> 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 4
## Square-Anchored Obstruction Residue Field, Atomic Reduction, and Exact Sieve Identity
### Square-Anchored Obstruction Residue Field, Atomic Reduction, and Exact Sieve Identity

**Version:** v0.1  
**Date:** 2026-09-30  
**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:** Phase 1–3  
**Status:** Phase 4 formalization + known-mathematics boundary audit + finite machine audit; not a completed proof of Legendre's conjecture

---

# 0. Objective of This Phase

Phase 3 compressed the core Legendre obligation to:

$$
\gcd(F_\square(n),H_n)=1.
$$

Phase 4 no longer uses the enormous product $H_n$ as its main interface, but decomposes it into the **forbidden residue class** imposed by each historical factor on the local offset $k$.

Let:

$$
q=n^2+k,
\qquad
1\le k\le2n.
$$

For any positive integer $d$:

$$
d\mid q
\iff
d\mid(n^2+k)
\iff
k\equiv-n^2\pmod d.
$$

Thus, every historical obstruction $d$ in the $n$th square cell blocks only one offset phase modulo $d$.

The core target becomes:

$$
\boxed{
\text{Construct a global offset section }
\kappa_{\mathcal F}(n)
\text{ that never falls into any historical forbidden phase.}
}
$$

---

# 1. The Local Offset Base Space

Continue to use:

$$
K_n
=
\{1,2,\dots,2n\}.
$$

Evaluation:

$$
E_n(k)=n^2+k.
$$

Thus:

$$
E_n(K_n)
=
\mathbb Z\cap(n^2,(n+1)^2).
$$

---

# 2. The $M6^*$ Presieved Offset Space

For $n\ge2$, define:

$$
K_n^{(6)}
=
\{
k\in K_n:
n^2+k\equiv\pm1\pmod6
\}.
$$

If:

$$
k\in K_n^{(6)},
$$

then:

$$
n^2+k
$$

already avoids the factors $2$ and $3$.

Subsequent work therefore needs to address only:

$$
M6^*
=
\{d>1:d\equiv\pm1\pmod6\}
$$

the multiplicative obstructions in this set.

---

# 3. The Atomic Obstruction Basis

Phase 3 uses all of:

$$
\mathcal D_n
=
\{
d\in M6^*:5\le d\le n
\}.
$$

However, many composite $d$ in this set are redundant.

Define:

$$
\boxed{
\mathfrak A
=
\operatorname{Fix}(T_{\mathrm{top}})
=
\{
a\in M6^*:T_{\mathrm{top}}(a)=a
\}.
}
$$

and define:

$$
\boxed{
\mathfrak A_n
=
\mathfrak A\cap[5,n].
}
$$

$\mathfrak A$ can be defined purely by the divisibility operator; by the fixed-point characterization in Phase 3, it corresponds exactly to primes $>3$.

---

# 4. The Atomic Reduction Theorem

## Theorem 4.1

Let:

$$
q\in(n^2,(n+1)^2)\cap M6^*.
$$

Then:

$$
q
\text{ is composite}
$$

if and only if there exists:

$$
a\in\mathfrak A_n
$$

such that:

$$
a\mid q.
$$

## Proof

If an atom $a\le n$ divides $q$, then:

$$
1<a<q,
$$

so $q$ is composite.

Conversely, if $q$ is composite, the Phase 3 historical obstruction theorem gives:

$$
T_{\mathrm{top}}(q)\le n.
$$

Let:

$$
a=T_{\mathrm{top}}(q).
$$

Since:

$$
T_{\mathrm{top}}(a)=a,
$$

we have:

$$
a\in\mathfrak A_n
$$

and:

$$
a\mid q.
$$

This completes the proof.

---

# 5. The Atomic Historical Kernel

Define:

$$
\boxed{
Q_n
=
6
\prod_{a\in\mathfrak A_n}
a.
}
$$

Then $Q_n$ is the squarefree historical kernel of all arithmetic atoms not exceeding $n$, together with $2,3$.

By Theorem 4.1, for:

$$
q\in(n^2,(n+1)^2),
$$

we have:

$$
\boxed{
q\text{ is prime}
\iff
\gcd(q,Q_n)=1.
}
$$

This is smaller and more canonical than Phase 3's $H_n$, because all composite obstructions have been removed.

---

# 6. Forbidden Residue Phases

For:

$$
a\in\mathfrak A_n,
$$

define:

$$
\boxed{
b_a(n)
=
[-n^2]_a,
}
$$

Here $[\cdot]_a$ denotes the canonical residue class modulo $a$.

Define the obstruction stripe:

$$
\boxed{
B_{n,a}
=
\{
k\in K_n:
k\equiv b_a(n)\pmod a
\}.
}
$$

Then:

$$
\boxed{
k\in B_{n,a}
\iff
a\mid(n^2+k).
}
$$

Thus $B_{n,a}$ consists of all offset positions blocked by the atom $a$ in the $n$th square cell.

---

# 7. A Square-Anchored Phase Is Not an Arbitrary Residue

For fixed $a$:

$$
b_a(n)
=
-n^2\pmod a.
$$

Hence:

$$
b_a(n+a)=b_a(n).
$$

Moreover:

$$
b_a(-n)=b_a(n).
$$

The obstruction phase therefore has, as a function of $n$:

- Period $a$;
- Reflection symmetry;
- Quadratic-residue structure.

It is not an arbitrarily specified collection of residue classes in a general covering system.

This square anchor:

$$
\boxed{
-n^2
}
$$

is the additional structure that this program must preserve.

---

# 8. The Local Survivor Set

Define:

$$
\boxed{
\mathcal S_n
=
K_n^{(6)}
\setminus
\bigcup_{a\in\mathfrak A_n}
B_{n,a}.
}
$$

Then $\mathcal S_n$ is the set of offsets in the $n$th square cell that are not divisible by any historical atom.

---

# 9. The Survivor–Prime Equivalence Theorem

## Theorem 9.1

For all:

$$
n\ge2,
$$

there is the bijective correspondence:

$$
\boxed{
k\in\mathcal S_n
\iff
n^2+k
\in
\mathbb P\cap(n^2,(n+1)^2).
}
$$

Thus:

$$
\boxed{
|\mathcal S_n|
=
\pi((n+1)^2-1)-\pi(n^2).
}
$$

and Legendre's conjecture is equivalent to:

$$
\boxed{
\mathcal S_n\neq\varnothing
\quad
\forall n\ge2.
}
$$

The case $n=1$ is handled directly by $2,3\in(1,4)$.

---

# 10. The Fit-Section Version

If the fitting operator outputs:

$$
\kappa_{\mathcal F}(n)\in K_n^{(6)},
$$

then:

$$
F_\square(n)
=
n^2+\kappa_{\mathcal F}(n).
$$

A sufficient condition for Legendre's conjecture is:

$$
\boxed{
\kappa_{\mathcal F}(n)
\in
\mathcal S_n
\quad
\forall n\ge2.
}
$$

That is:

$$
\boxed{
\kappa_{\mathcal F}(n)
\not\equiv
-n^2
\pmod a
\quad
\forall a\in\mathfrak A_n.
}
$$

---

# 11. Prime-Fit Obstruction Residue Field

Define:

$$
\boxed{
\mathfrak R_{\mathcal F}(n,a)
=
[n^2+\kappa_{\mathcal F}(n)]_a.
}
$$

The domain is:

$$
\boxed{
\mathcal O
=
\{
(n,a):
n\ge2,\,
a\in\mathfrak A,\,
a\le n
\}.
}
$$

Then:

$$
\boxed{
\mathfrak R_{\mathcal F}(n,a)=0
}
$$

means that the atom $a$ hits the $n$th fitted candidate.

The final fixed-point condition is exactly equivalent to:

$$
\boxed{
\mathfrak R_{\mathcal F}(n,a)\neq0
\quad
\forall(n,a)\in\mathcal O.
}
$$

---

# 12. Zero-Obstruction Count

Define:

$$
\boxed{
Z_{\mathcal F}(n)
=
\sum_{a\in\mathfrak A_n}
\mathbf 1
\{
\mathfrak R_{\mathcal F}(n,a)=0
\}.
}
$$

Then:

$$
\boxed{
Z_{\mathcal F}(n)=0
\iff
F_\square(n)\text{ is prime}.
}
$$

The primality question for the fitting operator can therefore become:

$$
\boxed{
Z_{\mathcal F}(n)\equiv0.
}
$$

---

# 13. CRT Collapse of Multi-Atom Intersections

Let:

$$
S
\subseteq
\mathfrak A_n
$$

be a finite set of atoms, and let:

$$
M_S
=
\prod_{a\in S}a.
$$

Distinct atoms are pairwise coprime.

For all:

$$
a\in S,
$$

require:

$$
k\equiv-n^2\pmod a.
$$

By the Chinese remainder theorem, these conditions are not merely compatible, but combine exactly into:

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

Thus:

$$
\boxed{
\bigcap_{a\in S}B_{n,a}
=
\{
k\in K_n:
k\equiv-n^2\pmod{M_S}
\}.
}
$$

This is a major simplification provided by the square anchor.

---

# 14. The Interval-Multiple Counting Kernel

For any:

$$
d\ge1,
$$

define:

$$
\boxed{
C_n(d)
=
\left\lfloor
\frac{n^2+2n}{d}
\right\rfloor
-
\left\lfloor
\frac{n^2}{d}
\right\rfloor.
}
$$

It counts exactly the integers in:

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

that are divisible by $d$.

By the previous section:

$$
\left|
\bigcap_{a\in S}B_{n,a}
\right|
=
C_n(M_S).
$$

---

# 15. Exact Square-Anchored Inclusion–Exclusion

Since $Q_n$ is squarefree, the Möbius function gives:

$$
\boxed{
L_\square(n)
=
\sum_{d\mid Q_n}
\mu(d)
C_n(d).
}
$$

That is:

$$
\boxed{
L_\square(n)
=
\sum_{d\mid Q_n}
\mu(d)
\left[
\left\lfloor
\frac{n^2+2n}{d}
\right\rfloor
-
\left\lfloor
\frac{n^2}{d}
\right\rfloor
\right].
}
$$

By survivor–prime equivalence:

$$
\boxed{
L_\square(n)
=
|\mathcal S_n|
=
\pi((n+1)^2-1)-\pi(n^2).
}
$$

Legendre's conjecture is therefore equivalent to:

$$
\boxed{
L_\square(n)>0
\quad
\forall n\ge2.
}
$$

---

# 16. The Main Term and Square-Phase Error

Write:

$$
C_n(d)
=
\frac{2n}{d}
+
\theta_n(d),
$$

where:

$$
\boxed{
\theta_n(d)
=
\left\{
\frac{n^2}{d}
\right\}
-
\left\{
\frac{n^2+2n}{d}
\right\},
}
$$

and:

$$
|\theta_n(d)|<1.
$$

Thus:

$$
L_\square(n)
=
2n
\sum_{d\mid Q_n}
\frac{\mu(d)}{d}
+
\sum_{d\mid Q_n}
\mu(d)\theta_n(d).
$$

And:

$$
\sum_{d\mid Q_n}
\frac{\mu(d)}d
=
\frac{\varphi(Q_n)}{Q_n}.
$$

This gives the exact decomposition:

$$
\boxed{
L_\square(n)
=
2n
\frac{\varphi(Q_n)}{Q_n}
+
\Delta_\square(n),
}
$$

where:

$$
\boxed{
\Delta_\square(n)
=
\sum_{d\mid Q_n}
\mu(d)\theta_n(d).
}
$$

---

# 17. The Actual Global Closure Obligation

The main term:

$$
2n
\frac{\varphi(Q_n)}{Q_n}
$$

is positive.

Thus, it suffices to prove:

$$
\boxed{
\Delta_\square(n)
>
-
2n
\frac{\varphi(Q_n)}{Q_n}
}
$$

to obtain:

$$
L_\square(n)>0.
$$

This compresses the problem to:

> Can square-anchored phase discrepancy completely cancel the positive sieve main term?

This is another canonical closure target obtained in this phase.

---

# 18. Why Is a Crude Termwise Absolute-Value Estimate Insufficient?

Although:

$$
|\theta_n(d)|<1,
$$

a direct estimate:

$$
|\Delta_\square(n)|
\le
\sum_{d\mid Q_n}1
=
2^{\omega(Q_n)},
$$

gives an enormously large bound.

What is actually needed is:

$$
\boxed{
\text{Möbius cancellation}
+
\text{square-anchor phase structure}.
}
$$

All the phase information must not be discarded.

---

# 19. The Known Relationship to Jacobsthal Covering

Define the general Jacobsthal function:

$$
j(m)
=
\min
\{
r:
\text{Every interval of }r\text{ consecutive integers contains an }m\text{-coprime integer}
\}.
$$

If we can prove:

$$
j(Q_n)\le2n,
$$

then every interval of length $2n$ contains an integer coprime to $Q_n$, yielding Legendre's conjecture in particular.

However, this proposition is much stronger than Legendre's conjecture.

The Jacobsthal problem in the literature studies the longest consecutive interval completely covered by a finite set of prime divisors; Iwaniec 1978 gives a sieve-type upper bound for general covering length, while Hagedorn 2009 performs large-scale exact computations for primorial cases.

References:

- Henryk Iwaniec, *On the Problem of Jacobsthal*, Demonstratio Mathematica 11(1), 1978, 225–232. DOI: 10.1515/dema-1978-0121.
- Thomas R. Hagedorn, *Computation of Jacobsthal's Function h(n) for n < 50*, Mathematics of Computation 78 (2009), 1073–1087. DOI: 10.1090/S0025-5718-08-02166-2.
- Mario Ziller, John F. Morack, *Algorithmic concepts for the computation of Jacobsthal's function*, arXiv:1611.03310.

---

# 20. A Concrete Example of the Generic Jacobsthal Bound Being Too Strong

For the primorial of the first $49$ primes, the computed value is:

$$
j(p_{49}\#)=742.
$$

And:

$$
p_{49}=227.
$$

Substituting:

$$
n=227
$$

into the general Jacobsthal sufficient condition would require:

$$
742
\le
2n
=
454,
$$

which clearly fails.

Thus:

$$
\boxed{
j(Q_n)\le2n
}
$$

is not a necessary route one can expect to use.

This does not refute Legendre's conjecture, because Jacobsthal asks for the **worst case over all translated intervals**, whereas Legendre's conjecture asks only about the special anchor:

$$
n^2.
$$

The square-anchored phase:

$$
-n^2\pmod d
$$

must therefore not be washed out by a general worst-case covering bound.

---

# 21. Current Frontier Audit

As of 2026, publicly available results have still not completed Legendre's conjecture.

However, Peter J. Campbell's 2026 result has proved:

> In every consecutive-square interval $(n^2,(n+1)^2)$, there exists an integer with at most three prime factors, counted with multiplicity.

That is, an explicit sieve has already reached a $3$-almost-prime in each pointwise square interval.

Thus, to actually reach primes, this program cannot merely reproduce a general sieve conclusion of ‘there is a survivor / a survivor with few factors’; it must exploit the stronger:

$$
\boxed{
\text{square phase}
+
\text{fitting section}
+
\text{exact fixed-point / residue avoidance}.
}
$$

Reference:

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.

---

# 22. A Fixed Polynomial Fitting Section Cannot Succeed Globally

## Theorem 22.1

There is no nonconstant integer-coefficient polynomial:

$$
F(n)\in\mathbb Z[n]
$$

such that:

$$
F(n)
$$

is prime for every sufficiently large positive integer $n$.

## Proof

Choose an integer $a$ such that:

$$
|F(a)|>1.
$$

Let:

$$
m=F(a).
$$

For any positive integer $t$:

$$
a_t
=
a+tm.
$$

Since integer-coefficient polynomials preserve congruences:

$$
F(a_t)
\equiv
F(a)
\equiv
0
\pmod m.
$$

When $t$ is sufficiently large:

$$
|F(a_t)|>|m|.
$$

Thus:

$$
F(a_t)
$$

has a proper divisor $|m|$, and is therefore composite.

This contradicts ‘prime for every sufficiently large $n$’.

This completes the proof.

---

# 23. Implications for This Program

If we require:

$$
F_\square(n)
=
n^2+\kappa_{\mathcal F}(n)
$$

to output a prime for every $n$, then $F_\square$ cannot merely be a fixed nonconstant integer-coefficient polynomial.

A genuinely globally valid prime-fitting section must therefore have at least one of:

- Adaptivity;
- Branching;
- State dependence;
- Nonpolynomial structure;
- Residue feedback;
- Operator fixed-point correction;
- Or another mechanism escaping the fixed-polynomial congruence trap.

This is consistent with the original research direction of an ‘adaptive prime-fitting operator’.

---

# 24. Finite Machine Audit

This round uses an actual sieve for finite checks up to:

$$
n\le3000.
$$

The results are:

- Prime counts are positive in all tested square intervals;
- The smallest observed prime count is: **2**;
- The earliest $n$ attaining that minimum are: **1, 2, 3, 5**;
- For sampled $n$, the survivors obtained by ‘removing all multiples of $p\le n$’ coincide exactly, element by element, with the actual prime set in the square interval.

Sampled results:

| $n$ | survivors | primes | identity | first survivor | first offset |
|---:|---:|---:|:---:|---:|---:|
| 10 | 5 | 5 | PASS | 101 | 1 |
| 30 | 8 | 8 | PASS | 907 | 7 |
| 100 | 23 | 23 | PASS | 10007 | 7 |
| 227 | 44 | 44 | PASS | 51539 | 10 |
| 500 | 71 | 71 | PASS | 250007 | 7 |
| 1000 | 152 | 152 | PASS | 1000003 | 3 |
| 3000 | 340 | 340 | PASS | 9000011 | 11 |

This machine experiment verifies only the finite equivalent implementation of Phase 4 and cannot carry the infinite quantifier.

---

# 25. The Two Main Routes of Phase 4

There are now two equivalent research routes with different styles.

## Route A — Section Avoidance

Construct:

$$
\kappa_{\mathcal F}(n)
$$

and prove:

$$
\boxed{
\mathfrak R_{\mathcal F}(n,a)\neq0
\quad
\forall a\in\mathfrak A_n.
}
$$

This is ‘one point passing through all obstruction stripes’.

## Route B — Survivor Count

Prove directly:

$$
\boxed{
L_\square(n)
=
2n
\frac{\varphi(Q_n)}{Q_n}
+
\Delta_\square(n)
>
0.
}
$$

This is ‘the entire local fiber cannot be completely covered’.

---

# 26. Which Route Better Matches the Original Idea?

The initial idea of this series is:

> Use a prime-fitting operator to produce the same fitted section in the global / local base spaces.

The main line should therefore remain:

$$
\boxed{
\text{Route A}.
}
$$

Route B serves as an independent certificate and comparison with known sieves.

In other words:

$$
\text{We do not merely want to prove survivor count > 0,}
$$

but hope to find a survivor specified by the fitting operator:

$$
\boxed{
\kappa_{\mathcal F}(n)\in\mathcal S_n.
}
$$

---

# 27. The Entry Point to Phase 5

The next phase should formally extract:

$$
\boxed{
\kappa_{\mathcal F}(n)
}
$$

and must distinguish two things:

### Generator

Generate a candidate center from the existing geometry / skewness / average-slope models, without scanning all historical atoms until a prime is found.

### Verifier

Use:

$$
T_{\mathrm{top}}
$$

or:

$$
\mathfrak R_{\mathcal F}(n,a)
$$

to check whether the candidate is a fixed point.

If the generator itself tests every $a\le n$ before selecting a survivor, that merely restates primality search; it is not the ‘fitting law’ this program seeks to prove.

What Phase 5 actually seeks to study is:

> Why does $\kappa_{\mathcal F}(n)$ generated by an independent fitting law systematically avoid every square-anchored forbidden phase?

---

# 28. Phase 4 Status

```text
Legendre Base-Space Program
Phase 4 — Square-Anchored Obstruction Residue Field

M6 pre-sieved offset space:              DEFINED
Atomic obstruction basis:                DEFINED
Composite obstruction redundancy:        PROVED
Atomic historical kernel Q_n:            DEFINED
Forbidden residue phase -n^2 mod a:      DEFINED
Square-anchor periodicity:               PROVED
Local survivor set:                      DEFINED
Survivor-prime equivalence:              PROVED
Prime-fit residue field:                 DEFINED
Zero-obstruction count:                  DEFINED
CRT intersection collapse:               PROVED
Exact anchored inclusion-exclusion:      DERIVED
Main term + phase discrepancy:           DERIVED
Generic Jacobsthal shortcut:             IDENTIFIED AS TOO STRONG
Fixed polynomial global fitter:          RULED OUT
Finite implementation audit:             PASS to n=3000
Current sieve frontier:                   AUDITED
Legendre proof status:                    OPEN
Next phase:                               CANONICAL FIT-SECTION EXTRACTION
```