Seems like it's possible for all pair $(m, n)$ such that $2|mn$, is it right? It's hard to make configuration.

CBMaster wrote: Seems like it's possible for all pair $(m, n)$ such that $2|mn$, is it right? It's hard to make configuration. Yes, it is right

Redacted The answer is $2|mn$. (It is hard to think of the proof if not) Without loss of generality let $m$ be even and let $m=2k$. Throughout the whole solution, the diagonals will all be top left to bottom right Case1. $n$ is even (let $n=2l$) Proposition. When $k=l$ there exists a configuration that is filled with $2 \times 1$ at the first and the last row, and the remaining squares of the first and the last column are filled with $1 \times 2$. Induct on $k$ When $k=2$ Let Line 1: two $2 \times 1$ and Line 2: $1 \times 2$ at two edge, and fill with two $2 \times 1$ Repeat Lines 1 2,and 1 Assume the proposition is true for all $k' \leq k-1$ Fill the first and the last row with $2k$ $2 \times 1$ dominoes Fill the remaining first column and the last column with $2k-2$ $1 \times 2$ dominoes. Lastly, fill last $(2k-2) \times (2k-2)$ grid according to the induction. Enumerate the $m \times (m-1)$ (Get rid of the first row). And it suffices to show only for $m \times n$ when $n < m$ and $n$ is even (by adding the first row) Proposition. When $l<k$ there exists a configuration that is filled with $1 \times 2$ at the first and the last column, and the remaining squares of the first and the last row are filled with $2 \times 1$. Induction on $k$ Obvious in $k\leq3$ Assume the proposition is true for all $k' \leq k-1$ Fill the first and the last column with $2l$ $1 \times 2$ dominoes Fill the remaining first column and the last rows with $2k-2$ $2 \times 1$ dominoes. Lastly, fill last $(2k-2) \times (2l-2)$ grid according to the induction.

Lleeya wrote: The answer is $2|mn$. (It is hard to think of the proof if not) Without loss of generality let $m$ be even and let $m=2k$. Then, think of three cases about $n$ with$\pmod 3$, however, the configuration is basically the same. The solution will only use the diagonals from left top to right bottom in both dominos. Let Line 1 $m \times 1$ filled with $k$ $2 \times 1$dominoes. Let Line 2 $m \times 2$ filled with $2k$ dominoes with two $1 \times 2$ dominoes at the end (one on each), and fill $2 \times 1$ dominoes at the rest. $\text{Case }I\text{)}$ $n \equiv 1 \pmod 3$ Repeat Line $1,2,1,2,\,\,\, \cdots \,\,\, ,2,1$ $\text{Case }II\text{)}$ $n \equiv 2 \pmod 3$ Repeat Line $2,1,2,1 \,\,\, \cdots \,\,\, ,1,2$ $\text{Case }III\text{)}$ $n \equiv 0 \pmod 3$ Repeat Line $1,2,1,2,\,\,\, \cdots \,\,\, ,2$ But if $k \geq 3$, line $2$ must have $4$ $2 \times 1$ dominoes in below configuration in it, and we can't draw non-vertex sharing diagonals in each dominoes. [asy][asy] add(grid(4,2)); draw((0,0)--(1,0), blue); draw((1,0)--(2,0), blue); draw((2,0)--(3,0), blue); draw((3,0)--(4,0), blue); draw((0,1)--(1,1), blue); draw((1,1)--(2,1), blue); draw((2,1)--(3,1), blue); draw((3,1)--(4,1), blue); draw((0,2)--(1,2), blue); draw((1,2)--(2,2), blue); draw((2,2)--(3,2), blue); draw((3,2)--(4,2), blue); draw((0,0)--(0,1), blue); draw((0,1)--(0,2), blue); draw((2,0)--(2,1), blue); draw((2,1)--(2,2), blue); draw((4,0)--(4,1), blue); draw((4,1)--(4,2), blue); draw((0,1)--(2,0), red); draw((0,2)--(2,1), red); draw((2,1)--(4,0), red); draw((2,2)--(4,1), red); [/asy][/asy] Is I understand it wrong?

CBMaster wrote: Lleeya wrote: The answer is $2|mn$. (It is hard to think of the proof if not) Without loss of generality let $m$ be even and let $m=2k$. Then, think of three cases about $n$ with$\pmod 3$, however, the configuration is basically the same. The solution will only use the diagonals from left top to right bottom in both dominos. Let Line 1 $m \times 1$ filled with $k$ $2 \times 1$dominoes. Let Line 2 $m \times 2$ filled with $2k$ dominoes with two $1 \times 2$ dominoes at the end (one on each), and fill $2 \times 1$ dominoes at the rest. $\text{Case }I\text{)}$ $n \equiv 1 \pmod 3$ Repeat Line $1,2,1,2,\,\,\, \cdots \,\,\, ,2,1$ $\text{Case }II\text{)}$ $n \equiv 2 \pmod 3$ Repeat Line $2,1,2,1 \,\,\, \cdots \,\,\, ,1,2$ $\text{Case }III\text{)}$ $n \equiv 0 \pmod 3$ Repeat Line $1,2,1,2,\,\,\, \cdots \,\,\, ,2$ But if $k \geq 3$, line $2$ must have $4$ $2 \times 1$ dominoes in below configuration in it, and we can't draw non-vertex sharing diagonals in each dominoes. [asy][asy] add(grid(4,2)); draw((0,0)--(1,0), blue); draw((1,0)--(2,0), blue); draw((2,0)--(3,0), blue); draw((3,0)--(4,0), blue); draw((0,1)--(1,1), blue); draw((1,1)--(2,1), blue); draw((2,1)--(3,1), blue); draw((3,1)--(4,1), blue); draw((0,2)--(1,2), blue); draw((1,2)--(2,2), blue); draw((2,2)--(3,2), blue); draw((3,2)--(4,2), blue); draw((0,0)--(0,1), blue); draw((0,1)--(0,2), blue); draw((2,0)--(2,1), blue); draw((2,1)--(2,2), blue); draw((4,0)--(4,1), blue); draw((4,1)--(4,2), blue); draw((0,1)--(2,0), red); draw((0,2)--(2,1), red); draw((2,1)--(4,0), red); draw((2,2)--(4,1), red); [/asy][/asy] Is I understand it wrong? Thanks, I edited my solution

Lleeya wrote: Thanks, I edited my solution It's wrong, too. Your proposition not even work for $k=3$, $l=1$ because of contradiction I've commented in #5. Moreover, it can be proved that your method of covering outermost cell of board by dominoes and cover inner $(n-2) \times (m-2)$ doesn't work here. If you trying to cover $2k \times (2k-4)$ for large $k$ by your induction, in final step, you must cover $6 \times 2$ in last. But whenever you cover inner $6 \times 2$ with dominoes and draw diagonals, some two of diagonals must met. For example, in $8 \times 4$ board, $2 \times 4$ board covered with four $2 \times 1$ domino(horizontal domino) cannot appear. This is because whenever you draw diagonals, the dominoes neighbor(sharing one of edges) to this $2 \times 4$ board cannot draw diagonal that don't have same endpoint. With this fact, you can prove when you use induction inner $6 \times 2$ board must covered by two $1 \times 2$ domino in leftmost and rightmost, and $4$ $2 \times 1$ dominoes in middle part(else, contradiction occurs in any way. This is just messy casework, so I'll left it for you.). This is contradiction as in #5.