Given two positive integers n and m and a function f:Z×Z→{0,1} with the property that
\begin{align*}
f\left(i, j\right) = f\left(i+n, j\right) = f\left(i, j+m\right) \qquad \text{for all } \left(i, j\right) \in \mathbb{Z} \times \mathbb{Z} .
\end{align*}
Let [k]={1,2,…,k} for each positive integer k.
Let a be the number of all (i,j)∈[n]×[m] satisfying
\begin{align*}
f\left(i, j\right) = f\left(i+1, j\right) = f\left(i, j+1\right) .
\end{align*}
Let b be the number of all (i,j)∈[n]×[m] satisfying
\begin{align*}
f\left(i, j\right) = f\left(i-1, j\right) = f\left(i, j-1\right) .
\end{align*}
Prove that a=b. combinatoricsEnumerative Combinatoricsfunction