TITLE: Is it possible to tile a $13 \times 13$ board with $4 \times 1$ dominoes such that the center square is left untiled?
QUESTION [5 upvotes]: 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?

REPLY [9 votes]: 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).