Normalized Type-II signed-divisor target — Free Computation Foundation

Corridor

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Status: proved exact sufficient criterion; complete for prime Type-II solutions after canonical normalization

Date: 2026-08-15

External framework: Bello-Hernández, Benito, Fernández, A Divisor Parametrization for the Erdős--Straus Conjecture, arXiv:2606.10922v1

Depends on: FAB-COPRIME-DIVISOR-CRITERION.md, FAB-UNBOUNDED-DIVISOR-RATIO-CERTIFICATE.md

Claim boundary: this note does not prove Erdős--Straus. It identifies a second exact target in the same fixed-k signed divisor box used by the strong Type-I/FAB lane.

1. Fixed-k signed divisor box

Let p be a prime with

p\equiv1\pmod4,

and let

k\equiv3\pmod4, \qquad \gcd(p,k)=1.

Put

C=\frac{p+k}{4} =\prod_i r_i^{e_i}.

Because gcd(C,k)=1, define

\boxed{ \mathcal R_k(C) = \left\{ \prod_i r_i^{z_i}\pmod k: -e_i\le z_i\le e_i \right\} \subseteq(\mathbb Z/k\mathbb Z)^\times. }

The existing strong fixed-k FAB theorem asks whether

-p^{-1}\in\mathcal R_k(C).

There is a second natural target.

2. Type-II target theorem

Theorem

\boxed{-1\in\mathcal R_k(C),}

then p satisfies the Erdős--Straus equation.

Proof

Choose exponents z_i with

-e_i\le z_i\le e_i

such that

\prod_i r_i^{z_i}\equiv-1\pmod k.

Split the prime powers of C into three positive integers A,B,T prime by prime:

if z_i<0, put r_i^{-z_i} into A;
if z_i>0, put r_i^{z_i} into B;
put the remaining r_i^{e_i-|z_i|} into T.

Then

\boxed{ABT=C,\qquad \gcd(A,B)=1,}

and

BA^{-1}\equiv-1\pmod k.

Since A is a unit modulo k, this is equivalent to

\boxed{k\mid A+B.}

Put

Q=\frac{A+B}{k}.

Also

p+k=4C=4ABT.

Therefore

\begin{aligned} \frac1{ABT} +\frac1{pAQT} +\frac1{pBQT} &= \frac{pQ+B+A}{pABQT}\\ &= \frac{Q(p+k)}{pABQT}\\ &= \frac4p. \end{aligned}

Hence

\boxed{ \frac4p = \frac1{ABT} + \frac1{pAQT} + \frac1{pBQT}. }

This has the standard Type-II shape: two displayed denominators carry a factor p. QED.

3. Exact normalized Type-II lane

The theorem can be stated without the box notation.

For p≡1 mod4, a normalized Type-II certificate consists of positive integers

A,B,T,Q,k

satisfying

\boxed{ A+B=kQ, \qquad p+k=4ABT, \qquad k\equiv3\pmod4. }

The associated identity is

\boxed{ \frac4p = \frac1{ABT} + \frac1{pAQT} + \frac1{pBQT}. }

At fixed k, the first two equations say exactly

AB\mid C=\frac{p+k}{4}

and

B/A\equiv-1\pmod k.

Every signed exponent vector in [-e_i,e_i] is exactly a coprime choice of the ratio B/A with the unused prime-power mass assigned to T. Hence the Type-II target is precisely

\boxed{\tau_{II}=-1.}

4. Completeness for prime Type-II solutions

The recent divisor-parametrization theorem is complete for Erdős--Straus decompositions after scaling by 4. Its canonical construction can be normalized so that every prime Type-II solution lands in the target above.

Start from a Type-II solution and scale it to

\frac1p=\frac1X+\frac1Y+\frac1Z, \qquad 4\mid X,Y,Z,

choosing X to be the unique denominator not divisible by p, and Y,Z the two denominators divisible by p.

Set

k=X-p, \qquad g=\gcd(X,Y), \qquad b=\frac Xg, \qquad q=\frac Yg, \qquad a=\frac{kY-pX}{g}=kq-pb.

The completeness proof of Bello-Hernández--Benito--Fernández gives these as admissible FAB data.

Because p\nmid X, one has p\nmid k and p\nmid g. Also gcd(b,q)=1. Further,

\gcd(a,b) = \gcd(kq-pb,b) = \gcd(kq,b) = \gcd(k,b).

Any common divisor of k and b divides both k and X=gb, hence divides X-k=p. Since p\nmid k,

\boxed{\gcd(a,b)=1.}

Because p\mid Y and p\nmid g, write

q=pQ.

Then kq=a+bp forces p\mid a; write

a=pA.

The divisor equation becomes

\boxed{kQ=A+b.}

The third scaled denominator is

Z=\frac{pq(p+k)}{a}.

Since p\mid Z, while p\nmid k, the normalized factor A cannot contain p: if p\mid A, then gcd(A,Q)=1 and p\nmid(p+k), so cancelling a=pA would remove the only remaining required p-factor from Z (and higher p-valuation in A would violate integrality). Hence

\boxed{p\nmid A.}

Write

c=\frac{p+k}{4}.

The two FAB divisibility conditions reduce, using the displayed coprimalities, to

b\mid c, \qquad A\mid c.

Since gcd(A,b)=1,

\boxed{Ab\mid c.}

Put

T=\frac c{Ab}.

Then

A+b=kQ, \qquad p+k=4AbT,

which is exactly the normalized Type-II lane above. Therefore every prime Type-II solution supplies a fixed-k signed-divisor hit at

\boxed{-1\in\mathcal R_k(C).}

5. Classical parameter match

The normalized equations are the standard Type-II surface in divisor coordinates. Eliminating k from

A+B=kQ, \qquad p+k=4ABT

gives

\boxed{(4BQT-1)A=pQ+B.}

Thus, with the standard Type-II parameters

(A_{\rm std},B_{\rm std},C_{\rm std},D_{\rm std}) =(Q,B,T,A),

this is

(4A_{\rm std}B_{\rm std}C_{\rm std}-1)D_{\rm std} =A_{\rm std}p+B_{\rm std}.

So the new point is not a new Type-II parametrization. The useful observation is that Type II and the strong fixed-k FAB/Type-I lane live in the same signed divisor box.

6. Two targets in one box

At fixed k and C=(p+k)/4, we now have

\boxed{ \begin{array}{rcl} \tau_I&=&-p^{-1}\pmod k,\\[2mm] \tau_{II}&=&-1\pmod k. \end{array}}

Thus the same multiplicative expansion machinery can attack both classical solution types simultaneously.

This observation is the input for FAB-TWO-TARGET-KNESER.md.