Exact `4p+1` factor filter — Free Computation Foundation

Corridor

Let p be an odd prime. If 4p+1 has a divisor

Status: proved elementary sufficient family / necessary counterexample restriction

Date: 2026-08-15

Claim boundary: this does not prove Erdős–Straus. It adds one exact shifted-factor restriction to the hard-prime counterexample sieve.

Theorem

Let p be an odd prime. If 4p+1 has a divisor

F\equiv3\pmod4

with complementary divisor

G=\frac{4p+1}{F},

then G≡3 mod 4 as well, and p has an Erdős–Straus decomposition.

In particular, if 4p+1 has any prime factor 3 mod 4, then p is solved.

Proof

Because

4p+1\equiv1\pmod4,

if F≡3 mod4, then the complementary factor G is also 3 mod4.

Write

F=4u-1, \qquad G=4v-1

with positive integers u,v. Then

4p+1=(4u-1)(4v-1) =16uv-4u-4v+1,

\boxed{p=4uv-u-v.}

Therefore

4uv=p+u+v.

Dividing by puv gives the exact identity

\boxed{ \frac4p = \frac1{uv} + \frac1{pv} + \frac1{pu}. }

This is a positive three-unit-fraction decomposition. QED.

Corollary — exact counterexample restriction

A prime counterexample must satisfy

\boxed{ q\mid(4p+1),\ q\text{ prime} \Longrightarrow q\equiv1\pmod4. }

Indeed, since 4p+1≡1 mod4, any occurrence of a 3 mod4 prime factor has even total 3 mod4 valuation parity; taking one such factor (with odd exponent contribution) supplies a divisor F≡3 mod4 and the complement is also 3 mod4.

Equivalently, 4p+1 must lie in the multiplicative semigroup generated by primes 1 mod4.

Quadratic-residue interpretation on hard primes

If q | 4p+1, then

p\equiv-4^{-1}\pmod q.

When q≡1 mod4,

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

For hard primes p≡1 mod4, reciprocity gives

\boxed{\left(\frac{q}{p}\right)=+1.}

Thus failure of this filter means every prime factor of 4p+1 is a quadratic residue modulo p.

This matches the earlier exact hard-prime restrictions:

factors of (p+1)/2 are 1 mod4, hence quadratic residues of p;
factors of (p+3)/4 are 1 mod3, hence quadratic residues of p;
factors of (3p+1)/4 are 1 mod3, hence quadratic residues of p;
factors of p+2 are 1 or 3 mod8, hence quadratic residues of p;
now factors of 4p+1 are 1 mod4, hence quadratic residues of p.

The all-prime wall can therefore be restated more sharply: a hard-prime counterexample forces several nearby linear forms to factor entirely over primes in the quadratic-residue half of (Z/pZ)^*, even though FAB-HARD-NONRESIDUE-BRIDGE.md proves that any coprime fab certificate must import an external quadratic nonresidue.

Finite regression signal

On the exact four-filter survivor population among the 20,513 Mordell-hard primes through 10^7, this additional theorem resolves 866 of the 2,173 survivors.

This count is finite evidence only; the theorem itself is universal.