TITLE: How can I find the minimum number of tiles?
QUESTION [5 upvotes]: Consider an $n\times n$ chessboard whose top-left corner is colored white. But Alice likes darkness, so she wants you to cover those white cells for her. The only tool you have are black L-shaped tiles each of which covers $3$ unit cells.
Formally, each tile covers unit cells satisfying the following:

Two of the cells are adjacent to the third (shares a side).
All three of the cells do not lie on the same row or same column.
No two tiles should overlap (cover the same cell) or go outside the board.

Since these tiles cost a lot, you have to cover all the white cells using the minimum number tiles.
Example: $1\times 1$
Answer: Impossible, there's a single cell which is white. Since one tile needs $3$ empty cells, there's no way to cover this cell.
Example: $4\times 4$
Answer: $4$ ($4$ tiles can be placed as shown)

Example: $7 \times 7$
If each tile can be represented by a number, and each uncovered piece of board can be represented by 'zero', then the answer for a $7 \times 7$ board is $16$:
$$
\begin{bmatrix}
16& 16& 15& 15& 14& 14& 13 \\
16& 12& 15& 11& 14& 13& 13 \\
12& 12& 11& 11& 10& 10& 9 \\
8& 8& 7& 6& 10& 9& 9 \\
8& 7& 7& 6& 6& 2& 2 \\
5& 5& 4& 3& 3& 1& 2 \\
5& 0& 4& 4& 3& 1& 1\\
\end{bmatrix}
$$
Question
For any given $n$, what will be the minimum number of tiles?
(Note: Answer exists for odd value of $n \geq 7$)

REPLY [2 votes]: An image of the 7×7 solution as mentioned by OP:

And one possible way to extend this to a 9×9 solution (and further):

This isn't a complete proof, but since we can extend this pattern by moving the extra tiles diagonally and adding two more on either side — we have $(5,16)$, $(7,16+9)$, $(9,16+9+11)$, $(11,16+9+11+13)$... and you'll find that they are square numbers.
$$
k = \begin{cases}
n^2/4 & n \text{ even} \\
(n+1)^2/4 & n \text{ odd} \wedge n\geq 7 \\
\text{not possible} & \text{otherwise}
\end{cases}
$$