Chess/Puzzles/Placement/32 Knights/Solution

Here's a possible solution to the puzzle:

a b c d e f g h
8 a8 b8 c8 d8 e8 f8 g8 h8 8
7 a7 b7 c7 d7 e7 f7 g7 h7 7
6 a6 b6 c6 d6 e6 f6 g6 h6 6
5 a5 b5 c5 d5 e5 f5 g5 h5 5
4 a4 b4 c4 d4 e4 f4 g4 h4 4
3 a3 b3 c3 d3 e3 f3 g3 h3 3
2 a2 b2 c2 d2 e2 f2 g2 h2 2
1 a1 b1 c1 d1 e1 f1 g1 h1 1
a b c d e f g h

Proof of maximality edit

Pair up the squares of the board, demonstrated by the pairs of chess pieces on the board below.

a b c d e f g h
8                 8
7                 7
6                 6
5                 5
4                 4
3                 3
2                 2
1                 1
a b c d e f g h

Continue this pairing onto the rest of the board. On each pair of such squares, only one knight may be located. Since there are 32 such pairs, it's impossible to place more than 32 knights on the board. Since placing 32 knights is possible, 32 is the maximum number of knights that can be placed on a chessboard so no two attack each other.

This solution is only best if we assume the board cannot have other pieces besides knights. Given that no such presumption is stated, by using king pins at least a few more than 32 knights may be placed with none able to legally capture each other. Here is a simple example of a 35 knight solution by adding 11 on-color-kings, 9 off-color-queens, and 3 new knights to the board state in image 1 of this page (though more optimal answers likely exist).

(edit) Wiki won't allow any picture uploads, but surely the advancements are not hard to imagine if you begin with the 32 single color square solutions and from there apply on-color-kings and off-color-queens to secure safe and immobile knight squares to occupy.

The necessity(?) of multiple same-color kings to this method can be argued to its invalidity, but of course, 32 knights––let alone 32 white knights as shown above––is an equally impossible board state given standard starts.