Mirror character obstruction for fixed-k fab rescue

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Status: proved universal theorem on the Mordell-hard prime lane

Date: 2026-08-15

Depends on: FAB-UNBOUNDED-DIVISOR-RATIO-CERTIFICATE.md, FAB-HARD-NONRESIDUE-BRIDGE.md

Claim boundary: this rules out a broad class of tempting fixed-k constructions. It does not by itself prove that a rescue always exists and therefore does not prove Erdős-Straus.


1. Setup

Let p be a Mordell-hard prime. In particular

p\equiv1\pmod8.

Let C>0 satisfy

\gcd(C,p)=1

and put

\boxed{k=4C-p.}

Assume k>1. Then

\boxed{k\equiv3\pmod4}

and

\gcd(C,k)=1.

The fixed-k divisor-square theorem says that this k supplies a sufficient fab certificate exactly when there is a divisor

u\mid C^2

such that

\boxed{4u\equiv-1\pmod k.}

We now show that a C assembled entirely from quadratic residues modulo p can never do this.


2. Reciprocity transfer lemma

Lemma

For every odd prime r|C,

\boxed{\left(\frac r k\right)=\left(\frac r p\right),}

where the left symbol is Jacobi when k is composite.

Proof

Because r|C,

k=4C-p\equiv-p\pmod r.

Since k\equiv3 mod4, quadratic reciprocity gives

\left(\frac r k\right) = \left(\frac{-1}{r}\right) \left(\frac k r\right).

But

\left(\frac k r\right) = \left(\frac{-p}{r}\right) = \left(\frac{-1}{r}\right) \left(\frac p r\right).

The two (-1/r) factors cancel, so

\left(\frac r k\right)=\left(\frac p r\right).

Finally p\equiv1 mod4, hence reciprocity between p and r contributes no sign:

\left(\frac p r\right)=\left(\frac r p\right).

QED.

The factor 2

If 2|C, then C is even and

k=4C-p\equiv-p\equiv7\pmod8.

Therefore

\left(\frac2k\right)=+1.

On the hard-prime lane p\equiv1 mod8, so also

\left(\frac2p\right)=+1.

Thus the same residue sign is preserved for the dyadic factor as well.


3. Mirror obstruction theorem

Theorem

Assume every prime factor r of C is a quadratic residue modulo p:

\boxed{ r\mid C\Longrightarrow \left(\frac r p\right)=+1. }

Then there is no divisor u|C^2 satisfying

4u\equiv-1\pmod k, \qquad k=4C-p>1.

Consequently this k cannot supply a fixed-k sufficient fab certificate.

Proof

By the reciprocity-transfer lemma, every prime divisor of C is also Jacobi-positive modulo k. Hence every divisor

u\mid C^2

satisfies

\boxed{\left(\frac u k\right)=+1.}

If instead

4u\equiv-1\pmod k,

then, since 4 is a square modulo odd k,

\left(\frac u k\right) = \left(\frac{-4^{-1}}k\right) = \left(\frac{-1}k\right).

But k\equiv3 mod4, so

\boxed{\left(\frac{-1}k\right)=-1.}

This contradicts (u/k)=+1. Therefore no such divisor exists. QED.


4. Structural interpretation

The theorem says that the fixed-k rescue mechanism cannot be obtained by simply taking a nearby integer C whose complete prime support has already been forced onto the quadratic-residue side of p, and then reflecting it through

\boxed{k=4C-p.}

That construction preserves the residue sign of every prime factor of C, while the fixed-k target

-4^{-1}\pmod k

is Jacobi-negative.

Thus any successful interior divisor-square rescue must import genuine nonresidue support:

\boxed{ \exists r\mid C: \left(\frac r p\right)=-1. }

This is the fixed-k mirror of the external-nonresidue theorem in FAB-HARD-NONRESIDUE-BRIDGE.md.


5. Exact corollaries for the current counterexample sieve

The theorem kills several seductive but structurally impossible one-line constructions.

Corollary A — mirror of the p+1 spine

Let

C=\frac{p+1}{2}, \qquad k=p+2.

If the simplest p+1 filter has failed, every odd prime factor of C is 1 mod4; hence every such factor is a quadratic residue modulo hard p. The factor 2, when present, is also a residue because p\equiv1 mod8.

Therefore the reflected choice

\boxed{k=p+2}

cannot rescue a hard-prime survivor through the fixed-k divisor-square criterion.

Corollary B — mirror of the Eisenstein neighbour

Let

A=\frac{p+3}{4}.

If the exact k=3 filter has failed, every prime factor of A is 1 mod3. For hard p, reciprocity gives those factors quadratic-residue sign relative to the corresponding mirror construction. Hence taking a fixed-k construction obtained merely by reflecting this already-residue-safe support cannot supply the missing nonresidue target.

The same principle applies to the other shifted-factor filters whenever their failure theorem has already forced every prime divisor of the chosen C to be a quadratic residue modulo p.


6. Research consequence

This theorem prunes an entire family of false proof strategies.

The next successful construction must not be a mirror of an already-safe shifted factor. It must deliberately incorporate a prime or factor carrying

\boxed{\left(\frac r p\right)=-1.}

For Mordell-hard primes the small shield gives

\left(\frac2p\right) = \left(\frac3p\right) = \left(\frac5p\right) = \left(\frac7p\right)=+1,

so the first possible genuinely new prime support begins at the external boundary

\boxed{11,13,17,\ldots}

depending on p.

This explains why the one-shot search should now construct k from external nonresidue data, rather than reflect any of the already-proved residue-safe neighbouring forms.