Consider a $n\times n$ checkerboard. Each cell can be colored white or black. $n$ is even. How many configurations are there so that each row and each column have an odd number of white cells?
[Math] Checkerboard coloring problem
combinatoricsmatrices
Related Solutions
Here is a solution for $n=12$, that you can adapt for any (odd or even) $n>4$. Draw a checkerboard, but make sure the four inner corners stay white.
Let $X$ be the set of rows, $Y$ the set of columns: $|X|=m\lt n=|Y|$.
Let $E\subseteq X\times Y$ be the set of red cells.
For $x\in X$ let $d(x)=|\{y\in Y:(x,y)\in E\}|$, the number of red cells in row $x$.
For $y\in Y$ let $d(y)=|\{x\in X:(x,y)\in E\}|$, the number of red cells in column $y$.
Let $X_0=\{x\in X:d(x)\ne0\}\subseteq X$ and $Y_0=\{y\in Y:d(y)\ne0\}=Y$.
I have to prove that $d(x)\gt d(y)$ for some $(x,y)\in E$. Assume for a contradiction that $d(x)\le d(y)$ whenever $(x,y)\in E$; in other words, $\frac1{d(y)}\le\frac1{d(x)}$ whenever $(x,y)\in E$. (Note that $d(x),d(y)\gt0$ when $(x,y)\in E$.) Then $$\sum_{(x,y)\in E}\frac1{d(y)}\le\sum_{(x,y)\in E}\frac1{d(x)}.$$ Since $$\sum_{(x,y)\in E}\frac1{d(x)}=\sum_{x\in X_0}\frac{d(x)}{d(x)}=|X_0|\le|X|=m$$ and $$\sum_{(x,y)\in E}\frac1{d(y)}=\sum_{y\in Y_0}\frac{d(y)}{d(y)}=|Y_0|=|Y|=n$$ it follows that $n\le m$, contradiction the assumption that $m\lt n$.
Remark. This puzzle is a lemma from matching theory in disguise. Namely, this is how you show that, if $G$ is a bipartite graph with partite sets $X,Y$, if there are no isolated vertices in $Y$, and if $\deg(x)\le\deg(y)$ whenever $x\in X$, $y\in Y$, and $(x,y)\in E(G)$, then thee is a matching in $G$ which covers $Y$.
Best Answer
Fill the $(n-1)\times(n-1)$ board arbitrary with the black and white.now you should just set the parity with the last row and column. the last cell($a_{n,n}$) will be same color for both last row and last column because of the parity of the $n-1$ first row is equal to parity of the $n-1$ first column and that is because your board is $n\times n$ and $n\equiv n \bmod{2}$ and if rows parity differ from column parity it is contradiction.
Finally the total answer is $2^{(n-1)\times (n-1)} $.
Update I asked generalized version of this question before.