For which odd $n$ can all the black squares of an $n \times n$ chessboard, whose corners are black, be covered with L-shaped pieces of 3 squares

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Let $m$ be such that $n=2 m+1$. If we color the black squares in odd columns gray, we notice that an L covers at most one gray square. Therefore, to cover all the black (and gray) squares, we need at least as many Ls as there are gray squares. There are $(m+1)^{2}$ gray squares, so we need at least $3(m+1)^{2}$ squares covered by Ls in a valid covering. This shows that for $n=1, 3$ or 5, there is no tiling. For $n=7$, there is the covering on the left.
![](https://cdn.mathpix.com/cropped/2024_05_10_ae651857c6c918274013g-68.jpg?height=620&width=1110&top_left_y=562&top_left_x=518)

If we can cover the black squares for the $n \times n$ chessboard, then we can construct a covering of the black squares for the $(n+2) \times (n+2)$ chessboard by adding Ls at the bottom and to the right as shown in the figure on the right. Therefore, we can construct a covering if and only if $n \geq 7$.