p\neq r).
}
$$

and every semiprime survivor lies in a unique row.

Therefore:

$$
\boxed{
\mathcal E_2(n)
=
\bigsqcup_{
\substack{
p\in\mathbb P\\
z_3<p\le n
}
}
\mathcal R_p(n).
}
$$

Why:

$$
p\le n?
$$

Because:

$$
p\le\sqrt q<n+1,
$$

and $p$ is an integer.

---

# 11. Exact Cofactor Switching

If:

$$
q\in\mathcal R_p(n),
$$

then:

$$
q=pr
$$

where:

$$
r
$$

must be prime.

And:

$$
n^2<pr<(n+1)^2.
$$

Thus:

$$
\boxed{
\frac{n^2}{p}
<
r
<
\frac{(n+1)^2}{p}.
}
$$

Conversely, any prime:

$$
r
$$

in this interval gives:

$$
q=pr
$$

in the square cell, with:

$$
r>
\frac{n^2}{p}
\ge n
\ge p.
$$

Thus $p$ is the smaller prime factor, and:

$$
q
$$

is automatically cube-root RootRough.

Hence:

## Theorem 11.1

Let:

$$
L_p(n)
=
\left\lfloor
\frac{n^2}{p}
\right\rfloor,
$$

$$
U_p(n)
=
\left\lfloor
\frac{(n+1)^2-1}{p}
\right\rfloor.
$$

Then:

$$
\boxed{
|\mathcal R_p(n)|
=
\pi(U_p(n))
-
\pi(L_p(n)).
}
$$

---

# 12. Square $P_2$ Boundary Identity

Combining the preceding statements:

$$
\boxed{
|\mathcal P_\square(n)|
=
|\mathcal S_3(n)|
-
\sum_{
\substack{
p\in\mathbb P\\
z_3<p\le n
}
}
\left[
\pi(U_p(n))
-
\pi(L_p(n))
\right].
}
$$

This is the main exact identity of Phase 12.

It decomposes Legendre into:

### unsigned rough bulk

$$
|\mathcal S_3(n)|
$$

minus:

### bilinear prime-cofactor capacity

$$
\sum_p
\left[
\pi(U_p)-\pi(L_p)
\right].
$$

---

# 13. The Version without a Prime-Cofactor Predicate

Although the switched formula uses:

$$
\pi,
$$

the original row:

$$
\mathcal R_p
$$

uses only:

- RootRough;
- Divisibility;
- Square comparison.

Thus the semiprime row mass has two exactly equivalent representations:

$$
\boxed{
\text{unsigned RootRough row}
}
$$

and:

$$
\boxed{
\text{prime-cofactor short interval}.
}
$$

This is precisely the one-dimensional version of the cofactor-switching structure of SRBRD-13–17 for the square-cell problem.

---

# 14. Square-Anchor Row Offset

For a fixed row prime:

$$
p,
$$

any row candidate can be written as:

$$
q=n^2+k.
$$

The divisibility condition:

$$
p\mid q
$$

is equivalent to:

$$
\boxed{
k
\equiv
-n^2
\pmod p.
}
$$

The row location therefore retains the square-anchor residue phase of Phase 4.

Thus:

$$
\boxed{
\text{least-factor row}
=
\text{cube-root boundary}
+
\text{square-anchor residue}.
}
$$

---

# 15. Sub-$z$ Window Matching Theorem

Consider a shorter physical window:

$$
n^2<q\le n^2+K
$$

with:

$$
K<z_3.
$$

Since every semiprime least factor satisfies:

$$
p>z_3>K,
$$

the spacing between multiples of a fixed $p$ is:

$$
p>K.
$$

so every row:

$$
\mathcal R_p
$$

has at most one candidate.

Therefore:

$$
\boxed{
|\mathcal R_p|\le1.
}
$$

Furthermore, the same cofactor prime:

$$
r>n>K
$$

cannot appear in two different row products either, because two odd primes:

$$
p_1\neq p_2
$$

differ by at least $2$, so:

$$
r|p_1-p_2|
\ge2r
>
2n
>
K.
$$

Thus sub-$z$ semiprime pairs form a genuine matching.

But:

> Proving that, for every $n$, the sub-$z$ window must contain a prime is stronger than Legendre.

This matching is therefore retained only as a strong-route structure, not a necessary target of the main line.

---

# 16. Full-Cell Factor Exponent Band

Return to the full Legendre cell.

Let:

$$
X=n^2.
$$

The cube-root least factor:

$$
p>X^{1/3}.
$$

Also:

$$
p\le\sqrt X.
$$

Thus write:

$$
\boxed{
p=X^t,
\qquad
\frac13<t\le\frac12.
}
$$

The cofactor:

$$
r
\asymp
\frac Xp
=
X^{1-t}.
$$

The row-interval length:

$$
\boxed{
\frac Hp
\asymp
X^{1/2-t}.
}
$$

Thus:

- At $t=1/3$, the row length is about $X^{1/6}$;
- As $t\to1/2$, the row length falls to the constant scale.

This:

$$
\boxed{
t\in[1/3,1/2]
}
$$

is precisely the parity-hard bilinear band.

---

# 17. Ideal Bilinear Semiprime Coefficient

Temporarily use the prime-density main term as a diagnostic:

For:

$$
p=X^t,
$$

the density of primes $p$ is:

$$
\frac{dp}{\log p}.
$$

For a fixed $p$, the cofactor-prime interval has length:

$$
\frac Hp,
$$

Its expected prime count is:

$$
\frac{
H/p
}{
\log(X/p)
}.
$$

Thus the semiprime main density is:

$$
\frac{dp}{\log p}
\cdot
\frac{H/p}{\log(X/p)}.
$$

Let:

$$
p=X^t,
\qquad
\frac{dp}{p}
=
\log X\,dt.
$$

This gives:

$$
\frac H{\log X}
\frac{dt}{t(1-t)}.
$$

Integrate:

$$
\int_{1/3}^{1/2}
\frac{dt}{t(1-t)}
=
\boxed{
\log2
}.
$$

Thus the ideal semiprime coefficient is:

$$
\boxed{
|\mathcal E_2(n)|
\sim
(\log2)
\frac H{\log X}.
}
$$

---

# 18. Prime Main Coefficient

At the:

$$
X<n<X+H
$$

scale, the prime-density main term is:

$$
\boxed{
|\mathcal P_\square(n)|
\sim
\frac H{\log X}.
}
$$

Thus the ideal prime / semiprime ratio is:

$$
\boxed{
\frac{|\mathcal E_2|}{|\mathcal P_\square|}
\sim
\log2
\approx0.693147.
}
$$

The prime share among cube-root survivors:

$$
\boxed{
\frac1{1+\log2}
\approx
0.590616.
}
$$

---

# 19. Buchstab $u=3$ Coefficient

The Buchstab function:

$$
\omega(u)
$$

satisfies:

$$
u\omega(u)=1
\quad
(1\le u\le2),
$$

and:

$$
(u\omega(u))'
=
\omega(u-1).
$$

For:

$$
2\le u\le3,
$$

we have:

$$
\omega(u-1)=\frac1{u-1}.
$$

Hence:

$$
u\omega(u)
=
1+\log(u-1).
$$

Substitute:

$$
u=3,
$$

to obtain:

$$
\boxed{
3\omega(3)
=
1+\log2.
}
$$

Thus the classical global rough-number benchmark:

$$
\Phi(X,X^{1/3})
\sim
\boxed{
(1+\log2)
\frac X{\log X}.
}
$$

decomposes exactly as:

$$
\boxed{
1
+
\log2
}
$$

that is:

$$
\boxed{
\text{prime main coefficient}
+
\text{semiprime-row coefficient}.
}
$$

---

# 20. Square-Local Buchstab Benchmark

If global Buchstab density could be transferred to the square interval without error:

$$
(X,X+H],
\qquad
H\asymp X^{1/2},
$$

we would expect:

$$
|\mathcal S_3(n)|
\approx
(1+\log2)
\frac H{\log X}.
$$

Subtract the semiprime rows:

$$
\approx
(\log2)
\frac H{\log X},
$$

and this leaves exactly:

$$
\boxed{
\frac H{\log X}.
}
$$

Thus there is no deficit in the algebraic main term:

it **leaves exactly the full prime main term**.

---

# 21. Error-Budget Form

Define the exact errors:

$$
\boxed{
E_S(n)
=
|\mathcal S_3(n)|
-
(1+\log2)\frac H{\log X},
}
$$

and:

$$
\boxed{
E_{SP}(n)
=
|\mathcal E_2(n)|
-
(\log2)\frac H{\log X}.
}
$$

By the exact $P_2$ identity:

$$
\boxed{
|\mathcal P_\square(n)|
=
\frac H{\log X}
+
E_S(n)
-
E_{SP}(n).
}
$$

Thus a sufficient condition is:

$$
\boxed{
E_{SP}(n)-E_S(n)
<
\frac H{\log X}.
}
$$

---

# 22. One-Sided Constant Crossing Criterion

If we can prove:

$$
|\mathcal S_3(n)|
\ge
\left(
1+\log2-\eta_S
\right)
\frac H{\log X},
$$

and:

$$
|\mathcal E_2(n)|
\le
\left(
\log2+\eta_{SP}
\right)
\frac H{\log X},
$$

where:

$$
\boxed{
\eta_S+\eta_{SP}<1,
}
$$

then:

$$
\boxed{
|\mathcal P_\square(n)|>0.
}
$$

This is the cleanest analytic closure gate of Phase 12.

---

# 23. Standard Linear-Sieve No-Go

For a consecutive interval:

$$
(X,X+H],
\qquad
H\asymp X^{1/2},
$$

local congruence counting gives:

$$
|A_d|
=
\frac Hd
+
O(1).
$$

With the standard linear sieve, the naturally available level is roughly no greater than:

$$
D
=
X^{1/2-o(1)}.
$$

The cube-root cutoff:

$$
z=X^{1/3}.
$$

Thus:

$$
s
=
\frac{\log D}{\log z}
=
\frac32+o(1).
$$

The standard lower linear sieve has:

$$
f(s)=0
\quad
(s\le2).
$$

Therefore:

$$
\boxed{
\text{standard one-dimensional lower linear sieve}
}
$$

cannot even directly give a positive RootRough lower bound at the cube-root cutoff.

This is the first analytic barrier at the square $P_2$ boundary.

---

# 24. The Generic Jacobsthal Route Is Insufficient Too

General rough-number covering can be expressed using the Jacobsthal function.

The generic upper bound from Iwaniec's shifted-sieve result controls the longest fully covered segment through the number of selected primes; at a primorial cutoff, its scale is still far larger than the present problem's:

$$
H\asymp X^{1/2}
$$

when:

$$
z=X^{1/3}.
$$

Thus:

$$
\boxed{
\text{generic arbitrary-translate rough-number guarantee}
}
$$

is still insufficient to handle a square interval directly.

This again shows:

$$
\boxed{
\text{The square anchor must genuinely enter the estimates.}
}
$$

---

# 25. Exact Alignment with SRBRD-13–20

The sibling program SRBRD has established:

1. Cube-root RootRough survivor $\Rightarrow$ prime / semiprime;
2. Least-factor rows can be defined using RootRough + divisibility;
3. A semiprime row can be switched exactly to a prime cofactor;
4. Cofactorization itself does not automatically break the parity barrier;
5. Naive separate row sieving has a quantitative constant no-go;
6. An aggregate bilinear / switched provider is actually needed.

Phase 12 shows that the Legendre square-cell problem has the same skeleton.

But there is one crucial difference.

The additive / complement rows of SRBRD usually have a long cofactor range.

Here the row length is only:

$$
\boxed{
H/p
=
X^{1/2-t},
}
$$

and, even as:

$$
t\to1/2
$$

the length falls to $O(1)$.

Thus the Maynard / Bombieri–Vinogradov uniform-residue provider used in SRBRD **cannot be transferred directly as a black box**.

What is needed here is:

$$
\boxed{
\text{short hyperbolic prime-convolution control}.
}
$$

---

# 26. Alignment with the Current Public Frontier

As of 2026-10:

### Every square interval

Dudek–Johnston have proved that every:

$$
(n^2,(n+1)^2)
$$

contains a $P_4$.

Campbell 2026 further proves that each cell contains:

$$
\boxed{
P_3.
}
$$

Thus the current unconditional all-square frontier has reached $P_3$, but not $P_2$ or primes.

### Almost all short intervals

Matomäki proves that almost all very short intervals contain $P_2$; Matomäki–Teräväinen also obtain strong results for exact $E_2$ in almost-all polylogarithmic short intervals.

Therefore:

$$
\boxed{
\text{P2 bilinear technology itself exists,}
}
$$

but:

$$
\boxed{
\text{every square anchor}
}
$$

and its pointwise uniformity are the genuinely distinctive requirement of this line.

---

# 27. Finite Audit: The Exact $P_2$ Partition

This round's machine audit:

$$
100\le n\le3000.
$$

Totals:

$$
|\mathcal S_3|
=
\boxed{
999937
},
$$

of which:

$$
|\mathcal P_\square|
=
\boxed{
601600
},
$$

$$
|\mathcal E_2|
=
\boxed{
398337
}.
$$

Deeper composite survivors:

$$
\boxed{
0.
}
$$

Thus the finite audit of the cube-root $P_2$ partition gives:

$$
\boxed{
\text{PASS}.
}
$$

---

# 28. Finite Audit: Cofactor Switching

For every finite semiprime row, compare row by row:

### RootRough/divisibility row count

with:

### switched prime-cofactor interval count

Mismatches:

$$
\boxed{
0.
}
$$

Thus, in the finite implementation:

$$
\boxed{
\mathcal R_p
\leftrightarrow
\text{prime cofactor interval}
}
$$

agree completely.

---

# 29. Finite Aggregate Ratio

In the cumulative data for:

$$
100\le n\le3000
$$

this range:

The prime share among cube-root survivors:

$$
\boxed{
0.601638
}
$$

The ideal Buchstab / bilinear value:

$$
\boxed{
\frac1{1+\log2}
\approx
0.590616.
}
$$

The finite aggregate:

$$
\frac{SP}{P}
=
\boxed{
0.662129
},
$$

The ideal value:

$$
\log2
\approx
0.693147.
$$

This is only diagnostic agreement, not an asymptotic proof.

---

# 30. Selected Finite Cells

| $n$ | $z_3$ | survivors | primes | semiprimes | deeper | prime share | $P\log X/H$ | $SP\log X/H$ | $S\log X/H$ |
|---:|---:|---:|---:|---:|---:|---:|---:|---:|---:|
| 100 | 22 | 36 | 23 | 13 | 0 | 0.6389 | 1.0592 | 0.5987 | 1.6579 |
| 300 | 45 | 87 | 56 | 31 | 0 | 0.6437 | 1.0647 | 0.5894 | 1.6541 |
| 1000 | 101 | 241 | 152 | 89 | 0 | 0.6307 | 1.0500 | 0.6148 | 1.6648 |
| 2000 | 159 | 437 | 267 | 170 | 0 | 0.6110 | 1.0147 | 0.6461 | 1.6608 |
| 3000 | 209 | 610 | 340 | 270 | 0 | 0.5574 | 0.9074 | 0.7206 | 1.6280 |

The ideal normalized coefficients:

$$
\boxed{
P\frac{\log X}{H}
\to1,
}
$$

$$
\boxed{
SP\frac{\log X}{H}
\to\log2,
}
$$

$$
\boxed{
S\frac{\log X}{H}
\to1+\log2.
}
$$

The finite samples already fluctuate around the same scales.

---

# 31. Prime Majority Is Not an Available Theorem

Although the aggregate prime share exceeds $1/2$, a prime majority is not guaranteed cell by cell.

In the finite audit:

$$
P\le SP
$$

occurs at:

$n=119$ ($P=19,\ SP=19$), $n=425$ ($P=56,\ SP=57$), $n=696$ ($P=85,\ SP=88$)

Thus:

$$
\boxed{
P>SP
}
$$

cannot be treated as an intermediate theorem for all square cells.

What is actually needed is only:

$$
\boxed{
S-SP>0.
}
$$

---

# 32. The Actual Closure Target of Phase 12

Legendre can now be compressed to:

$$
\boxed{
L_{\mathrm{rough}}(n)
-
U_{\mathrm{SP}}(n)
>0,
}
$$

where:

$$
L_{\mathrm{rough}}
\le
|\mathcal S_3(n)|,
$$

and:

$$
|\mathcal E_2(n)|
\le
U_{\mathrm{SP}}.
$$

This does not require a positive lower bound directly for the prime count.

---

# 33. Provider A: A Square-Anchored Rough Lower Bound

We need:

$$
\boxed{
|\mathcal S_3(n)|
\text{ a pointwise square-anchored lower bound for this quantity}.
}
$$

It cannot use only a generic translated sieve.

It must exploit:

$$
n^2+k
$$

and the:

$$
\boxed{
k\equiv-n^2\pmod p
}
$$

square phase it produces.

This brings the square-anchor obstruction field of Phase 4 back into the main line.

---

# 34. Provider B: A Bilinear Semiprime Upper Bound

We need to control:

$$
\boxed{
\sum_{z_3<p\le n}
\left[
\pi(U_p)-\pi(L_p)
\right].
}
$$

This is a pointwise upper bound for a prime × prime convolution in the hyperbolic strip:

$$
X<pr<X+H
$$

as specified above.

Because the row band is:

$$
p=X^t,
\qquad
1/3<t\le1/2,
$$

this is precisely a Type-II / bilinear structure.

But the row width:

$$
H/p
=
X^{1/2-t}
$$

is very short, especially near the balanced edge where it approaches constant length.

Thus we need:

$$
\boxed{
\text{aggregate bilinear switching}
}
$$

rather than a row-by-row short-interval PNT.

---

# 35. The Entry Point to Phase 13

For the next phase, it is recommended to tackle only one thing:

## Square-Anchored $P_2$ Constant Ledger

Construct available providers for:

$$
L_{\mathrm{rough}},
\qquad
U_{\mathrm{SP}}
$$

and normalize them all to the:

$$
\frac H{\log X}
$$

scale.

The target is directly a constant ledger:

$$
\boxed{
C_{\mathrm{rough}}^{-}
-
C_{\mathrm{SP}}^{+}
>0?
}
$$

Divide into bands:

$$
t\in
[1/3,2/5],
$$

$$
[2/5,1/2-\varepsilon],
$$

and:

$$
t\approx1/2.
$$

Check band by band:

- Classical switched sieves;
- Maynard uniform-residue convolution;
- Yang 2026 convolution-BV;
- Which almost-all short-interval $E_2$ techniques can / cannot become square-anchor pointwise providers.

---

# 36. Phase 12 Status

```text
Legendre Base-Space Program
Phase 12 — Square P2 Boundary

Full-cell scale restored as main target:          YES
Cube-root RootRough cutoff:                       DEFINED
Prime-or-semiprime boundary theorem:              PROVED
Exact P / E2 partition:                           PROVED
Least-factor semiprime rows:                      DEFINED
Disjoint row partition:                           PROVED
Prime-cofactor switching:                         PROVED
Square P2 boundary identity:                      PROVED
Square-anchor residue retained:                   YES
Sub-z semiprime matching:                         PROVED
Sublinear route classified as stronger:           YES
Bilinear factor band [1/3,1/2]:                  DERIVED
Ideal semiprime coefficient log 2:                DERIVED
Buchstab u=3 coefficient 1+log 2:                 DERIVED
Exact error-budget identity:                      DERIVED
One-sided constant-crossing gate:                 DEFINED
Standard linear-sieve cube-root lower bound:      NO-GO
Generic Jacobsthal shortcut:                      INSUFFICIENT
SRBRD cofactor-switching transfer:                ALIGNED
Finite P2 partition audit:                        PASS
Finite switched-row audit:                        PASS
Legendre proof status:                            OPEN
Next phase:                                       SQUARE-ANCHORED P2 CONSTANT LEDGER
```

---

# 37. References

1. Peter J. Campbell, *On the Existence of Integers with at Most 3 Prime Factors Between Every Pair of Consecutive Squares*, arXiv:2603.10356, 2026; Journal of Number Theory, DOI 10.1016/j.jnt.2026.07.014.
2. Adrian W. Dudek, Daniel R. Johnston, *Almost primes between all squares*, Journal of Number Theory **278** (2026), 726–745; arXiv:2501.18048.
3. Kaisa Matomäki, *Almost primes in almost all very short intervals*, Journal of the London Mathematical Society **106** (2022), 1061–1097.
4. Kaisa Matomäki, Joni Teräväinen, *Almost primes in almost all short intervals II*, Transactions of the AMS **376** (2023); arXiv:2207.05038.
5. Henryk Iwaniec, *On the problem of Jacobsthal*, Demonstratio Mathematica **11** (1978), 225–232.
6. James Maynard, *Primes in arithmetic progressions to large moduli III: Uniform residue classes*, Memoirs of the AMS **306** (2025), no. 1544.
7. Zhiyuan Yang, *Convolution-type Bombieri-Vinogradov theorem with well-factorable weights, and its applications*, arXiv:2608.13299v2, 2026.
8. SRBRD-13–20, EveMissLab internal research series, 2026-09-30.