Is it possible to tile a $13 \times 13$ board with $4 \times 1$ dominoes such that the center square is left untiled?

Question

Problem Is it possible to tile a $13 \times 13$ board with $4 \times 1$ dominoes such that the center square is left untiled?

I was not able to find a tiling so I am trying to prove that it is no possible.

I tried the usual way of coloring the board with $4$ colors using a chessboard style alternating coloring. Lets say the colors are $1, 2, 3, 4$ then I find that we have $43$ $1$’s, $42$ $2$‘s, $42$ $3$‘s, $42$ $4$’s and the center ($7^{\text{th}}$ row and $7^{\text{th}}$ column) cell has color $1$. But then this meets the demands of the $4 \times 1$ dominoes, so there is no contradiction.

Any hint will be helpful. Do I need to do a different kind fo coloring?

Answer 1

Any $4\times 4$ Latin square gives a colouring by repeating across the board, starting from the top left corner. You just need a Latin square where the $(1,1)$ entry is different to the $(3,3)$ entry.

For example $$ \begin{array}{c} 1234\\2143\\3421\\4312 \end{array} $$

which gives: $$\small\begin{array}{|c|c|c|c|} \hline 1234&1234&1234&1 \\ 2143&2143&2143&2 \\ 3421&3421&3421&3 \\ 4312&4312&4312&4 \\ \hline 1234&1234&1234&1 \\ 2143&2143&2143&2 \\ 3421&34{\tiny\fbox{2}}1&3421&3 \\ 4312&4312&4312&4 \\ \hline 1234&1234&1234&1 \\ 2143&2143&2143&2 \\ 3421&3421&3421&3 \\ 4312&4312&4312&4 \\ \hline 1234&1234&1234&1\\ \hline \end{array} $$

Any $4\times 1$ (or $1\times 4)$ tetromino will cover one square of each colour, so a collection of tetrominos will cover the same number of squares of each colour. Thus they cannot cover all but the center square (boxed), as you have an extra $1$ (e.g. bottom right square of board) and one less $2$ (center square).