Exact two-target filter at the prime shift `k=19`

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Corridor

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Source in the repository

Status: proved exact group-theoretic filter

Date: 2026-08-15

Depends on: ES-TWO-TARGET-SIGNED-BOX-EQUIVALENCE.md, K15-TWO-TARGET-FILTER.md, Q11-TYPE-I-COMPANION.md

Claim boundary: classifies both exact targets at the next corridor prime after 15. It does not prove that a 3,7,11,15 survivor must hit at 19, and therefore does not prove Erdős--Straus.


1. Setup

Let p be Mordell-hard and p\neq19. Put

C=\frac{p+19}{4}.

Then C is an integer. Because every hard class satisfies p\equiv1\pmod8,

p+19\equiv20\equiv4\pmod8,

so C is odd. Also p\not\equiv0\pmod{19}, hence 19\nmid C and every prime factor of C is a unit modulo 19.

Unlike q=7 (forced factor 2) and q=11 (forced factor 3), there is no prime that divides C for every hard class. The six classes give

\begin{array}{c|c|c} p\bmod840 & C\bmod210 & \text{forced primes}\\ \hline 1 & 5 & 5\\ 121 & 35 & 5,7\\ 169 & 47 & \text{none}\\ 289 & 77 & 7\\ 361 & 95 & 5\\ 529 & 137 & \text{none} \end{array}

The classification is therefore by the residue support of C, not by a single forced generator.


2. The unit group and the two targets

The group \((\mathbb Z/19\mathbb Z)^\times\) is cyclic of order eighteen. The residue 2 is a primitive root. Quadratic residues are

\boxed{ Q=\{1,4,5,6,7,9,11,16,17\}. }

The complementary nonresidues are

N=\{2,3,8,10,12,13,14,15,18\}.

Since 19\equiv3\pmod4, -1 is a nonresidue:

-1\equiv18\pmod{19}.

That is the Type-II target. The Type-I target is -p^{-1}\bmod{19}. Quadratic reciprocity of the Jacobi symbol gives

\Bigl(\frac{-p^{-1}}{19}\Bigr) = \Bigl(\frac{-1}{19}\Bigr) \Bigl(\frac p{19}\Bigr) = -\Bigl(\frac p{19}\Bigr).

Thus the Type-I target lies in Q if and only if p is a nonresidue modulo 19. Explicitly:

\begin{array}{c|c|c} p\bmod19 & -p^{-1}\bmod19 & \text{side}\\ \hline 1 & 18 & N\\ 4 & 14 & N\\ 5 & 15 & N\\ 6 & 3 & N\\ 7 & 8 & N\\ 9 & 2 & N\\ 11 & 12 & N\\ 16 & 13 & N\\ 17 & 10 & N\\ 2 & 9 & Q\\ 3 & 6 & Q\\ 8 & 7 & Q\\ 10 & 17 & Q\\ 12 & 11 & Q\\ 13 & 16 & Q\\ 14 & 4 & Q\\ 15 & 5 & Q\\ 18 & 1 & Q \end{array}

The nine residue classes in the first block are exactly Q. On those classes both exact targets are nonresidues.


3. The QR-trap

Write \(\mathcal R_{19}(C)\) for the signed divisor box of C modulo 19. Write K for the signed box of the quadratic-residue part of C (so K=\{1\} if every prime factor is a nonresidue).

Theorem — QR-trap misses Type II

If every prime factor of C lies in Q, then \(\mathcal R_{19}(C)\subseteq Q\), hence

\boxed{-1\notin\mathcal R_{19}(C).}

Proof

Q is a subgroup. Every generator is already in Q, so the signed box they generate stays in Q. The Type-II target 18 lies outside Q. QED.

Theorem — QR-trap Type I

Under the same hypothesis,

\boxed{ -p^{-1}\notin\mathcal R_{19}(C) \quad\text{whenever}\quad \Bigl(\frac p{19}\Bigr)=+1. }

If instead \(\bigl(\frac p{19}\bigr)=-1\), the Type-I target lies in Q, and Type I hits if and only if that residue already lies in the (possibly thin) box K.

Proof

On a residue class the Type-I target is a nonresidue, by the table in Section 2, hence cannot lie in a box contained in Q. On a nonresidue class the target is in Q, so membership is exactly membership in K. QED.

In particular, if the QR box is the full subgroup Q, then a QR-trap combined miss occurs if and only if p itself is a residue modulo 19. Type I then contributes extra coverage precisely on the nine nonresidue classes of p.


4. Forced 5 and 7 fill Q on one hard class

Both 5 and 7 are quadratic residues modulo 19. Their simple local sets are

\{5^{-1},1,5\}=\{1,4,5\}, \qquad \{7^{-1},1,7\}=\{1,7,11\}.

The product is already the whole subgroup:

\{1,4,5\}\cdot\{1,7,11\} = Q.

Theorem — class 121

If p\equiv121\pmod{840}, then 5\mid C and 7\mid C, so K=Q. Consequently:

  1. if every prime factor of C is a quadratic residue modulo 19, then Type II misses, and Type I hits if and only if \(\bigl(\frac p{19}\bigr)=-1\);
  2. if some prime factor of C is a nonresidue, then Q translated by that nonresidue fills the nonresidue coset, so Type II hits and p satisfies Erdős--Straus at k=19.

Combined miss on this class is therefore exactly

\boxed{ \text{every prime factor of }C\text{ is in }Q \quad\text{and}\quad \Bigl(\frac p{19}\Bigr)=+1. }

Proof

p=840t+121 gives C=210t+35=5\cdot7\cdot(6t+1). The product of the two local sets is Q as computed above. A nonresidue factor multiplies Q onto N, which contains 18. The QR-trap case is Section 3 with K=Q. QED.

On the other four classes that force only 5 or only 7, the QR box contains {1,4,5} or {1,7,11} respectively and need not be all of Q.


5. A single nonresidue inverse pair

The nonresidues pair under inversion:

\{2,10\},\quad\{3,13\},\quad\{8,12\},\quad\{14,15\},\quad\{18\}.

A prime r\equiv18\pmod{19} contributes -1 itself, so Type II hits.

For each of the other four pairs, a single such prime to the first power has local set of size three, none of which is 18. Type II then hits if and only if the QR box already contains a definite companion residue.

Theorem — one inverse pair

Suppose every nonresidue prime factor of C is congruent to r or r^{-1} modulo 19, for a single r\in N\setminus\{18\}, and write K for the QR box. Then

-1\in\mathcal R_{19}(C) \quad\Longleftrightarrow\quad \gamma(r)\in K,

where

\begin{array}{c|c} r\bmod19 & \gamma(r)\\ \hline 2 & 9\\ 10 & 17\\ 3 & 6\\ 13 & 16\\ 8 & 7\\ 12 & 11\\ 14 & 4\\ 15 & 5 \end{array}

Proof

The nonresidue elements of the box are {r,r^{-1}}\cdot K. This set contains 18 if and only if K contains 18\cdot r^{-1} or 18\cdot r. Those two companions are the displayed values of \gamma (they coincide with the two rows of each inverse pair). QED.

Two distinct inverse pairs, with no extra QR mass, produce only four nonresidue residues and still miss 18, because a product of two nonresidues is a residue. Extra QR mass or a factor 18\bmod{19} is required to finish Type II in that case.


6. Combined miss theorem

Theorem

For a Mordell-hard prime p\neq19, both exact targets miss at k=19 if and only if -1 and -p^{-1} both lie outside \(\mathcal R_{19}(C)\). Structurally this is one of the following:

  1. QR-trap, residue side: every prime factor of C is in Q, and \(\bigl(\frac p{19}\bigr)=+1\);
  2. QR-trap, thin nonresidue side: every prime factor of C is in Q, \(\bigl(\frac p{19}\bigr)=-1\), and the Type-I target does not lie in the thin QR box K;
  3. nonresidue packet that misses 18: the nonresidue support of C does not place 18 in the signed box (Section 5), and the Type-I target also misses that box.

In every other case at least one target hits, and p satisfies Erdős--Straus.

On the single hard class p\equiv121\pmod{840}, cases 2 and 3 with a nonresidue factor cannot occur: case 2 is absorbed into a full-Q Type-I hit, and a nonresidue factor is a Type-II hit. Combined miss on that class is exactly case 1.


7. Corridor position

The integer (p+19)/4 is the next neighbour after the four already-classified forms

A,\quad A+1,\quad A+2,\quad A+3 \qquad\bigl(A=(p+3)/4\bigr).

A hypothetical counterexample that has escaped q=3,7,11 and k=15 must place this fifth consecutive integer into a QR-trap of residue type, a thin QR box that misses the Type-I target, or a nonresidue packet that misses both 18 and -p^{-1}.

Type I does contribute extra coverage at k=19, unlike q=3 and q=7, and like q=11. The extra coverage is exactly the QR-trap on nonresidue classes of p for which the QR box contains the Type-I target, together with those nonresidue packets whose box happens to contain -p^{-1} but not 18.


8. Finite signal

Through 400{,}000 there are 1005 Mordell-hard primes above 19. Of these, 331 are QR-traps and 500 are a single nonresidue inverse pair. The QR-trap Type-II lemma, the class-121 filling of Q, and the companion table \gamma have no exceptions in that range. The identities are proved above; the count is only a census.

The next exact corridor target is the Type-I companion to the existing q=23 Type-II theorem.

Independent checks live in verify_two_target_companions.py.