On a chessboard with 8 × 8 squares a rook threatens all the chess pieces in the the same line or the same column in which it stands, regardless of whether another rook stands in between or not!!
How many rooks can be placed on the chessboard maximally so that each tower threatens at most two more rooks?
The solution is 16 pieces I guess, I tried it with a real chess board, I do just not know how to prove this mathematically.
Maybe any ideas how I can transfer this into the mathematical language? I thought of like 8x8 dots of numbers like (1;1) up to (8;8) but i do not know how to go on
In this problem we ignore the rules of chess and say that two rooks "attack" each other if they are in the same row or the same column, regardless of whether the intervening squares are occupied.
Suppose you had more than $16$ rooks. Then you would have to have $3$ rooks on some row; without loss of generality (since nothing is affected if you swap rows or swap columns) you have rooks at squares a1, b1, and c1. Likewise, you must have $3$ rooks on some column; say you have rooks on d2, d3, d4. Then you must have $11$ more rooks, and they have to be placed on the $4\times4$ chessboard in the upper right corner.
So now you've got $11$ rooks on a $4\times4$ board. So you've got $3$ rooks on one row (otherwise you couldn't have more than $8$ rooks). s those $3$ rooks can't be attacking any additional rooks, that leaves only $3$ squares for the rest of the rooks. But then you've got only $6$ rooks on the board, a contradiction.
More generally, for $n\ge1,$ the maximum number of rooks you can put on an $n\times n$ chessboard with no rook attacking more than $2$ other rooks is $2n.$ This is easily proved by induction.