Pure-dyadic first denominators are universally obstructed on the hard-prime lane

Geometry

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Status: proved universal obstruction

Date: 2026-08-15

Depends on: BINARY-R-DIVISOR-COLLISION.md, FAB-MIRROR-CHARACTER-OBSTRUCTION.md

Claim boundary: this rules out an apparently attractive smooth-denominator strategy. It does not prove Erdős-Straus.

1. Setup

Let p be Mordell-hard. Then

\boxed{p\equiv1\pmod8.}

Choose any n>=1 such that

2^{n+2}>p

and put

\boxed{A=2^n,\qquad r=4A-p=2^{n+2}-p.}

Then r>0, and because p==1 mod 8,

\boxed{r\equiv7\pmod8.}

In particular r is odd and r==3 mod 4.

The one-denominator subtraction is

\frac4p-\frac1A =\frac r{pA}.

Put

N=pA=p2^n.

By the binary divisor-collision theorem, the remainder can split into two unit fractions only if

-1\in D_r(N)D_r(N)^{-1}.

2. Every divisor ratio is dyadic modulo r

From

r=2^{n+2}-p

we have

\boxed{p\equiv2^{n+2}\pmod r.}

Every divisor of N=p2^n has the form

p^\epsilon2^j, \qquad \epsilon\in\{0,1\},\quad0\le j\le n.

Therefore every quotient of two divisors is, modulo r, a power of 2:

\boxed{ D_r(N)D_r(N)^{-1} \subseteq \langle2\rangle. }

Indeed replacing every occurrence of p by 2^(n+2) makes this immediate.

3. Jacobi obstruction

For every odd positive integer

r\equiv7\pmod8,

the Jacobi symbols satisfy

\boxed{\left(\frac2r\right)=+1}

and

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

Hence every power of 2 has Jacobi sign +1 modulo r, while -1 has sign -1.

Thus

\boxed{-1\notin\langle2\rangle\pmod r.}

Consequently

\boxed{ D_r(N)\cap(-D_r(N))=\varnothing, }

and the binary remainder does not split.

Theorem

For every Mordell-hard prime p and every positive dyadic first denominator A=2^n with 4A>p, the associated binary numerator

r=4A-p

cannot rescue p.

QED.

4. Why this matters

The obstruction is not lack of divisor count. It is a squareclass obstruction.

The maximally smooth choice A=2^n collapses the entire signed divisor box into one Jacobi-positive subgroup, while the binary target -1 is Jacobi-negative.

Therefore a successful smooth-denominator construction must import at least one factor carrying the opposite character.

On the Mordell-hard lane, the frozen small shield satisfies

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

So the missing ingredient cannot be built solely from 2,3,5,7. It must include a genuine external nonresidue prime or squareclass.

This independently matches:

FAB-HARD-NONRESIDUE-BRIDGE.md;
FAB-MIRROR-CHARACTER-OBSTRUCTION.md;
FAB-GCD-SURFACE-REFORMULATION.md.

All three routes now point to the same reduced design principle:

\boxed{ \text{large smooth/square bulk} +\text{one unavoidable external nonresidue defect}. }