External-nonresidue factor cycle — Free Computation Foundation

Theorem

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Status: proved universal elementary theorem

Date: 2026-08-15

Depends on: FAB-HARD-NONRESIDUE-BRIDGE.md, SHIFTED-NONRESIDUE-TRANSFER.md, FAB-KNESER-DIVISOR-DEFECT.md

Claim boundary: constructs a finite descent/cycle graph of external quadratic nonresidue primes for every Mordell-hard prime. It does not by itself force a FAB divisor placement and therefore does not prove Erdős-Straus.

1. Hard-prime external nonresidue set

Let p be a Mordell-hard prime. Then

p\equiv1\pmod8

and

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

Define

\boxed{ E_p= \left\{ q<p:q\text{ prime and }\left(\frac qp\right)=-1 \right\}. }

Lemma — E_p is nonempty and begins beyond the hard shield

The set E_p is nonempty, and every member satisfies

\boxed{q\ge11.}

Proof

The quadratic character modulo p is nontrivial, so there is a positive integer n<p with

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

Factor n. The Legendre symbol is multiplicative, so at least one prime divisor q|n has odd nonresidue contribution:

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

Then q<=n<p, so q in E_p.

The primes 2,3,5,7 are quadratic residues modulo every Mordell-hard p, hence no member of E_p can lie in the hard shield. QED.

2. Positive shifted factor attached to every external nonresidue

For q in E_p, define

\boxed{ \sigma(q)= \begin{cases} 1,&q\equiv3\pmod4,\\ 3,&q\equiv1\pmod4, \end{cases}}

and

\boxed{ A_q=\frac{p+\sigma(q)q}{4}. }

This is always a positive integer:

if q=3 mod4, then p+q=0 mod4;
if q=1 mod4, then p+3q=0 mod4.

Because q<p, we also have

\boxed{A_q<p.}

Indeed,

A_q<\frac{p+p}{4}=\frac p2

in the 3 mod4 case, while

A_q<\frac{p+3p}{4}=p

in the 1 mod4 case.

Also

\boxed{\gcd(A_q,q)=1,}

because

4A_q\equiv p\pmod q

and q!=p.

3. Every vertex has a distinct outgoing nonresidue factor

Modulo p,

4A_q\equiv\sigma(q)q.

Since 4 is a square and the hard prime satisfies

\left(\frac3p\right)=+1,

we obtain

\boxed{ \left(\frac{A_q}{p}\right) = \left(\frac{\sigma(q)}p\right) \left(\frac qp\right) =-1. }

Therefore the prime factorization of A_q contains at least one prime r with odd valuation contribution and

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

Because r|A_q<p,

\boxed{r<p.}

Because gcd(A_q,q)=1,

\boxed{r\ne q.}

Thus

\boxed{r\in E_p\setminus\{q\}.}

Theorem — external nonresidue factor descent

For every

q\in E_p,

the shifted integer

A_q=\frac{p+\sigma(q)q}{4}

has a prime divisor

\boxed{r\in E_p,\qquad r\ne q.}

No search bound or density statement is used.

4. Directed graph and cycle theorem

Create a directed graph on the finite vertex set E_p by choosing, for each vertex q, one prime divisor

f(q)\mid A_q

with

\left(\frac{f(q)}p\right)=-1.

The theorem above guarantees

f(q)\in E_p

and

f(q)\ne q.

Hence every vertex has outdegree one and there are no self-loops.

A finite functional digraph always contains a directed cycle. Since self-loops are absent, every cycle has length at least two.

Corollary — external nonresidue factor cycle

Every Mordell-hard prime admits distinct external nonresidue primes

q_1,\ldots,q_m, \qquad m\ge2,

with indices understood cyclically such that

\boxed{ q_{i+1}\mid \frac{p+\sigma(q_i)q_i}{4} }

and

\boxed{ \left(\frac{q_i}{p}\right)=-1 \quad\text{for every }i. }

Equivalently, there are positive integers c_i satisfying

\boxed{ p+\sigma(q_i)q_i=4c_iq_{i+1}.}

This is an exact finite cyclic system attached to every hard prime.

5. Edge character when the source is 3 mod 4

Suppose

q\equiv3\pmod4

and r is a nonresidue factor chosen from

A_q=\frac{p+q}{4}.

SHIFTED-NONRESIDUE-TRANSFER.md proves factorwise that

\left(\frac rp\right) = \left(\frac rq\right).

Since the edge was chosen with (r/p)=-1,

\boxed{ q\equiv3\pmod4 \Longrightarrow \left(\frac rq\right)=-1. }

Thus every outgoing edge from a 3 mod4 vertex lands on a quadratic nonresidue modulo the source as well as modulo p.

6. Edge character when the source is 1 mod 4

Now suppose

q\equiv1\pmod4

and

r\mid A_q=\frac{p+3q}{4}

is chosen with (r/p)=-1.

Modulo r,

p\equiv-3q.

Because p=1 mod4, reciprocity gives

\left(\frac rp\right) = \left(\frac pr\right) = \left(\frac{-3q}{r}\right).

Since q=1 mod4,

\left(\frac qr\right)=\left(\frac rq\right).

Therefore

-1 = \left(\frac{-3}{r}\right) \left(\frac rq\right),

\boxed{ q\equiv1\pmod4 \Longrightarrow \left(\frac rq\right) =-\left(\frac{-3}{r}\right). }

This is the exact reciprocity rule on the second edge type.

7. Two-cycle obstruction in the all-3-mod-4 sector

Suppose two distinct primes

q,r\equiv3\pmod4

formed a two-cycle:

q\to r\to q.

The edge rule would give

\left(\frac rq\right)=-1

and

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

But quadratic reciprocity for two 3 mod4 primes gives

\left(\frac rq\right) =-\left(\frac qr\right),

a contradiction.

Therefore:

Corollary

\boxed{ \text{No directed 2-cycle can consist of two }3\bmod4\text{ vertices.} }

So the shortest possible cycles are already constrained by reciprocity.

8. Relation to the Kneser divisor defect

At a 3 mod4 vertex q, the same shifted integer

A_q=\frac{p+q}{4}

is precisely the fixed-k FAB factor box

C_q=\frac{p+q}{4}.

Therefore every such vertex carries two simultaneous structures:

a signed divisor product box in G_q whose failure has a Kneser quotient defect;
an outgoing external-nonresidue prime factor r|C_q leading to another vertex of the finite cycle graph.

This gives the desired entropy-or-descent framework:

\boxed{ \begin{array}{c} \text{external nonresidue vertex }q\\ \downarrow\\ \text{FAB divisor box at }C_q\\ \begin{cases} \text{target hit} &\Rightarrow \text{ES certificate},\\ \text{target miss}&\Rightarrow\text{proper Kneser quotient defect} \end{cases}\\ \downarrow\\ \text{nonresidue factor edge }q\to r\\ \downarrow\\ \text{finite directed cycle.} \end{array} }

A universal proof would follow if one can show that the quotient defect cannot persist consistently around such a cycle.

9. Next exact target

The first nontrivial defect is the cubic case from FAB-KNESER-DIVISOR-DEFECT.md.

At a 3 mod4 vertex with index-three failure:

every prime factor of (p+q)/4 is a cubic residue modulo q;
2 is not a cubic residue modulo q;
every chosen nonresidue factor edge q->r therefore has

r\in G_q^3 \quad\text{and}\quad \left(\frac rq\right)=-1.

So the next theorem target is concrete:

\boxed{ \text{prove that cubic-defect edge labels cannot persist around an external-nonresidue cycle,} }

or show that persistence forces a higher-index defect with strictly smaller Kneser room.

That is now an exact finite-cycle obstruction problem rather than an unbounded search over unrelated auxiliary parameters.