> 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 1
## Canonicalizing the Algebraic Consecutive-Square Base Space
### Canonical Algebraic Consecutive-Square Base Space

**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  
**Status:** Phase 1 formalization; not a proof of Legendre's conjecture

---

# 0. Objective of This Phase

The previous version of the program rewrote Legendre's conjecture as:

$$
\text{Square base space}
\longrightarrow
\text{Prime-fitting operator}
\longrightarrow
\text{Local / global compatibility}
\longrightarrow
\text{Cell internality}
\longrightarrow
\text{Primality}
\longrightarrow
\text{Legendre}.
$$

The task of Phase 1 is to fully canonicalize the first object:

$$
\text{‘Algebraic consecutive-square base space’}
$$

as a canonical structure.

The key correction is:

> The prime objects of Legendre's conjecture lie in the integer domain. Thus, the base space actually carrying prime existence should not consist solely of continuous real intervals $(n^2,(n+1)^2)$, but must include the discrete integer fiber in each square interval.

This paper therefore adopts a two-layer description, ‘continuous carrier + discrete arithmetic fiber’, and further unifies all local square cells into a global triangular integer lattice.

---

# 1. The Original Legendre Proposition

Legendre's conjecture is:

$$
\forall n\in\mathbb Z_{>0},
\qquad
\exists p\in\mathbb P
\quad
\text{such that}
\quad
n^2<p<(n+1)^2.
$$

For fixed $n$, the integers between the two squares are:

$$
n^2+1,\,
n^2+2,\,
\dots,\,
n^2+2n.
$$

Since:

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

the number of integers lying strictly between the two squares is exactly:

$$
2n
$$

integers.

---

# 2. The Continuous Carrier and Arithmetic Fibers

## 2.1 Continuous Square Cells

Define:

$$
C_n
=
(n^2,(n+1)^2)
\subset\mathbb R.
$$

Call it the $n$th continuous square cell.

## 2.2 Discrete Arithmetic Square Cells

Define:

$$
A_n
=
C_n\cap\mathbb Z.
$$

Thus:

$$
\boxed{
A_n
=
\{n^2+k:1\le k\le2n\}.
}
$$

This is the local arithmetic space on which Legendre's conjecture acts directly.

Its cardinality is:

$$
\boxed{
|A_n|=2n.
}
$$

---

# 3. Local Displacement Coordinates

Let:

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

Define the local coordinate map:

$$
\psi_n:K_n\to A_n,
$$

$$
\boxed{
\psi_n(k)=n^2+k.
}
$$

Its inverse map is:

$$
\psi_n^{-1}(m)=m-n^2.
$$

Thus:

$$
\boxed{
A_n\cong K_n.
}
$$

The Legendre proposition is therefore equivalent to:

$$
\forall n\ge1,
\qquad
\exists k\in K_n
\quad
\text{such that}
\quad
n^2+k\in\mathbb P.
$$

---

# 4. The Global Algebraic Consecutive-Square Base Space

Unify all local offset fibers and define:

$$
\boxed{
\mathcal T_{\square}
=
\{(n,k)\in\mathbb Z_{>0}^2:
1\le k\le2n\}.
}
$$

Call it:

## The Algebraic Consecutive-Square Base Space

### Algebraic Consecutive-Square Base Space

The local fiber for fixed $n$ is:

$$
\mathcal T_{\square,n}
=
\{n\}\times K_n.
$$

Hence:

$$
\boxed{
\mathcal T_{\square}
=
\bigsqcup_{n\ge1}
\mathcal T_{\square,n}.
}
$$

This is the rigorous form of ‘the global base space consists of all local square base spaces’.

---

# 5. The Evaluation Map

Define:

$$
E:
\mathcal T_{\square}
\to
\mathbb Z_{>1},
$$

$$
\boxed{
E(n,k)=n^2+k.
}
$$

For a local fiber:

$$
E\bigl(\mathcal T_{\square,n}\bigr)
=
A_n.
$$

---

# 6. The First Formal Theorem: Unique Square-Cell Decomposition

## Theorem 6.1

Let:

$$
\mathbb N_{\mathrm{ns}}
=
\{m\in\mathbb Z_{>1}:m\text{ is not a perfect square}\}.
$$

Then:

$$
\boxed{
E:
\mathcal T_{\square}
\longrightarrow
\mathbb N_{\mathrm{ns}}
}
$$

is a bijection.

## Proof

### Surjectivity

Take any:

$$
m\in\mathbb N_{\mathrm{ns}}.
$$

Let:

$$
n=\lfloor\sqrt m\rfloor.
$$

Since $m$ is not a square:

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

let:

$$
k=m-n^2.
$$

Then:

$$
1\le k\le2n.
$$

Hence:

$$
(n,k)\in\mathcal T_{\square},
$$

and:

$$
E(n,k)=n^2+k=m.
$$

Thus $E$ is surjective.

### Injectivity

If:

$$
E(n,k)=E(n',k'),
$$

then:

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

But:

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

and:

$$
n'^2<n'^2+k'<(n'+1)^2.
$$

Intervals between different consecutive squares are disjoint, so necessarily:

$$
n=n'.
$$

and consequently:

$$
k=k'.
$$

Thus $E$ is injective.

Therefore:

$$
\boxed{
E
\text{ is a bijection.}
}
$$

This completes the proof.

---

# 7. Canonical Coordinates

Theorem 6.1 gives:

$$
\boxed{
m
\longleftrightarrow
\left(
\lfloor\sqrt m\rfloor,\,
m-\lfloor\sqrt m\rfloor^2
\right)
}
$$

for all nonsquare integers greater than $1$.

That is:

$$
\boxed{
m=n^2+k,
\qquad
1\le k\le2n,
}
$$

with $(n,k)$ unique.

Thus, ‘between consecutive squares’ can be rigorously rewritten as a two-dimensional algebraic base space.

---

# 8. The Base Projection

Define:

$$
\pi_{\square}:
\mathcal T_{\square}
\to
\mathbb Z_{>0},
$$

$$
\boxed{
\pi_{\square}(n,k)=n.
}
$$

Then:

$$
\pi_{\square}^{-1}(n)
=
\mathcal T_{\square,n}.
$$

and:

$$
|\pi_{\square}^{-1}(n)|
=
2n.
$$

---

# 9. The Prime Pullback Subspace

Using $E$, define:

$$
\boxed{
\mathcal P_{\square}
=
E^{-1}(\mathbb P).
}
$$

That is:

$$
\mathcal P_{\square}
=
\{(n,k)\in\mathcal T_{\square}:
n^2+k\in\mathbb P\}.
$$

Its $n$th local fiber is:

$$
\mathcal P_{\square,n}
=
\mathcal P_{\square}
\cap
\mathcal T_{\square,n}.
$$

---

# 10. The Canonical Base-Space Form of Legendre's Conjecture

Legendre's conjecture is equivalent to:

$$
\boxed{
\forall n\ge1,
\qquad
\mathcal P_{\square,n}\neq\varnothing.
}
$$

Equivalently:

$$
\boxed{
\pi_{\square}
\bigl(
\mathcal P_{\square}
\bigr)
=
\mathbb Z_{>0}.
}
$$

That is:

> Legendre's conjecture is equivalent to the projection of the genuine prime subspace onto the square-index axis being surjective.

---

# 11. The Prime Occupancy Function

Define:

$$
\Omega_{\mathbb P}(n)
=
|\mathcal P_{\square,n}|.
$$

Then:

$$
\Omega_{\mathbb P}(n)
=
\pi((n+1)^2-1)-\pi(n^2).
$$

Legendre's conjecture is equivalent to:

$$
\boxed{
\Omega_{\mathbb P}(n)\ge1
\quad
\forall n\ge1.
}
$$

Subsequent work, however, does not use counting as the primary proof interface, but views $\Omega_{\mathbb P}$ as prime occupancy of the square fiber.

---

# 12. The Section Form of the Prime-Fitting Operator

Define the offset-fitting function:

$$
\kappa_{\mathcal F}:
\mathbb Z_{>0}
\to
\mathbb Z_{>0}.
$$

Require:

$$
1\le
\kappa_{\mathcal F}(n)
\le
2n.
$$

Then define:

$$
s_{\mathcal F}:
\mathbb Z_{>0}
\to
\mathcal T_{\square},
$$

$$
\boxed{
s_{\mathcal F}(n)
=
\left(
n,\kappa_{\mathcal F}(n)
\right).
}
$$

Then:

$$
\boxed{
\pi_{\square}
\circ
s_{\mathcal F}
=
\operatorname{id}_{\mathbb Z_{>0}}.
}
$$

Thus $s_{\mathcal F}$ is a section of $\pi_{\square}$.

---

# 13. Recovering Integer Candidates from a Section

Define:

$$
F_{\mathbb P}
=
E\circ s_{\mathcal F}.
$$

Then:

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

If:

$$
1\le
\kappa_{\mathcal F}(n)
\le2n,
$$

we automatically have:

$$
n^2
<
F_{\mathbb P}(n)
<
(n+1)^2.
$$

Square-cell internality is therefore compressed to:

$$
\boxed{
1\le
\kappa_{\mathcal F}(n)
\le2n.
}
$$

---

# 14. Prime-Valued Section

If we can prove that:

$$
F_{\mathbb P}(n)\in\mathbb P
$$

holds for every $n$, that is:

$$
n^2+\kappa_{\mathcal F}(n)\in\mathbb P,
$$

then:

$$
s_{\mathcal F}(n)\in\mathcal P_{\square}.
$$

Thus:

$$
\boxed{
s_{\mathcal F}:
\mathbb Z_{>0}
\to
\mathcal P_{\square}
}
$$

becomes a prime-valued section.

---

# 15. The Second Formal Theorem: Equivalence Between a Section and Legendre's Conjecture

## Theorem 15.1

The following statements are equivalent:

### (A)

$$
\forall n\ge1,
\qquad
\exists p\in\mathbb P:
n^2<p<(n+1)^2.
$$

### (B)

There exists:

$$
\kappa:
\mathbb Z_{>0}
\to
\mathbb Z_{>0}
$$

such that:

$$
1\le\kappa(n)\le2n
$$

and:

$$
n^2+\kappa(n)\in\mathbb P
$$

for every $n$.

### (C)

There exists a section:

$$
s:
\mathbb Z_{>0}
\to
\mathcal P_{\square}
$$

satisfying:

$$
\pi_{\square}\circ s
=
\operatorname{id}_{\mathbb Z_{>0}}.
$$

## Proof

If (A) holds, select for each $n$ a:

$$
p_n\in\mathbb P\cap A_n,
$$

and let:

$$
\kappa(n)=p_n-n^2.
$$

This gives (B).

If (B) holds, define:

$$
s(n)=(n,\kappa(n)).
$$

Then:

$$
s(n)\in\mathcal P_{\square}
$$

and:

$$
\pi_{\square}(s(n))=n,
$$

which gives (C).

If (C) holds, write:

$$
s(n)=(n,k_n).
$$

Then:

$$
1\le k_n\le2n
$$

and:

$$
n^2+k_n\in\mathbb P.
$$

Thus:

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

giving (A).

This completes the proof.

---

# 16. The Logical Boundary

Theorem 15.1 is an equivalent reformulation, not a proof of Legendre's conjecture.

If the section $s$ is selected cell by cell only after a prime is already known to exist in every square interval, it does not reduce the proof obligations at all.

The genuinely new research must be:

$$
\boxed{
\text{independent operator construction}
+
\text{prime-image proof}.
}
$$

That is, construct $s_{\mathcal F}$ directly from a global formula that neither assumes Legendre's conjecture nor queries the target prime.

---

# 17. The Prime-Fitting Graph Base Space

Define:

$$
\boxed{
\mathcal G_{\mathcal F}
=
\{
(n,\kappa_{\mathcal F}(n)):
n\ge1
\}.
}
$$

If:

$$
1\le
\kappa_{\mathcal F}(n)
\le2n
$$

holds for every $n$, then:

$$
\mathcal G_{\mathcal F}
\subset
\mathcal T_{\square}.
$$

If we further prove:

$$
n^2+\kappa_{\mathcal F}(n)\in\mathbb P,
$$

then:

$$
\boxed{
\mathcal G_{\mathcal F}
\subset
\mathcal P_{\square}.
}
$$

---

# 18. The Third Formal Theorem: The Fitting-Graph Inclusion Theorem

## Theorem 18.1

If an independently constructed fitting operator $\mathcal F$ satisfies:

$$
\mathcal G_{\mathcal F}
\subset
\mathcal T_{\square}
$$

and:

$$
\mathcal G_{\mathcal F}
\subset
\mathcal P_{\square},
$$

Legendre's conjecture holds.

## Proof

This is because $\mathcal G_{\mathcal F}$ selects exactly one point for each $n$:

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

The first inclusion gives:

$$
1\le
\kappa_{\mathcal F}(n)
\le2n.
$$

The second inclusion gives:

$$
n^2+\kappa_{\mathcal F}(n)\in\mathbb P.
$$

Hence:

$$
n^2
<
n^2+\kappa_{\mathcal F}(n)
<
(n+1)^2.
$$

Thus, there is at least one prime between every two consecutive squares.

This completes the proof.

---

# 19. The Actual Proof Target after Phase 1

At this point, the problem has been compressed to:

$$
\boxed{
\mathcal G_{\mathcal F}
\subset
\mathcal P_{\square}.
}
$$

And $\mathcal G_{\mathcal F}$ must be generated independently from the existing prime-fitting formula.

What must actually be proved next is:

$$
\boxed{
1\le
\kappa_{\mathcal F}(n)
\le2n
}
$$

and:

$$
\boxed{
n^2+\kappa_{\mathcal F}(n)
\in\mathbb P.
}
$$

The former is cell internality; the latter is the prime-image property.

---

# 20. Normalized Local Coordinates

Define:

$$
u_n(k)
=
\frac{k}{2n+1}.
$$

Then:

$$
0<u_n(k)<1.
$$

The local discrete standard cell is:

$$
U_n
=
\left\{
\frac1{2n+1},
\frac2{2n+1},
\dots,
\frac{2n}{2n+1}
\right\}
\subset(0,1).
$$

For the candidate fitting operator, define:

$$
\boxed{
\gamma_{\mathcal F}(n)
=
\frac{
\kappa_{\mathcal F}(n)
}{
2n+1
}.
}
$$

Cell internality is equivalent to:

$$
\boxed{
0<
\gamma_{\mathcal F}(n)
<
1.
}
$$

---

# 21. Unifying the Global and Local

We currently have:

$$
\mathcal T_{\square}
$$

as the global base space,

$$
\mathcal T_{\square,n}
$$

as the local fiber,

and:

$$
\mathcal G_{\mathcal F}
$$

as the global fitting section graph.

We also have:

$$
\mathcal G_{\mathcal F}
\cap
\mathcal T_{\square,n}
=
\{
(n,\kappa_{\mathcal F}(n))
\}.
$$

The ‘same global / local operator’ actually needed in subsequent work should be interpreted as:

> $\kappa_{\mathcal F}$ must be generated by a single global rule. It cannot be refitted, retrained, assigned new parameters, or supplied with queried answers for different $n$.

This will serve as the global provenance invariance of Phase 2.

---

# 22. The Core Results of Phase 1

The canonical global base space:

$$
\boxed{
\mathcal T_{\square}
=
\{(n,k):
n\ge1,\,
1\le k\le2n
\}.
}
$$

The evaluation map:

$$
\boxed{
E(n,k)=n^2+k.
}
$$

The bijection:

$$
\boxed{
E:
\mathcal T_{\square}
\overset{\sim}{\longrightarrow}
\mathbb N_{\mathrm{ns}}.
}
$$

The genuine prime subspace:

$$
\boxed{
\mathcal P_{\square}
=
E^{-1}(\mathbb P).
}
$$

Legendre's conjecture is equivalent to:

$$
\boxed{
\pi_{\square}
(
\mathcal P_{\square}
)
=
\mathbb Z_{>0}.
}
$$

The candidate prime-fitting graph:

$$
\boxed{
\mathcal G_{\mathcal F}
=
\{
(n,\kappa_{\mathcal F}(n))
\}.
}
$$

And if we can prove:

$$
\boxed{
\mathcal G_{\mathcal F}
\subset
\mathcal P_{\square},
}
$$

Legendre's conjecture holds.

---

# 23. The Entry Point to Phase 2

The task of Phase 2 is:

> Extract a canonical prime-fitting section from the existing prime-geometry predictor.

That is, use:

$$
\varepsilon_{\mathrm{full}}(N),
$$

$$
\widehat p_N,
$$

the dynamic-average slope, and the additive-anchoring mechanism to construct:

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

The first bridging question is:

> The natural input of the old formula is the prime-sequence index $N$, whereas the natural input of the new base space is the square-cell index $n$. How can we establish the canonical map without assuming Legendre's conjecture or directly querying the target prime:

$$
n
\longmapsto
N(n)
\longmapsto
\widehat p_{N(n)}
\longmapsto
\kappa_{\mathcal F}(n)?
$$

This will be the first formal question of Phase 2.

---

# 24. Phase 1 Status

```text
Legendre Base-Space Program
Phase 1 — Canonical Algebraic Consecutive-Square Base Space

Continuous square cells:             DEFINED
Arithmetic square fibers:            DEFINED
Triangular algebraic base space:     DEFINED
Evaluation map E(n,k)=n^2+k:         DEFINED
E bijection to nonsquare integers:   PROVED
Prime pullback subspace:             DEFINED
Legendre as fiber-surjectivity:      PROVED EQUIVALENT
Prime-fitting section formalism:     DEFINED
Prime-fitting graph criterion:       PROVED
Circularity boundary:                IDENTIFIED
Normalized local coordinate:         DEFINED

Phase 1 verdict:                     CLOSED
Legendre proof status:               OPEN
Next phase:                          OPERATOR EXTRACTION
```