Wilbrink's orbit matrix with diagonal entries restricted to $\{0,2,4\}$

Wilbrink's orbit matrix, with the diagonal already classified.

There is no $9\times9$ matrix $C=(c_{ij})$ over $\mathbb{N}$ that is symmetric, has every row summing to $14$, satisfies $C^{2}+C=12I_9+22J_9$, and has every diagonal entry equal to $0$, $2$ or $4$.

This is Conway99.conway_99_no_orbit_matrix with the last condition added. That condition is not an extra assumption in any real sense — it is forced by the other three, and the companion reduction proves exactly that. Its purpose is to cut the search: without it a diagonal entry could a priori be anything up to $14$, and there are $15^{9}$ diagonals to consider; with it there are $3^{9}$, a factor of about $2600$ smaller, and each of the three values pins down the row it sits in very tightly.

How the constraint arises. The diagonal case $i=j$ of the matrix identity reads

Squaring preserves parity termwise, so $\sum_k c_{ik}^2 \equiv \sum_k c_{ik} = 14 \equiv 0 \pmod 2$, and therefore $c_{ii}$ is even. Separating the diagonal term and applying Cauchy–Schwarz to the eight off-diagonal entries of the row, which sum to $14-c_{ii}$ and whose squares sum to $34-c_{ii}-c_{ii}^{2}$, gives

that is $9c_{ii}^{2}-20c_{ii}-76\le 0$, so $c_{ii}\le 4$.

What is left. The three admissible diagonal values are each rigid. If $c_{ii}=4$ the eight off-diagonal entries of row $i$ sum to $10$ with squares summing to $14$, which is exactly the minimum possible, so that row is $\{2,2,1,1,1,1,1,1\}$ up to order and nothing else. The values $0$ and $2$ leave a little more freedom. Combining that rigidity with the off-diagonal equations $\sum_k c_{ik}c_{jk} = 22 - c_{ij}$ is what remains, and it is the content of Wilbrink's Theorem 5.