We take away four corners of a $n\times n$ square. For which $n$ can we fill it with $L$ tetrominoes?
My attempt: Because we only have tetrominoes then $4 \mid n^2-4$ then we have $2 \mid n$, which means $n$ is even. If we color it line by line it shows that the black and white colors are the same number. Any $L$-tetromino can fill $1$ white and $3$ blacks or $3$ whites and $1$ black. From that, we can know that the two kinds of $L$_tetrominos are equal. Then we can know that the number of tetrominos are even. From that we know that $8 \mid n^2-4$.Which gives us:
$n^2\equiv 4\pmod 8 \Rightarrow n \equiv 2,6 \pmod 8 \Rightarrow n \equiv 2 \pmod 4$
Now we have to show that for any $n=4k+2 (k \in \mathbb{N})$ it is possible to fill or give another condition for $n$.
As you have shown in the original post, such tilings must be of the form $4k+2$ for some positive integer $k$ (if $k=0$ the resulting shape is empty). So we just have to show that all such squares are possible.
Here is a diagram illustrating one possible tiling of a $(4k+2)\times (4k+2)$ rectangle for all $k\ge 1$ (in the image I have used $k=3$):
We tile the left and right three columns with a sort of overlapping arrangement, as shown; each of these arrangements uses $4k$ tiles. Then, since two L-tetrominoes form a $4\times 2$ rectangle, we can tile any rectangle with even dimensions and one side a multiple of $4$; this gets us the green and blue regions, with $2k$ and $4k^2-6k$ tiles respectively. (The total is $4k^2+4k$ tiles, as desired.)