{2,3,5,7}-smooth forced factors cannot produce a Type-II hit

Status: proved obstruction, with one explicit external-prime rescue family

Date: 2026-08-15

Depends on: ES-TWO-TARGET-SIGNED-BOX-EQUIVALENCE.md, FAB-HARD-NONRESIDUE-BRIDGE.md

Claim boundary: this rules out a uniform Type-II covering that uses only the hard-class small primes as forced factors. It does not prove or disprove Erdős--Straus. The explicit 13-family is an identity for one arithmetic progression inside a Mordell-hard class; no literature-priority claim is made for that residue without a separate prior-art check against Salez/Webb-type modular lists.

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1. Independent question

A tempting covering strategy is:

pick a fixed multiplier M, force M to divide C=(p+k)/4 by taking k ≡ -p (mod 4M), and hope that the signed box of M already contains -1.

If this worked for a single M on every reduced residue class of the six Mordell-hard progressions modulo 4M, it would prove prime Erdős--Straus by Type II alone.

The strategy is formally decidable for each M. It fails completely when M is supported on {2,3,5,7}, and the failure is a Jacobi obstruction rather than a computational accident.

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2. Aligned shift

Let p be a Mordell-hard prime and let M be a positive integer coprime to p. Write

Then k ≡ -p \pmod{4M} and, because p ≡ 1 \pmod 4,

The corresponding shifted integer is

and the defining congruence forces

Hence

A Type-II hit from the forced factors alone is exactly

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3. Hard-class Jacobi calculus

Mordell-hard primes satisfy

If a prime r\in\{2,3,5,7} divides M, then 4r divides 4M, so the aligned shift inherits a rigid residue modulo 4r.

r=2

If 2\mid M then 8\mid 4M, hence

and

r=3

If 3\mid M then

Quadratic reciprocity and k\equiv3\pmod4 give

r=5

If 5\mid M then k\equiv-p\pmod5 lies in {1,4}, both squares, and 5\equiv1\pmod4, so

r=7

If 7\mid M then p is a quadratic residue modulo 7 and k\equiv-p\pmod7. Thus

Since 7\equiv3\pmod4 and k\equiv3\pmod4,

Coprimality

For r\in\{2,3,5,7\} and p\neq r one has gcd(r,k)=gcd(r,p)=1, so each local factor is a unit modulo k.

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4. The obstruction

Theorem — hard-smooth Type-II obstruction

Let p be Mordell-hard and let M be a positive integer whose prime factors all lie in {2,3,5,7}, with gcd(M,p)=1. Let k be the aligned shift of Section 2. Then every element of the signed box R_k(M) is a Jacobi residue modulo k, while

Therefore

In particular the forced {2,3,5,7}-smooth part of C cannot by itself produce a Type-II solution at this aligned shift.

Proof

The signed box is generated by the prime factors of M and their inverses. Each such prime is Jacobi-positive modulo k by Section 3, so the whole box lies in the Jacobi-positive units. The Type-II target is Jacobi-negative because k\equiv3\pmod4. QED.

The same argument applies to any subset of {2,3,5,7}. Special cases recover the forced-factor geometry of the small corridor:

• M=2 produces k=7 on every hard class, and {2^{-1},1,2} cannot contain -1;
• M=6 produces k=23 on every hard class, and the forced {2,3}-box cannot contain -1.

A Type-II hit at those shifts always requires a cofactor prime outside {2,3,5,7}.

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5. Covering consequence

Corollary — no {2,3,5,7}-smooth Type-II cover

There is no {2,3,5,7}-smooth multiplier M such that the aligned shift k\equiv-p\pmod{4M} is a Type-II hit from the forced factors of M for every Mordell-hard prime p.

A finite scan of every such M with 210\mid M and M\le120120 found zero complete covers and, for M=210, zero hits among the six hard classes. The theorem explains the scan: those misses are forced, not sparse.

Any uniform Type-II arithmetic-progression cover of a hard class must therefore import an external prime

into M. This is the covering-language form of the same external-nonresidue boundary recorded in FAB-HARD-NONRESIDUE-BRIDGE.md.

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6. Positive counterpart: an explicit 13-family

Once an external prime is allowed, some refined hard subclasses do become uniform Type-II hits. The smallest clean example uses

The residue

is reduced modulo 10920. Its aligned shift is

The signed box of M modulo 551 contains -1. An explicit pair of inverse witnesses is

and its reciprocal. Taking the first witness as a Type-II ratio B/D gives

Theorem — uniform Type-II family

Let t\ge0 be an integer and put

Then

In particular every prime in the progression 10920t+10369 satisfies Erdős--Straus by a Type-II solution at shift 551.

Proof

Set A=1, B=546, D=5. Then

Dividing the identity 4ABDT=Ap+B+D by ABDTp yields the displayed decomposition. Dirichlet's theorem supplies infinitely many such primes because

The seed t=0 is itself prime:

QED.

The same construction, with 11 or 13 included in M, produces further isolated subclasses (for example p\equiv18169\pmod{18480} at shift 311). They are thin: a few dozen hits among thousands of reduced subclasses. They do not cover any full Mordell-hard class modulo 840.

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7. What this does not do

The obstruction does not prevent:

• a Type-II hit at a non-aligned shift;
• a Type-I hit at the aligned shift, using the cofactor of C/M;
• a Type-II hit that uses primes of C outside M.

It only kills the hope that a fixed {2,3,5,7}-smooth packet of forced factors, placed by CRT at k\equiv-p\pmod{4M}, can finish the hard classes.

The remaining all-prime problem is unchanged: some later two-target shift, necessarily importing an external nonresidue into the signed box, must still be shown to exist for every hard prime.

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8. Finite checks

verify_hard_smooth_typeii.py independently checks:

1. the four Jacobi evaluations of Section 3 on a range of {2,3,5,7}-smooth M and all six hard residues;
2. absence of -1 from R_k(M) on those aligned shifts;
3. the exponent witness 2\cdot3\cdot5^{-1}\cdot7\cdot13\equiv-1\pmod{551};
4. the master identity 4ABDT=Ap+B+D for 0\le t\le 20.