Coprime fab certificates force an external hard-class nonresidue

Corridor

Let p be a Mordell-hard prime:

Status: proved structural theorem

Date: 2026-08-15

Depends on: FAB-COPRIME-DIVISOR-CRITERION.md

Claim boundary: this gives a necessary condition for coprime fab certificates on the Mordell-hard prime classes. It does not prove existence of such a certificate for every prime and therefore does not prove Erdős-Straus.

1. Setup

Let p be a Mordell-hard prime:

p\bmod840\in\{1,121,169,289,361,529\}.

In particular

\boxed{p\equiv1\pmod8}

and p is a quadratic residue modulo 3, 5, and 7.

Suppose a,b are coprime positive integers below p, and k is a coprime fab certificate as in FAB-COPRIME-DIVISOR-CRITERION.md:

k\mid a+bp, \qquad k\equiv-p\pmod{4ab}.

Write

\boxed{p+k=4abc}

with c>0, and put

q=\frac{a+bp}{k}.

2. Auxiliary factorization

From

kq=a+bp

and

p=4abc-k,

we get

kq =a+b(4abc-k) =a+4ab^{2}c-bk.

Hence

k(q+b)=a(1+4b^{2}c).

The coprime criterion gives gcd(a,k)=1, so

\boxed{k\mid1+4b^{2}c.}

Write

\boxed{kd=1+4b^{2}c.}

Since the right side is 1 mod4 and k=3 mod4, necessarily

\boxed{d\equiv3\pmod4.}

Also

\boxed{\gcd(k,c)=1,}

because a common divisor would divide 1.

3. Nonresidue theorem

Theorem

For a Mordell-hard prime p, the leftover factor c satisfies

\boxed{\left(\frac{c}{p}\right)=-1,}

where the symbol is the Legendre/Jacobi symbol after removing square factors in the numerator.

Proof for odd c

Modulo k,

4b^{2}c\equiv-1.

Since 4b^2 is a square modulo k,

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

because k≡3 mod4.

On the other hand

p\equiv-k\pmod c.

For odd c, Jacobi reciprocity and k≡3 mod4 give

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

Therefore

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

Because p≡1 mod4, reciprocity introduces no sign when reversing p and the odd part of c, hence

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

Even c

Write c=2^e c_0 with c_0 odd.

Since the hard classes satisfy p≡1 mod8,

\left(\frac2p\right)=1.

If e>0, the identity p+k=4abc forces p+k≡0 mod8, so

k\equiv7\pmod8

and therefore

\left(\frac2k\right)=1.

Removing the factor 2^e from the reciprocity calculation leaves the same sign as in the odd case. Thus again

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

QED.

4. External-prime corollary

For every hard class,

\left(\frac2p\right) = \left(\frac3p\right) = \left(\frac5p\right) = \left(\frac7p\right) =1.

The first equality follows from p=1 mod8; the others follow by quadratic reciprocity from the fact that the six hard residues are quadratic residues modulo 3, 5, and 7.

If every prime factor of c belonged to {2,3,5,7} or occurred only through square contributions, then (c/p)=1, contradicting the theorem.

Therefore:

Corollary

Every coprime fab certificate for a Mordell-hard prime contains an odd prime factor

\boxed{\ell\mid c,\qquad \ell\ge11}

with odd valuation contribution and

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

Since p≡1 mod4, equivalently

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

5. Why this matters

The Type A/B / shadow program reached residual small-prime coordinates beginning at 11 and 13 after the 3,5,7 hard shield was imposed.

The divisor-parametrization framework reaches the same boundary from the opposite direction: a coprime certificate cannot live entirely inside the 2,3,5,7 squareclass support. It must import a genuine external nonresidue prime.

Thus a promising all-prime strategy is to coordinate:

the first external quadratic nonresidue of the hard prime;
the factorization of the linear forms a+bp;
the divisor class k=-p mod 4ab from the coprime criterion.

This is a structural bridge between the two research languages, not a claim that the remaining existence theorem is closed.