The answer is 8 bishops minimum. The hard part is finding all possible arrangements. 

1st, prove "8": Take White squares alone. There are 14 white squares on the outer border of the board, and each white square bishop can cover only as many as 4. (A single bishop may cover 2,3 or 4 border squares). So a minimum of 4 bishops are needed. Putting all four in a row along one of the middle two ranks works as a solution, for example. 

2nd, What are all the possible arrangments? With the board oriented with the bottom right white square on your right ("white on the right"), consider all the white diagonals of slope +1. Starting from one white corner (a diagonal of length 1), they have lengths 1,3,5,7,7,5,3,1. Call the set (5,7,7,5) the "major +1 diagonals". Consider one of the diagonals of length 3 and call its two end squares the "Two Ends". 

The necessary and sufficient positioning for the 4 white bishops, I think, is this: place one bishop on each of the 4 major +1 diagonals such that: 1) no bishop covers another, 2) one of these bishops is on the board's longest diagonal (8 squares, slope -1) and 3) the Two Ends are covered. These are all the White solutions. I will need a computer program to count them. The total, White and Black, will be that number squared.