I know that Walkup published a paper stating that $m\times n$ table can be tiled with T-tetrominoes iff $4\mid m$ and $4\mid n$. The converse is clearly true because we can tile a $4\times 4$ table with 4 T-tetrominoes. I am not able to prove the direction $\implies$. I can prove that $8\mid mn$.
Here is my argument. If $m\times n$ can be tiled, clearly $mn$ is divisible by $4$. Then we color each $1\times 1$ squares in the checkerboard pattern. A T-tetromino can occupy $1$ white square and $3$ black squares (and we call it type-A), or $3$ white squares and $1$ black square (and we call it type-B). If there are $a$ type-A pieces and $b$ type-B pieces, then the number of white squares is $\frac{mn}{2}=a+3b$ and the number of black squares is $\frac{mn}{2}=3a+b$. Therefore $a=b$ and $mn=4(a+b)=8a$.
But I tried to find Walkup's paper to read online. I couldn't find a pdf available to read. I'm only a highschool student and dont have access to https://www.jstor.org/stable/2313337?seq=1. If you know Walkup's proof, can you post it/give me a sketch? Thank you.
I answered my question so this question is answered and accepted. (unless other users post something here, i'll accept my answer when I'm allowed to) Walkup's argument was the following.
Let the table be given by $\{0,1,2,...,m\}\times \{0,1,2,...,n\}$. The idea was to prove that points of the form $(2r,2s)$ where $r\not\equiv s\pmod2$ can't be a corner of any T-tetromino and any edge with an endpoint of the form $(2r,2s)$ where $r\equiv s\pmod2$ has to be a boundary of a T-tetromino. Walkup proved this by strong induction claiming the statement is true for all such points $(x,y)$ with $x+y\le 4\lambda$ and use $\lambda$ to do math induction.