Infinite knights and the Chessboard

Notice that we can decompose a $8 \times 8$ chessboard into $8$ non intersecting $2 \times 4$ rectangles. And on each of these rectangles we can put at most $4$ knights that do not capture each other. Follow the image:

Imgur

So, now since in $8$ of these $2 \times 4$ rectangles we can at most $4 \times 8 =32$ knights on a $8 \times 8$ board hat satisfies our condition .

But wait! We know that it isn't possible to put more than $32$ knight on a $8 \times 8$ chess board in such a way that satisfies our condition. But it doesn't mean that $32$ is attainable. So, we have show that it is.

If we arrange the knights in only in black squares we can get $32$ knights.

So, $32$ is our answer.

Its clear that a chess board has 32 black squares and 32 white squares. Now, any knight in white square goes to black square in next move, and vice-versa. Now if we place all the knights in white(or black), then they will be in non-attacking position, since their next move will be to a black(or white) square. Thus, the no. of knights is 32. Now, we'll show that its not possible to place more than 32 knights. Let if possible, at least 1 more knight can be placed. Thus, it must be placed in a black square and hence, it will be in an attacking position with some other knight in white. Thus, the max. no. of knights is 32.

This is one of the examples of pigeonhole principle ...

I made a discussion problem about a month ago that builds on this, which no one has answered yet: https://brilliant.org/discussions/thread/no-knights-tour/

Knights only attack the opposite color squares which they are on. Since there are 32 white squares and 32 black squares, the answer is 32.