Every entry is at most $4$ in the no-four orbit-matrix case

Let $C$ be a $9\times 9$ matrix over $\mathbb{N}$ satisfying the Conway99 orbit-matrix equations (symmetric, row sums $14$, $(C^2)_{ij}+c_{ij}$ equal to $34$ on the diagonal and $22$ off it), with every diagonal entry in $\{0,2,4\}$ and no diagonal entry equal to $4$. Then every entry of $C$ is at most $4$. Indeed each row has diagonal $0$ or $2$, and the row-profile census gives fiber counts with no $5$s in every case, while a direct square-sum argument excludes entries above $5$. This makes the entry domains finite and explicit for the trace-$10$, no-four branch of Wilbrink's Theorem 5.