On a chessboard of a finite size, we want to place kings and knights such that each king attacks exactly 2 kings and 2 knights, and each knight attacks exactly 2 kings and 2 knights.

If some smallest, finite board exists that fulfills the requirements, we know the following of that board:

- The board has edges
- Every edge has at least one piece on it

In the boards, I’m going to use black pawns to mark squares that have to be empty, and white pawns to mark squares that need to have either a king or a knight, but it’s not yet clear which of the two.

The piece on the edge can be either a king, or a knight, which means at least one of the following 20 formations (or a vertical reflection of one of those) has to be somewhere on the board:

- King 1 (impossible)
- King 2 (impossible)
- King 3 (impossible)
- King 4 (impossible)
- King 5 (impossible)
- King 6 (impossible)
- King 7 (still plausible)
- King 8 (still plausible)
- King 9 (impossible)
- King 10 (still plausible)
- King 11 (still plausible)
- King 12 (impossible)
- King 13 (impossible)
- King 14 (impossible)
- King 15 (impossible)
- King 16 (still plausible)
- Knight 1 (still plausible)
- Knight 2 (still plausible)
- Knight 3 (still plausible)
- Knight 4 (still plausible)

If it’s possible to prove all of these 20 formations are impossible, it’s proven no piece can be on the edge. That would contradict the idea a finite, smallest board exists, which in turn would prove there is no finite, valid board.

So far I’ve been able to proof that 11 of these 20 formations are impossible. I tried to prove the others to be impossible as well, but I was unable to do it. I suspect there's no short and simple proof for any of those.

Below are the proofs why the other 11 formations are impossible. All of these boards only have an edge on the left. The edges on the top, right, and bottom are only there to make the images fit on the screen.

- King 1

Then how about a king? That king would then attack 2 knights, and the unknown pieces on c3 and d4. D3 and d5 would then have to be empty. However, the knight on b4 would then only be able to attack 3 non-empty squares. Therefore c4 has to be empty.

If the unknown piece on a3 would be a knight, it would attack 3 non-empty squares, so it has to be a king.

That king on a3 attacks only 4 non-empty squares, so a2, b2, and b3 need to have 2 kings and 1 knight on them. The piece on b3 isn’t allowed to be one of the kings, because it would then attack 5 pieces. Therefore the knight has to go on b3, and a2 and b2 are kings. We’ve now reached a contradiction. The king on b2 attacks 4 pieces, so a1 and b1 have to be empty. However, the king on a2 is only attacking 3 pieces, so either a1 or b1 needs to have a knight. Therefore the formation king 1 isn’t allowed.- King 2

- King 3

- King 4

- King 5

- King 6

- King 7

I haven’t been able to prove this is impossible yet.

- King 8

I haven’t been able to prove this is impossible yet.

- King 9

- King 10

I haven’t been able to prove this is impossible yet.

- King 11

I haven’t been able to prove this is impossible yet.

- King 12

- King 13

- King 14

- King 15

- King 16

I haven’t been able to prove this is impossible yet.

- Knight 1

I haven’t been able to prove this is impossible yet.

- Knight 2

I haven’t been able to prove this is impossible yet.

- Knight 3

I haven’t been able to prove this is impossible yet.

- Knight 4

I haven’t been able to prove this is impossible yet.

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