Exclusion of the double-even complementary branch

Exact statement

No normalized primitive proper solution with $a=0<g$ and $f=2$ has $v_2(B+C)=v_2(D+E)=1$ . Indeed, the two odd parts must have opposite character and hence residues $5$ and $17\pmod{24}$ , a pair excluded by FPRD-IC-T22.

StatusProved without exponent bounds; independently reconstructed

External reviewNo documented external or specialist review of these FPRD results is recorded.

Context

When f=2 splits as one factor of two from each inner sum, both sums are 1+3^b with even b. Their odd parts are 5 or 17 modulo 24 according to b modulo 4.

Proof or evidence

Opposite character forces the forbidden residue pair (5,17), so the entire infinite subbranch is empty.

Double-even branch exclusion →

Verification notes

The full exponent period and the absent residue row were reconstructed independently in Python and C++; a bounded canonical census supplies controls only.

Limitations

The alternative f=2 split (2,0), all f=1 rows, and f=0 in the complementary orientation remain open.
This closes one subbranch of one normalized three-addition grammar, not Selfridge's conjecture.
No external review or novelty determination is recorded.

Open work

Classify the three surviving f=1 residue rows, beginning with their unique-minimum and tied 3-adic strata.