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Shifted nonresidue transfer and the `p+4` mirror filter

Theorem · hosted from the CENTL repository

Research library · Theorem

Theorem

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Source in the repository

Status: proved exact elementary theorems

Date: 2026-08-15

Project: Free Computation Foundation / CENTL

Depends on: PRIME-MODULUS-BACKBONE.md, FAB-HARD-FIRST-FILTERS.md, FAB-FIXED-K-SQUARE-DIVISOR.md

Claim boundary: these results sharpen necessary conditions on a hypothetical hard-prime counterexample and the arithmetic of an adaptive nonresidue shift. They do not prove that a fixed-k divisor target is always hit and therefore do not prove Erdős–Straus.

1. The missing p+4 mirror of the p+1 spine

Let p be a prime and suppose an odd prime

q\equiv3\pmod4

divides

p+4.

Put

k=\frac{q+1}{4}.

Then

4k-1=q

and

p\equiv-4\pmod q.

Since 1|k, the Type A/B trap set at depth k contains the residue -4. Therefore p has an immediate Type A/B hit.

Theorem — p+4 mirror filter

A prime escaping every Type A/B solution must satisfy

\boxed{ q\mid p+4,\ q\text{ prime} \Longrightarrow q\equiv1\pmod4. }

For a Mordell-hard prime p≡1 mod4, this is equivalent to saying that

\boxed{p+4}

lies entirely in the multiplicative semigroup generated by primes 1 mod4.

This is the exact -4 mirror of the familiar -1 / p+1 prime-modulus spine.

2. Quadratic-residue consequence

Let p be Mordell-hard, hence p≡1 mod4, and suppose it survives the p+4 filter.

If

r\mid p+4

is prime, then r≡1 mod4 and

p\equiv-4\pmod r.

Therefore

\left(\frac p r\right) = \left(\frac{-4}{r}\right) =+1.

Because p≡1 mod4, quadratic reciprocity gives

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

Thus every prime factor of p+4 is a quadratic residue modulo a hypothetical hard-prime counterexample.

This joins the already-proved residue-only factor restrictions on nearby shifted forms.

3. Factorwise reciprocity transfer on an admissible 3 mod 4 shift

Let

p\equiv1\pmod4

be an odd prime and let

q\equiv3\pmod4

be an odd prime distinct from p.

Put

\boxed{N_q=\frac{p+q}{4}.}

Let r be an odd prime divisor of N_q.

Then

p+q\equiv0\pmod r,

so

\boxed{p\equiv-q\pmod r.}

Hence

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

Because q≡3 mod4, quadratic reciprocity gives

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

The two sign factors cancel, yielding

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

Since p≡1 mod4, reciprocity between p and r has no sign, so

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

Theorem — primewise sign transfer

For every odd prime

r\mid\frac{p+q}{4},
\boxed{ (r/p)=(r/q). }

This is a factorwise statement, stronger than comparing only the Jacobi symbol of the whole shifted integer.

4. The factor 2

Suppose now that p is Mordell-hard, so

p\equiv1\pmod8.

The factor 2 divides N_q=(p+q)/4 exactly when

p+q\equiv0\pmod8.

Since p≡1 mod8 and q≡3 mod4, this occurs exactly for

q\equiv7\pmod8.

Then

(2/p)=+1

and

(2/q)=+1.

So the same transfer identity holds for r=2 whenever 2 actually occurs in N_q.

Thus for hard p, every prime factor r of N_q obeys

\boxed{(r/p)=(r/q).}

5. Nonresidue factor corollary

Assume

\left(\frac q p\right)=-1.

Because 4 is a square modulo p,

\left(\frac{N_q}{p}\right) = \left(\frac q p\right) =-1.

Therefore the prime factorization of N_q contains at least one prime r with odd valuation contribution such that

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

By the primewise transfer theorem,

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

Hence an external nonresidue shift automatically manufactures a nonresidue prime factor in the shifted integer, and that factor is simultaneously a nonresidue for both moduli.

6. Least-3mod4-nonresidue ladder

Let q be the least prime satisfying

q\equiv3\pmod4, \qquad (q/p)=-1.

Apply the preceding corollary to

N_q=\frac{p+q}{4}.

There is a prime factor r|N_q with

(r/p)=(r/q)=-1.

If

r\equiv3\pmod4,

then the minimality of q forces

\boxed{r>q,}

because q itself cannot divide N_q (gcd(q,N_q)=1).

If instead

r\equiv1\pmod4,

then the process has reached a 1 mod4 external nonresidue.

Thus every least-3 mod4 nonresidue shift has the exact dichotomy

\boxed{ \text{strict ascent to a larger }3\bmod4\text{ nonresidue factor} \quad\text{or}\quad \text{arrival at a }1\bmod4\text{ nonresidue factor}. }

This is not yet a contradiction. Its value is that it upgrades the aggregate external-nonresidue condition to an explicit factor-level ladder inside (p+q)/4.

7. Relation to the fixed-k square-divisor target

For fixed k=q, FAB-FIXED-K-SQUARE-DIVISOR.md says a coprime certificate is equivalent (in its stated range) to

\exists D\mid N_q^2, \qquad D\equiv-4^{-1}\pmod q.

When (q/p)=-1, the target -4^{-1} is a quadratic nonresidue modulo q, and the theorem above guarantees that the factorization of N_q contains a prime factor on exactly that quadratic side.

Therefore quadratic character is no longer the obstruction at an external nonresidue shift. The remaining gap is strictly finer:

\boxed{ \text{upgrade one available nonresidue factor} \text{ to the exact target residue }-4^{-1}\pmod q \text{ using the bounded divisor box of }N_q^2. }

That is now the precise multiplicative-defect / subset-product theorem to attack.