Exact two-target signed-box equivalence for prime Erdős--Straus

Theorem

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Status: proved exact reformulation

Date: 2026-08-15

External background: the standard prime Type-I/Type-II parametrization, as recorded for example in Bello-Hernández, Benito, Fernández, A Divisor Parametrization for the Erdős--Straus Conjecture, arXiv:2606.10922v1, equations (13)--(16)

Depends on: FAB-TYPE-II-SIGNED-DIVISOR.md, FAB-UNBOUNDED-DIVISOR-RATIO-CERTIFICATE.md

Claim boundary: this is an exact reformulation of the prime Erdős--Straus problem in one family of finite multiplicative boxes. It does not prove that one of the two targets is always hit. No literature-priority claim is made without a separate prior-art review.

1. The signed box

Let p be a prime with

p\equiv1\pmod4.

Let k be a positive integer satisfying

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

Put

\boxed{C_k=\frac{p+k}{4}}

and factor

C_k=\prod_i r_i^{e_i}.

Since gcd(C_k,k)=1, define the signed divisor box

\boxed{ \mathcal R_k(C_k) = \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.}

It is symmetric under inversion:

\boxed{\mathcal R_k(C_k)^{-1}=\mathcal R_k(C_k).}

The two distinguished targets are

\boxed{ \tau_I=-p^{-1}\pmod k, \qquad \tau_{II}=-1\pmod k.}

2. Every signed ratio has a coprime factor realization

Let

\rho=\prod_i r_i^{z_i}\in\mathcal R_k(C_k).

Construct positive integers B,D,T prime by prime:

if z_i>0, place r_i^{z_i} in B;
if z_i<0, place r_i^{-z_i} in D;
place the remaining r_i^{e_i-|z_i|} in T.

Then

\boxed{BDT=C_k,\qquad \gcd(B,D)=1}

and

\boxed{BD^{-1}\equiv\rho\pmod k.}

Thus the whole box is exactly the set of ratios B/D obtained by assigning each prime-power unit of C_k to the numerator, denominator, or neutral leftover factor.

3. Type-I target gives a standard Type-I solution

Assume

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

Choose B,D,T as above so that

BD^{-1}\equiv-p^{-1}\pmod k.

Equivalently,

\boxed{k\mid D+pB.}

Put

\boxed{A=\frac{D+pB}{k}.}

Because

p+k=4BDT,

we have

k=4BDT-p.

Substituting into Ak=D+pB gives

A(4BDT-p)=D+pB.

Rearranging,

\boxed{(4ABT-1)D=(A+B)p.}

This is exactly the standard prime Type-I equation.

Therefore

\boxed{-p^{-1}\in\mathcal R_k(C_k)\Longrightarrow\text{a Type-I Erdős--Straus solution}.}

The corresponding unit-fraction decomposition is

\boxed{ \frac4p = \frac1{ABTp} + \frac1{BTD} + \frac1{ATD}.}

4. Every Type-I solution gives the first target

Conversely, suppose positive integers A,B,T,D satisfy the standard Type-I equation

\boxed{(4ABT-1)D=(A+B)p.}

Rearrange it as

\boxed{A(4BDT-p)=D+pB.}

Set

\boxed{k=4BDT-p.}

The right side is positive, so k>0. Also

k\equiv-p\equiv3\pmod4.

The non-p denominator BDT in the Type-I decomposition is not divisible by p, so

\gcd(k,p)=1.

Further,

C_k=\frac{p+k}{4}=BDT.

Modulo k,

D+pB\equiv0,

hence

BD^{-1}\equiv-p^{-1}\pmod k.

Because B and D both divide C_k, their ratio is represented by a signed exponent vector in the box (common prime powers cancel into the neutral exponent). Therefore

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

Thus the first target is exact for Type I.

5. Type-II target gives a standard Type-II solution

Assume

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

Choose B,D,T with

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

Then

\boxed{k\mid B+D.}

Put

\boxed{A=\frac{B+D}{k}.}

Again k=4BDT-p. Substituting into Ak=B+D gives

A(4BDT-p)=B+D,

hence

\boxed{(4ABT-1)D=Ap+B.}

This is exactly the standard prime Type-II equation.

Therefore

\boxed{-1\in\mathcal R_k(C_k)\Longrightarrow\text{a Type-II Erdős--Straus solution}.}

The associated decomposition is

\boxed{ \frac4p = \frac1{ABTp} + \frac1{BTD} + \frac1{ATDp}.}

6. Every Type-II solution gives the second target

Conversely, suppose

\boxed{(4ABT-1)D=Ap+B.}

Rearrange:

\boxed{A(4BDT-p)=B+D.}

Set

\boxed{k=4BDT-p.}

Then k>0, k≡3 mod4, and because BDT is the unique non-p denominator in the standard Type-II form,

\gcd(k,p)=1.

Moreover

C_k=BDT.

Modulo k,

B+D\equiv0,

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

As before, this ratio belongs to the signed divisor box. Therefore

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

Thus the second target is exact for Type II.

7. Exact prime equivalence

The standard Type-I/Type-II parametrization is complete for prime Erdős--Straus solutions. Combining the two directions above gives:

Theorem — exact two-target signed-box equivalence

For every prime

p\equiv1\pmod4,

the following are equivalent:

p satisfies the Erdős--Straus equation;
there exists a positive integer

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

such that, with <div class="math" role="math">C_k=(p+k)/4,</div>

one has <div class="math" role="math">\boxed{ \{-p^{-1},-1\} \cap \mathcal R_k(C_k) \ne\varnothing.}</div>

Equivalently,

\boxed{ \text{prime ES} \iff \exists k\equiv3\pmod4: \bigl(-p^{-1}\in\mathcal R_k(C_k) \ \lor\ -1\in\mathcal R_k(C_k)\bigr).}

This converts the prime conjecture into an exact two-target divisor-placement problem.

8. The inverse Type-I orientation

Since the signed box is inversion-symmetric,

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

The two residues are the two orientations of the same Type-I ratio.

Therefore an unsolved fixed shift must avoid the three natural residues

\boxed{-p^{-1},\quad -p,\quad -1.}

although the first two encode one solution type.

This three-residue exclusion is the source of the strengthened Kneser budget in FAB-TWO-TARGET-KNESER.md.

9. Strategic consequence

The direct prime Erdős--Straus problem is now exactly:

For every prime p≡1 mod4, prove that at least one admissible shift k≡3 mod4 has a signed divisor box of (p+k)/4 containing either -p^{-1} or -1 modulo k.

The external-nonresidue program is one structured way to choose such shifts. It is no longer merely a search inside a sufficient FAB subclass: the two-target box language is an exact reformulation of the complete standard Type-I/Type-II solution space.