Binary-r rescue after one exact greedy denominator

Theorem

Let

Status: proved exact criterion; finite theorem-mining evidence deposited separately

Date: 2026-08-15

Claim boundary: this gives an exact sufficient-and-necessary binary criterion after fixing the first denominator. It does not prove that a suitable r always exists and therefore does not prove Erdős–Straus.

1. One denominator leaves a binary problem

Let

p\equiv1\pmod4

be prime, and let

\boxed{r\equiv3\pmod4}

be a prime. Put

\boxed{A_r=\frac{p+r}{4}}.

Then

4A_r=p+r,

\frac4p-\frac1{A_r} = \frac{4A_r-p}{pA_r} = \boxed{\frac r{pA_r}}.

Define

\boxed{N_r=pA_r.}

Hence

\boxed{ \frac4p = \frac1{A_r} + \frac r{N_r}. }

The Erdős–Straus problem for this chosen first denominator is therefore exactly a two-unit-fraction problem.

2. Exact binary divisor criterion

Because p≡1 mod4 and r≡3 mod4, r!=p. Also

A_r\equiv p\,4^{-1}\pmod r,

\gcd(N_r,r)=1.

Let d be a positive divisor of N_r^2, and put

d_1=\frac{N_r^2}{d}.

\boxed{d\equiv-N_r\pmod r,}

then, since N_r is invertible modulo r,

d^{-1}\equiv-N_r^{-1}\pmod r

and therefore

d_1=N_r^2d^{-1}\equiv-N_r\pmod r.

Thus both

\boxed{Y=\frac{N_r+d}{r}}, \qquad \boxed{Z=\frac{N_r+d_1}{r}}

are positive integers.

Now

\frac1Y+\frac1Z = \frac{r}{N_r+d}+ \frac{r}{N_r+d_1}.

Since dd_1=N_r^2,

(N_r+d)(N_r+d_1) =N_r(2N_r+d+d_1),

while the numerator is

r(2N_r+d+d_1).

Hence

\boxed{ \frac1Y+\frac1Z=\frac r{N_r}. }

Combining with the first denominator gives

\boxed{ \frac4p = \frac1{A_r} + \frac1Y + \frac1Z. }

Conversely, the standard factorization of a two-unit-fraction equation shows that every split

\frac r{N_r}=\frac1Y+\frac1Z

produces

d=rY-N_r, \qquad d_1=rZ-N_r,

with

dd_1=N_r^2, \qquad d\equiv d_1\equiv-N_r\pmod r.

Therefore the divisor condition is exact for the fixed r.

3. The target is always a quadratic nonresidue

Modulo r,

N_r =p\frac{p+r}{4} \equiv\frac{p^2}{4}.

Thus the target class is

\boxed{ -N_r\equiv-\frac{p^2}{4}\pmod r. }

The factor p^2/4 is a square modulo r, while

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

because r≡3 mod4. Hence

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

So every binary-r rescue must select a quadratic-nonresidue divisor of N_r^2 in one exact residue class.

This is the cleanest direct bridge found so far between:

the external-nonresidue phenomenon in FAB-HARD-NONRESIDUE-BRIDGE.md;
the factorization of a nearby shifted integer (p+r)/4;
an actual three-term Erdős–Straus decomposition.

Unlike the coprime-fab route, this criterion does not require the three denominators of the associated 1/p decomposition to all be divisible by 4.

4. Relation to the r=3 filter

For

r=3, \qquad A_3=\frac{p+3}{4},

the unit group modulo 3 has only two classes. The target nonresidue is the unique class 2 mod3.

Thus the failure of the r=3 binary rescue forces the prime-factor support of A_3 into the 1 mod3 side, recovering the exact first filter already deposited in FAB-HARD-FIRST-FILTERS.md.

The significance of larger r is that they preserve the same exact binary mechanism while introducing a richer but still finite multiplicative residue group.

5. Finite theorem-mining signal

one_shot_es_probe.py independently replays this criterion.

Through p<=500,000:

Mordell-hard primes: 1,246;
survivors of the first four exact shifted-factor theorems: 202;
every one of those 202 has a binary-r rescue for a prime

r\equiv3\pmod4, \qquad r\le71;

the first successful r histogram is

r=7   : 77
r=11  : 73
r=19  : 27
r=23  : 10
r=31  : 13
r=59  : 1
r=71  : 1

There are zero unresolved candidates in that finite census when the probe is allowed primes r<=200.

This is finite evidence only. It is not a proof that r<=71, r<=200, or any fixed finite bound works universally.

6. New theorem target

The pointwise all-prime wall can now be attacked in the following form:

For every Mordell-hard prime p, prove that there exists a prime r≡3 mod4 such that the divisor box of <div class="math" role="math">> N_r^2

> =p^2\left(\frac{p+r}{4}\right)^2 ></div>

contains the exact nonresidue class <div class="math" role="math">> -N_r\pmod r.

></div>

Equivalently, classify failure of the exact target class for each small r as a multiplicative residue-support restriction on (p+r)/4, then prove that the simultaneous failure restrictions cannot persist for all r.

This route attacks Erdős–Straus directly and does not require resurrecting the false universal DSC conjecture.