Dyachenko ED2 lattice repair audit — Free Computation Foundation

Prior art

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Status: exact proof audit and corrected lattice lemma

Date: 2026-08-15

Primary source: E. Dyachenko, arXiv:2511.07465v1

Claim boundary: this note does not claim that no repair of the paper is possible. It proves that the stated Lemma 9.24 / Proposition 9.25 route is invalid, gives a correct replacement rectangle bound, and shows that replacing the false d' bound by that correct bound does not by itself establish Theorem 9.21.

1. The lattice in the claimed unconditional step

The paper considers

L=\{(u,v)\in\mathbb Z^2:b'u+c'v\equiv0\pmod g\},

with

\gcd(b',g)=\gcd(c',g)=1,

and defines

\alpha=\gcd(g,b'+c'), \qquad \boxed{d'=g/\alpha}.

The map

\phi:\mathbb Z^2\to\mathbb Z/g\mathbb Z, \qquad (u,v)\mapsto b'u+c'v

is surjective because b' is a unit modulo g. Hence

\boxed{[\mathbb Z^2:L]=g.}

So the geometric scale of the full kernel lattice is its index g, not the diagonal period d' alone.

2. Lemma 9.24 is false as stated

The paper states that if

w=(d',d')\in L

and p0=(u0,v0) in L, then

p_0+\mathbb Zw = \{(u,v)\in L:u\equiv u_0,\ v\equiv v_0\pmod{d'}\}.

This equality is false in general.

Smallest counterexample

Take

g=2,\qquad b'=c'=1.

Then

L=\{(u,v):u+v\equiv0\pmod2\},

and

\alpha=\gcd(2,2)=2, \qquad d'=1.

Choose p0=(0,0). Since congruence modulo 1 imposes no condition, the right-hand side of the claimed equality is all of L.

The left-hand side is only

\{(m,m):m\in\mathbb Z\},

a single diagonal line.

For example (2,0) in L lies on the right but not the left.

Therefore

\boxed{\text{Lemma 9.24 is false as stated.}}

The correct statement is only the inclusion

p_0+\mathbb Z(d',d') \subseteq \{(u,v)\in L:u\equiv u_0,\ v\equiv v_0\pmod{d'}\}.

The reverse inclusion generally contains multiple parallel diagonal cosets.

3. Proposition 9.25 is false as stated

The proposition claims that every axis-parallel rectangle

R=[x_0,x_0+H)\times[y_0,y_0+W)

with

H,W\ge d'

meets L.

The earlier g=2,b'=c'=1,d'=1 example gives

R=[0,1)\times[1,2).

Its only integer point is (0,1), and

0+1\not\equiv0\pmod2.

Hence

\boxed{L\cap R=\varnothing.}

So Proposition 9.25 is false.

The proof error is exact: it chooses an x-coordinate representative and a y-coordinate representative independently, then uses one diagonal shift parameter and assumes it realizes both choices simultaneously.

4. There is no replacement bound depending only on d'

The failure is not a missing factor of 2 or another small constant.

For any integer

g\ge2,

take

b'=1,\qquad c'=g-1.

Then both are units modulo g and

b'+c'=g,

\alpha=g, \qquad \boxed{d'=1}.

But

L=\{(u,v):u-v\equiv0\pmod g\}.

Its diagonal period is (1,1), yet the lattice has index g and consists of g distinct diagonal congruence layers inside Z^2.

Unit-size rectangles can plainly be placed on integer points outside L, no matter how large g is.

Therefore there is no universal theorem of the form

H,W\ge F(d') \Longrightarrow L\cap R\ne\varnothing

for a function F depending only on d'.

The missing geometric information is the transverse spacing / full index g.

5. Correct elementary rectangle-hitting theorem

Theorem

Let

L=\{(u,v)\in\mathbb Z^2:b'u+c'v\equiv0\pmod g\}

with

\gcd(c',g)=1.

Then every axis-parallel rectangle

R=[x_0,x_0+H)\times[y_0,y_0+W)

with

\boxed{H\ge1,\qquad W\ge g}

meets L.

Symmetrically, if gcd(b',g)=1, then

\boxed{H\ge g,\qquad W\ge1}

also suffices.

Proof

Assume H>=1. The half-open interval [x0,x0+H) contains an integer u.

For this fixed u, the lattice congruence is

c'v\equiv-b'u\pmod g.

Because c' is invertible modulo g, there is exactly one residue class

v\equiv r(u)\pmod g.

Every half-open interval of length at least g contains a representative of every residue class modulo g. Since W>=g, choose such a v in [y0,y0+W).

Then (u,v) in L cap R. QED.

This is crude but universal and exact.

6. Why the corrected bound does not repair Theorem 9.21

There are two independent issues.

A. The paper verifies only the weaker d'-scale condition

The Type-I discussion and Corollary 9.26 explicitly use

H,W\ge d'=g/\alpha.

The corrected elementary theorem needs the full g scale in one coordinate.

When alpha>1, these are genuinely different requirements. The family in section 4 has

d'=1

with arbitrarily large g.

Thus the stated box-size argument does not imply the corrected hitting condition.

In the special setup of Section 9.6 the paper says to fix alpha,d', so g=alpha d' is fixed and a polylogarithmic coordinate range eventually exceeds g for sufficiently large P. That observation can repair only the linear lattice hit for such a fixed parameter family and sufficiently large P. It does not prove that one fixed (alpha,d') family produces the required nonlinear ED2 point for every prime P.

B. A linear lattice hit is not the nonlinear ED2 existence theorem

The paper itself states after Proposition 9.25 that the lattice step provides a point satisfying the linear constraints.

To obtain an ED2 solution it invokes the normalized coordinates and the inverse test. In Appendix D, Lemma D.16 gives the necessary-and-sufficient conditions

m\mid u, \qquad u\equiv v\pmod2, \qquad \boxed{u^2-v^2=4M},

or equivalently requires the discriminant

\boxed{u^2-4M}

to be a perfect square.

That lemma is a verifier/equivalence theorem. It does not assert that an arbitrary lattice point furnished by a rectangle hit satisfies the square equation.

Therefore even a corrected proof that the rectangle contains a point of L would establish only a linear-congruence point. An additional existence theorem is still required to force the nonlinear discriminant/perfect-square condition.

7. Appendix status agrees with this boundary

The paper's own appendix summary distinguishes:

the unconditional algebraic core;
unconditional direct/back algorithms and their correctness;
a conditional finite covering scheme;
computational/counting criteria for windows.

This is consistent with the repair audit: the algorithms can correctly verify or construct a solution when the required arithmetic point exists, while universal existence still needs an independent covering/existence argument.

8. Conclusion for the CENTL/FCF proof hunt

The attempted shortcut

\text{replace false Proposition 9.25 by a correct lattice-radius lemma}

does not close Erdős-Straus.

What survives is useful but narrower:

the ED2 algebraic identities and direct/back tests are legitimate theorem-mining tools;
the kernel lattice has exact index g and a simple correct g-scale hitting theorem;
the diagonal period d'=g/alpha cannot control arbitrary rectangle hitting by itself;
after any corrected linear hit, the perfect-square/discriminant existence problem remains.

Accordingly, the highest-value CENTL route remains the internally derived exact divisor-placement / external-nonresidue program rather than attempting to resurrect Theorem 9.21 from its present lattice argument.