Subfield: Number Theory


Question: For all positive integers $n, k$, let $f(n, 2 k)$ be the number of ways an $n \times 2 k$ board can be fully covered by $n k$ dominoes of size $2 \times 1$. (For example, $f(2,2)=2$ and $f(3,2)=3$.)

Find all positive integers $n$ such that for every positive integer $k$, the number $f(n, 2 k)$ is odd.

Solution:
The integers $n$ with the requested property are exactly the ones of the form $2^{k}-1$.

In what follows, let $f(m, n)$ denote the number of domino tilings of an $m \times n$ grid. (For convenience, we also allow $m$ and $n$ to be 0 , in which case $f(m, n)=1$.)

Claim. $f(m, 2 n+1) \equiv f(m, n)(\bmod 2)$ for all $n$ and even $m$

Proof. Consider reflecting $m \times(2 n+1)$ tilings across the central column. In this way, the tilings are grouped into singletons and pairs, so modulo $2, f(m, 2 n+1)$ is congruent to the number of singletons.

If a tiling is invariant under reflection in the central column, then every domino with one cell in that column must have both cells in that column.

In other words, the central column is filled with $\frac{m}{2}$ vertical dominoes, splitting the remainder of the grid into two $m \times n$ grids. Obeying the symmetry constraint, there are $f(m, n)$ tilings: each of the tilings of the left grid fixes the tiling of the right grid.

Claim. $f(n, n) \equiv 0(\bmod 2)$ for all even $n \geq 2$. (Recall that $f(0,0)=1$ is odd.)

Proof. Consider reflecting $n \times n$ tilings across the diagonal. This groups the tilings into pairs, (no tiling is grouped with itself). Hence the number of $n \times n$ tilings is even.

We are ready to complete the solution.

- If $n$ is odd, the first claim shows that $n$ satisfies the property if and only if $\frac{1}{2}(n-1)$ does.
- If $n \geq 2$ is even, the second claim shows that $n$ does not satisfy the property.
- If $n=0$, then $n$ satisfies the property, as $f(m, 0)=1$ always.

This concludes the proof that the sought numbers are the ones of the form $2^{k}-1$.
Color the board as a chessboard. Consider the bipartite graph whose vertices are the squares and the neighbors are connected by an edge. Notice that a domino tiling described in the problem corresponds to a perfect matching in this bipartite graph, so we are interested in the number of perfect matchings.

And that is the permanent of the (bipartite) adjacency matrix. However we need to compute it $\bmod 2\left(\right.$ over $\left.F_{2}\right)$, so that is the determinant of that matrix.

So our goal is to decide whether the determinant is 0 or $1 \bmod 2$, or in other world, that matrix is singular or not. From now we compute everything mod 2. Now we understand what does it mean, that a vector $v$ is in the nullspace (has eigenvalue 0 ). But each entry of $v$ is $0-1$, so in other words we choose a subset of the black squares ( $v$ is the characteristic vector of that subset). So in the language of the board, there is a nontrivial $v$ in the nullspace iff we have a subspace of black square such a way that each white square has even number of black chosen neighbors.

If $n$ is even, then we can choose a subset such a way in the $n \times n$ square: the diagonal.

Now we prove similar connection between $n$ and $2 n+1$ : there is a construction for $n$ iff we have one for $2 n+1 . k$ is always even. If we have a construction in $n \times k$ then we have one in $(2 n+1) \times k$ : we mirror the construction to the middle row.

Now we prove that from a construction in $(2 n+1) \times k$ we can make a construction in $n \times k$. First, we don't have any chosen black square in the middle row. Then the chosen black squares in the upper (or lower) $n \times k$ rectangle is a construction for $n \times k$. Second, we have some chosen black square in the middle row. Because we can find a white square in the middle row such that is has only one chosen black neighbor in the middle row. Thus this construction isn't symmetric to the horizontal line the symmetric difference of the construction and its reflection (to the vertical line) is a nontrivial vector in the nullspace, thus we can use the first case.

Final answer: the ones of the form $2^k-1$