Exact r=7 binary rescue for Mordell-hard primes

Theorem

Let p be a Mordell-hard prime. Then

Status: proved iff classification

Date: 2026-08-15

Depends on: BINARY-R-RESCUE.md, BINARY-R-DIVISOR-COLLISION.md

Claim boundary: this closes the r=7 stage exactly. It does not prove that every hard prime is rescued at r=7, and therefore does not prove Erdős–Straus.

1. Setup

Let p be a Mordell-hard prime. Then

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

so in particular

\boxed{p\bmod7\in\{1,2,4\}.}

These are exactly the nonzero quadratic residues modulo 7.

Put

\boxed{A_7=\frac{p+7}{4}}, \qquad N_7=pA_7.

Since hard p == 1 (mod 8),

p+7\equiv0\pmod8,

hence

\boxed{2\mid A_7.}

By the binary-r collision theorem, p is rescued at r=7 iff the divisor residues of N_7 meet their negatives modulo 7.

2. Fixed quadratic-residue divisors already fill the QR half

The quadratic-residue subgroup modulo 7 is

Q_7=\{1,2,4\}.

We always have divisor residue 1.

We also have the divisor p, whose residue lies in Q_7, and because 2|A_7, the literal divisor 2 belongs to N_7.

Casewise:

if p == 1 (mod 7), divisors 1 and 2 are present, and A_7 == 2 (mod 7); products available inside the divisor lattice give the needed QR support;
if p == 2 (mod 7), the fixed divisors 1,p,A_7 have residues

1,2,4;

if p == 4 (mod 7), the fixed divisors 1,p,2 have residues

1,4,2.

More directly in the first case, if a nonresidue factor q occurs, multiplication by the divisor 2 supplies the missing negative whenever q == 3 (mod 7), while q == 5,6 collide with already present 2,1 respectively. Thus no separate assumption that all three QR residues are present is needed.

3. Any nonresidue prime factor forces rescue

The nonresidues modulo 7 are

\{3,5,6\}.

Let q|A_7 be prime and assume (q/7)=-1.

q == 5 mod 7

Then

-q\equiv2\pmod7.

The divisor 2|A_7 is present, so the divisor residue set contains both q and -q. Rescue follows.

q == 6 mod 7

Then

-q\equiv1\pmod7.

The divisor 1 is present, so rescue follows.

q == 3 mod 7

Then

2q\equiv6\equiv-1\pmod7.

Because q is odd and 2q|A_7, both 1 and the divisor 2q are present, so the collision follows.

Thus

\boxed{ \exists q\mid A_7\text{ prime with }(q/7)=-1 \Longrightarrow r=7\text{ rescues }p. }

4. Converse

If every prime factor of A_7 is a quadratic residue modulo 7, then every divisor of A_7 is a quadratic residue modulo 7.

The hard prime p is itself a quadratic residue modulo 7. Therefore every divisor of

N_7=pA_7

lies in Q_7.

But

-1\equiv6\pmod7

is a quadratic nonresidue, so negation maps the QR half to the NQR half. Hence

D_7(N_7)\cap(-D_7(N_7))=\varnothing.

No binary-r rescue exists at r=7.

Combining the two directions gives the exact classification:

\boxed{ \text{r=7 rescues }p \iff \frac{p+7}{4} \text{ has a prime factor that is a quadratic nonresidue mod }7. }

Equivalently, failure at r=7 is exactly

\boxed{ q\mid\frac{p+7}{4},\ q\text{ prime} \Longrightarrow \left(\frac q7\right)=+1. }

5. Simultaneous residue interpretation modulo p

By the reciprocity bridge in BINARY-R-DIVISOR-COLLISION.md, for every odd prime q|A_7,

\left(\frac qp\right)=\left(\frac q7\right).

Therefore failure of the r=7 rescue says that every prime factor of the neighboring integer

\boxed{A_7=A_3+1}

is simultaneously a quadratic residue modulo 7 and modulo p.

The preceding r=3 failure says every prime factor of

A_3=\frac{p+3}{4}

is 1 mod 3, equivalently quadratic-residue-side modulo 3.

Thus a hypothetical counterexample surviving both binary stages has two consecutive integers

A_3,\quad A_3+1

with exact split-prime restrictions in two different quadratic characters.