Exact `r=11` binary rescue classification for Mordell-hard primes

Theorem

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Status: proved iff classification

Date: 2026-08-15

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

Claim boundary: this completely closes the fixed binary numerator r=11. It does not prove that every Mordell-hard prime is rescued at r=11, and therefore does not prove Erdős-Straus.

1. Hard-prime normalization

Let p be Mordell-hard. Then

p\equiv1\pmod{24}.

For the binary numerator

\boxed{r=11}

put

A_{11}=\frac{p+11}{4}.

Because p==1 mod 3,

3\mid A_{11}.

Write

\boxed{A_{11}=3B,\qquad B=\frac{p+11}{12}.}

Modulo 11, since 12==1,

\boxed{B\equiv p\pmod{11}.}

The binary denominator is

N_{11}=pA_{11}=3pB.

By BINARY-R-DIVISOR-COLLISION.md, rescue at r=11 is equivalent to

\boxed{-1\in D_{11}(N_{11})D_{11}(N_{11})^{-1}.}

Equivalently, the signed divisor-exponent box of N_11 must hit -1 mod 11.

2. Discrete-log coordinates

Use 2 as a primitive root modulo 11:

\begin{array}{c|cccccccccc} e&0&1&2&3&4&5&6&7&8&9\\ \hline 2^e\bmod11&1&2&4&8&5&10&9&7&3&6. \end{array}

Write

\lambda(x)=\log_2(x)\in\mathbb Z/10\mathbb Z.

Then

\lambda(-1)=5, \qquad \lambda(3)=8=-2.

Put

\alpha=\lambda(p\bmod11).

Factor

B=\prod_q q^{e_q}

and put

\beta_q=\lambda(q\bmod11).

Because B==p mod 11,

\boxed{\sum_q e_q\beta_q\equiv\alpha\pmod{10}.}

The signed-divisor collision exists iff there are

z_3,z_p\in\{-1,0,1\}, \qquad -e_q\le z_q\le e_q

such that

\boxed{ 8z_3+\alpha z_p+\sum_q\beta_qz_q\equiv5\pmod{10}. }

This single finite cyclic-group equation gives the complete classification below.

3. Three residue classes are automatic

\alpha\in\{3,5,7\},

then the fixed factors 3 and p already hit exponent 5; no factorization information about B is needed.

These exponents correspond to

p\bmod11\in\{8,10,7\}.

Theorem A

\boxed{p\equiv7,8,10\pmod{11},}

then r=11 always rescues p.

Proof

For each alpha in {3,5,7}, the set

\{8z_3+\alpha z_p:z_3,z_p\in\{-1,0,1\}\}

contains 5 mod 10. QED.

4. Nonzero quadratic-residue classes

The nonzero even exponents are

\alpha\in\{2,4,6,8\},

corresponding to

p\bmod11\in\{4,5,9,3\}.

For every such alpha, the fixed factors 3 and p generate all five even exponents:

\boxed{ \{8z_3+\alpha z_p\} =\{0,2,4,6,8\}. }

A prime factor q|B is a quadratic nonresidue modulo 11 exactly when beta_q is odd. Adding one occurrence of any odd exponent to the full even set hits every odd exponent, in particular 5.

Conversely, if every prime factor of B is a quadratic residue modulo 11, every signed exponent contribution is even, so exponent 5 is impossible.

Theorem B

\boxed{p\equiv3,4,5,9\pmod{11},}

then

\boxed{ \text{`r=11` rescues `p`} \iff B=\frac{p+11}{12} \text{ has a prime factor that is a quadratic nonresidue mod }11. }

Equivalently, failure occurs iff every prime factor of B belongs to

\boxed{\{1,3,4,5,9\}\pmod{11}.}

5. The two `±1` exponent classes

Now

\alpha\in\{1,9\},

corresponding to

p\bmod11\in\{2,6\}.

The fixed factors 3,p produce

\boxed{\{0,1,2,3,7,8,9\},}

missing only 4,5,6.

Therefore the presence of any prime-factor exponent

\beta_q\in\{2,3,4,5,6,7,8\}

immediately forces rescue.

Suppose no such factor occurs. Then every nontrivial prime factor of B has exponent class 1 or 9=-1; factors of class 0 are inert in the signed box. If the total multiplicity of these ±1 occurrences is at least 2, their signed contribution can supply ±2, and the fixed exponent 3 then reaches 5. Thus failure requires exactly one nontrivial occurrence.

The total-product constraint

B\equiv p\pmod{11}

then fixes its direction.

Theorem C

\boxed{p\equiv2\pmod{11},}

then r=11 fails iff

exactly one prime-factor occurrence of B is 2 mod 11, with valuation exactly 1;
every other prime factor of B is 1 mod 11.

\boxed{p\equiv6\pmod{11},}

then r=11 fails iff

exactly one prime-factor occurrence of B is 6 mod 11, with valuation exactly 1;
every other prime factor of B is 1 mod 11.

All other factorizations rescue.

6. The class `p == 1 mod 11`

Finally let

\boxed{p\equiv1\pmod{11}.}

Then alpha=0, so the fixed factor p contributes nothing in log coordinates and the factor 3 contributes the symmetric set

\boxed{\{0,2,8\}=\{0,\pm2\}.}

Thus rescue is equivalent to the signed exponent box of B meeting

\boxed{\{3,5,7\}.}

because adding 0,±2 must reach target 5.

There are exactly two ways this can fail.

Failure family I — entirely quadratic-residue support

If every prime factor of B is a quadratic residue modulo 11, every signed exponent is even, so the odd target set {3,5,7} is unreachable.

Thus every factor may lie in

\boxed{\{1,3,4,5,9\}\pmod{11}.}

Failure family II — one cancelling `2/6` pair

The only nonresidue-support exception that still avoids {3,5,7} is:

one prime-factor occurrence 2 mod 11, valuation 1;
one prime-factor occurrence 6 mod 11, valuation 1;
every other prime factor 1 mod 11.

Indeed the two nontrivial logs are +1 and -1; their signed box is

\{0,\pm1,\pm2\},

which, after adding 0,±2, still avoids 5 exactly in this minimal cancelling configuration. Any additional nontrivial occurrence, or any factor from another residue class, expands the signed box into {3,5,7} and forces rescue.

The total-product constraint B==1 mod 11 is automatically respected by the cancelling pair.

Theorem D

\boxed{p\equiv1\pmod{11},}

then r=11 fails iff exactly one of the following holds:

every prime factor of B is a quadratic residue modulo 11; or
B has exactly one occurrence of a prime 2 mod 11 and exactly one occurrence of a prime 6 mod 11, each to valuation 1, and every remaining prime factor is 1 mod 11.

Every other factorization rescues.

7. Consolidated classification

Let

\boxed{B=\frac{p+11}{12}.}

For a Mordell-hard prime p, fixed binary numerator r=11 behaves as follows.

`p mod 11`	Exact `r=11` status
`7,8,10`	always rescued
`3,4,5,9`	rescued iff `B` has a quadratic-nonresidue prime factor mod `11`
`2`	fails iff exactly one `2 mod 11` prime occurrence and all others `1 mod 11`
`6`	fails iff exactly one `6 mod 11` prime occurrence and all others `1 mod 11`
`1`	fails iff all factors are QR mod `11`, or the unique minimal `2/6` cancelling pair occurs

This is an iff theorem, not a finite-range pattern.

8. Independent finite regression

The classification was independently compared against the exact binary-divisor collision test for every Mordell-hard prime below

10^6.

Population:

\boxed{2,370\text{ hard primes}.}

Result:

\boxed{0\text{ classification mismatches}.}

This finite regression is supporting evidence only; the theorem is the cyclic-group proof above.

9. Research consequence

The first three binary stages are now unusually explicit:

r=3: failure forces every prime factor of (p+3)/4 onto the 1 mod 3 side;
r=7: failure forces every prime factor of (p+7)/4 onto the quadratic-residue side modulo 7;
r=11: failure is now classified exactly by the table above.

Writing

P=\frac{p-1}{4},

these are the consecutive translates

\boxed{P+1,\quad P+2,\quad P+3.}

For hard primes P is divisible by 6. A hypothetical counterexample must therefore make three consecutive translates satisfy three different, exact multiplicative-compression laws.

The next theorem target is no longer “understand r=11.” It is to exploit the simultaneous incompatibility, if any, of the exact r=3, r=7, and r=11 failure patterns, and then incorporate r=19/23 only if genuine survivors remain.