How many knights can be placed on an n-by-m chessboard?
For an 8*8 chess board,


You need to observe that when 2 knights beat one another they lie on cells of different colors.
We can generate a solution by placing 32 knights in the 32 white cells so that no 2 knights beat each other. 
To prove that this is the optimal solution, we can group our rows into pairs (1,3),(2,4),(5,7),(6,8). Let us consider the row pair (1,3). Consider the pairs of cells (R1C1,R3C2) and (R1C2,R3C1). We can see that we can place only 1 knight in the cell pair (R1C1,R3C2) and only 1 knight in the cell pair (R1C2,R3C1). Continuing this idea, we can show that we cannot place more than 32 knights in an 8X8 chessboard.
We already have an arrangement of 32 knights from point 2. Hence, the solution described in point 2 is optimal.


Expanding the above idea, below are the maximum no. of knights that can places such that no 2 beat each other on any N*M chessboard.


If both N and M are odd, then the solution is ((N*M)+1)/2
If either N or M is even, then the solution is (N*M)/2