Dual descent system behind sufficient fab certificates

Corridor

---

Status: proved exact algebraic structure

Date: 2026-08-15

Project: Free Computation Foundation / CENTL

Depends on: FAB-UNBOUNDED-DIVISOR-RATIO-CERTIFICATE.md, FAB-HARD-NONRESIDUE-BRIDGE.md

Claim boundary: these identities expose a hidden duality and reciprocity package inside every certificate. They do not prove that a certificate exists for every hard prime.

1. Master equation

Let p be a prime with

p\equiv1\pmod4.

Suppose positive integers a,b,k give a sufficient certificate:

k\mid a+bp, \qquad 4ab\mid p+k.

Write

q=\frac{a+bp}{k}, \qquad c=\frac{p+k}{4ab}.

Then

kq=a+bp, \qquad k=4abc-p.

Substituting gives

(4abc-p)q=a+bp,

hence

\boxed{4abcq=a+p(b+q).}

This is the symmetric master equation.

2. Hidden prime cofactor

Rearrange the master equation as

a(4bcq-1)=p(b+q).

Set

D=4bcq-1.

For positive b,c,q,

4bcq-1>b+q.

Indeed

4bcq-1-(b+q) \ge4bq-1-b-q>0.

Because

aD=p(b+q)

and p is prime, p must divide D. Otherwise gcd(D,p)=1 would force

D\mid b+q,

contradicting D>b+q.

Therefore there is a positive integer s such that

\boxed{4bcq-1=ps.}

Cancelling p in the master equation gives

\boxed{b+q=as.}

Thus every certificate has the exact dual form

\boxed{ ps+1=4bcq, \qquad b+q=as. }

Since

ps=4bcq-1\equiv3\pmod4

and p≡1 mod4,

\boxed{s\equiv3\pmod4.}

Also D>b+q gives

ps>as,

\boxed{a<p.}

This is automatic; it need not be assumed for a sufficient certificate.

3. Automatic coprimalities

From

ps+1=4bcq,

no prime divisor of p or s can divide b, c, or q. Hence

\boxed{ \gcd(ps,bcq)=1. }

In particular

\gcd(p,b)= \gcd(p,c)= \gcd(p,q)=1

and the same holds with p replaced by s.

4. Swap duality

The master equation is symmetric in b and q.

Define

\boxed{k'=4acq-p.}

Then

k'b =(4acq-p)b =4abcq-bp =a+pq.

Thus

\boxed{b\mid a+pq.}

More precisely,

\boxed{k'b=a+pq.}

Also

p+k'=4acq,

so k' is a valid sufficient-certificate divisor for the swapped parameter pair (a,q).

Therefore every certificate has a dual certificate

\boxed{ (a,b;k,q) \longleftrightarrow (a,q;k',b). }

Both divisors satisfy

k\equiv k'\equiv3\pmod4.

5. Norm factorization

The two dual divisors multiply to

\boxed{ kk'=p^2+4a^2c. }

Proof:

\begin{aligned} kk' &=(4abc-p)(4acq-p)\\ &=16a^2b c^2q-4acp(b+q)+p^2. \end{aligned}

Using

4abcq=a+p(b+q),

we have

16a^2bc^2q =4ac(a+p(b+q)),

which cancels the mixed term and leaves

p^2+4a^2c.

Thus every certificate factors the quadratic norm-like quantity

\boxed{p^2+4a^2c}

into two positive 3 mod4 factors.

This is the exact algebraic bridge to the repository's quadratic-field / norm machinery.

6. Reciprocity consequence on the hard-prime coprime-fab lane

Now assume additionally that the certificate lies in the coprime hard-prime fab lane where FAB-HARD-NONRESIDUE-BRIDGE.md proves

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

From

ps+1=4bcq

and the automatic coprimalities,

\left(\frac bp\right) \left(\frac cp\right) \left(\frac qp\right)=1.

Therefore

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

So exactly one of the symmetric side factors b,q carries the nonresidue sign modulo p.

Odd-c dual nonresidue

If c is odd, then the same c is also a Jacobi nonresidue modulo the hidden cofactor s:

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

Indeed, from

ps\equiv-1\pmod c

we get

\left(\frac pc\right) \left(\frac sc\right) = \left(\frac{-1}{c}\right).

Because p≡1 mod4, reciprocity gives

\left(\frac pc\right) = \left(\frac cp\right) =-1.

Hence

\left(\frac sc\right) =-\left(\frac{-1}{c}\right).

Since s≡3 mod4, quadratic reciprocity between s and odd c contributes exactly the factor (-1/c), giving

\left(\frac cs\right)=-1.

Thus the external nonresidue factor is not attached only to the original prime p; it propagates across the dual system to the new 3 mod4 cofactor s.

7. Current descent target

The exact certificate geometry can now be written as

\boxed{ \begin{array}{rcl} 4abcq&=&a+p(b+q),\\ ps+1&=&4bcq,\\ b+q&=&as,\\ k&=&4abc-p,\\ k'&=&4acq-p,\\ kk'&=&p^2+4a^2c. \end{array} }

with

p\equiv1\pmod4, \qquad s,k,k'\equiv3\pmod4.

On the hard coprime-fab lane, c is a nonresidue modulo both p and, for odd c, s.

The next proof target is an actual descent/closure theorem on this dual system, not another large exact-depth scan. A successful route would show that a hypothetical all-prime failure cannot remain closed under the (b,q) duality and the (p,s) reciprocity transfer.