pigeonhole principle, for a chessboard problem. assume we have a chess-board 8 by 8 squares. lets define the "prince" play tool: the "prince" can move from any given square to any other square the Queen can move to or to any other square the Horse can move to. means that if from some square on the chess board the set off all squares the Queen can go to is Q = {$x_1,...,x_m$} and the set off all squares the Horse can go to is H = {$y_1,...,y_n$} than the set off all squares the Prince can go to is P = Q $\cup$ H. prove that for 7 pieces of "princes" on the board there must be 2 that will Threaten each other. * * * im' not sure if the statement is still correct for even 6 pieces but i know that for less than 6 pieces it isn't.

HINT

Note that a prince ends up attacking all of the 24 squares around it with itself being in the middle of a $5 \times 5$ square. So think of it how you can place such $5 \times 5$ squares on a board, noting of course that a prince placed along the side of the board threatens less than $24$ actual squares of the board (so it is still not an easy problem this way, but maybe a little easier nevertheless...)