Direct-shadow smoothness theorem — Free Computation Foundation

Shadow

Fix a candidate progression

Status: proved universal theorem

Date: 2026-08-15

Claim boundary: this theorem restricts which earlier moduli can directly shadow a fixed candidate progression. It does not imply that collective/union shadows collapse to direct shadows, and it does not prove Erdős-Straus.

Setup

Fix a candidate progression

x(s)=r+Ls.

For an earlier Type A/B layer i, write

m=4i-1, \qquad g=\gcd(L,m), \qquad q=m/g.

The attained fibre modulo m is

\{r+Ls\pmod m:s\in\mathbb Z/q\mathbb Z\}.

Every Type A/B trap residue is a unit modulo m, because each trap is -e or -4e with e|i and

\gcd(i,4i-1)=1.

Theorem

If layer i directly shadows the candidate, then every prime divisor of m already divides L:

\boxed{\operatorname{rad}(m)\mid\operatorname{rad}(L).}

Equivalently, every direct-shadow modulus is smooth over the prime support of L.

Proof

Suppose a prime p|m does not divide L.

Then p divides the quotient

q=m/\gcd(L,m),

and L is invertible modulo p.

As s runs over a complete parameter period modulo q, it runs over every class modulo p. Therefore there is an s satisfying

r+Ls\equiv0\pmod p.

That attained fibre point is not a unit modulo m, while every Type A/B trap is a unit modulo m. Hence that fibre point is not in T_i, so the layer cannot directly shadow the whole candidate.

Contradiction. QED.

Consequence for exact direct-shadow falsification

For a concrete target candidate with target modulus M=4k-1, every earlier direct-shadow modulus satisfies

m<M, \qquad m\equiv3\pmod4, \qquad \operatorname{rad}(m)\mid\operatorname{rad}(L).

Thus an exact direct-shadow check need not scan all i<k. It is enough to enumerate the finitely many L-smooth odd moduli below M that are 3 mod 4, and test their complete attained fibres.

This reduction is used by verify_dsc_counterexample.py.