Consider an $n \times n$ grid. A $T$-tetromino is a set of four cells of the grid in a "T" shape. Find the values of $n$ such that the $n \times n$ grid can be covered by $T$-tetrominoes without overlapping pieces.
If $4 \mid n$, then we can cover each 4-sided grid with the following tiling:

![](https://cdn.mathpix.com/cropped/2024_05_10_a954a7760d797f08770eg-368.jpg?height=420&width=416&top_left_y=2194&top_left_x=820)

Now let's show that for the grid to be tileable by T's, it is necessary that $4 \mid n$. We color the grid in a checkerboard pattern (so there are as many white squares as black squares). There are two types of T's: those that cover three white squares and one black square, and those that cover three black squares and one white square. Let $k$ be the number of T's of type 1. Since there are as many white squares as black squares, there must be as many T's of type 1 as T's of type 2. Therefore, there are also $k$ T's of type 2, which gives us $3k + k$ white squares, and similarly for the black squares. Thus, there are $4k + 4k = 8k$ squares in total. Therefore, $8 \mid n^2$, which implies that $4 \mid n$.