Exact `p+4` factor rescue — Free Computation Foundation

Corridor

Let p be an odd positive integer. Suppose q is a positive divisor of

Status: proved elementary sufficient family / necessary counterexample restriction

Date: 2026-08-15

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

Theorem

Let p be an odd positive integer. Suppose q is a positive divisor of

p+4

with

q\equiv3\pmod4.

Put

\boxed{m=\frac{q+1}{4}}.

Then

\boxed{v=\frac{mp+1}{q}}

is a positive integer and

\boxed{ \frac4p = \frac1v + \frac1{mp} + \frac1{mpv}. }

Thus any 3 mod 4 divisor of p+4 gives an Erdős–Straus decomposition.

In particular, any prime factor

q\mid p+4, \qquad q\equiv3\pmod4

solves p.

Proof

Since q=4m-1 and q | p+4,

4(mp+1) =(q+1)p+4 =qp+(p+4).

The right side is divisible by q. Because q is odd, q | mp+1, so v is an integer.

Now

qv=mp+1.

Substitute q=4m-1:

(4m-1)v=mp+1,

hence

\boxed{4mv=mp+v+1.}

Divide by mpv:

\frac4p = \frac1v + \frac1{mp} + \frac1{mpv}.

QED.

Corollary — counterexample restriction

If p is a counterexample, then p+4 has no divisor 3 mod 4. Since

p+4\equiv1\pmod4

for Mordell-hard p≡1 mod4, this implies

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

Thus a hard-prime counterexample forces p+4 into the multiplicative semigroup generated by primes 1 mod 4.

Quadratic-residue interpretation

For a hard prime p≡1 mod4, any prime q|p+4 with q≡1 mod4 obeys

p\equiv-4\pmod q.

Therefore

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

Since p≡1 mod4, quadratic reciprocity gives

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

So failure of the p+4 rescue means every prime factor of p+4 is a quadratic residue modulo p.

This aligns with FOUR-P-PLUS-ONE-FILTER.md and the earlier shifted-factor restrictions in FAB-HARD-FIRST-FILTERS.md: an increasingly large collection of nearby forms is forced to factor entirely over the quadratic-residue half modulo a hypothetical hard-prime counterexample.

Finite regression signal

On the exact four-filter survivor population among Mordell-hard primes through 10^7, this theorem resolves 856 of the 2,173 survivors.

Applied after FOUR-P-PLUS-ONE-FILTER.md, the first six exact factor restrictions leave 791 finite survivors through 10^7.

These counts are finite evidence only; the theorem above is universal.