Universal 3-adic valuation law in the complementary f=2 split

Exact statement

In every normalized primitive proper solution in the complementary $f=2$ , split- $(2,0)$ branch, let $V_b=(1+3^b)/4$ , where $b\ge3$ is odd, and write the other odd factor as $W=2^h+2^q3^y$ . Then $\min(b,y,g)=1+v_3(K+2-h)$ . All ten tied-minimum strata whose leading 3-adic digit could cancel are impossible: nine by complete periods modulo 7 and 73, and the sole residual class by incompatible complete periods modulo 487 and 2593.

StatusProved without exponent bounds; independently reconstructed

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

Context

In the remaining f=2 split, one inner sum contributes both factors of two. Its odd part cycles through four residues modulo 24, leaving eight placements after the quadratic-reciprocity gate.

Proof or evidence

Canonical parity reduces possible lowest-digit cancellation to ten families. Complete mod-7 and mod-73 periods eliminate nine; the last residue class has 343 survivors modulo 487 and four modulo 2593, with no CRT-compatible pair.

Complementary f=2 valuation law →

Verification notes

Independent Python and C++ implementations reconstruct every survivor count and the terminal CRT incompatibility. A bounded census checks 870,168 canonical presentations, finds six controls, and finds no valuation-law failure; it is not used for completeness.

Limitations

The theorem controls but does not exclude all complementary f=2 split-(2,0) solutions.
The controlled f=1 branch, the f=0 rows, and arbitrary expression trees remain open.
No external review or novelty determination is recorded.

Open work

Combine the f=1 and f=2 laws at minimum layers m=1,2, seeking a finite order-modulus cover before opening the f=0 rows.