Number of $A \in Mat[\{ +1 , -1\}]$ with positive row/column product. Consider matrices with $m \times n$ size from $\{+1,-1\}$. We want to determine number of matrices with positive product in each row and each column.
First observation is that there should be $2k$ of "$-1$". And the minimum case when there are only 4 of "-1"
So we should consider something like : $\displaystyle \sum_{k=0}^{(m+n)/2} P(m,n,2k)$. But how can we determine $P(m,n,2k)$ (numbers of matrices with following property and $2k$ of "$-1$")? 
 A: We shall call a $m\times n$ $\pm 1$-matrix positive, provided it has positive product in each row and each column. We claim that the number of positive matrices is $2^{(m-1)(n-1)}$. To show this it sufficiently to check that for each $(m-1)\times (n-1)$ $\pm 1$-matrix $A=(a_{ij})$ there exists a unique positive matrix $A'=(a_{ij})$ which has a submatrix $M$ in its left upper corner. The matrix $A'$ is positive iff for each $1\le i\le n$ and $1\le j\le m$ we have $a_{in}=\prod_{k=1}^{n-1} a_{ik}$ and   $a_{mj}=\prod_{k=1}^{m-1} a_{kj}$. These conditions for $1\le i\le n-1$ and $1\le j\le m-1$ uniquely determine $a_{in}$ and $a_{mj}$. The remaining two conditions are satisfied iff 
$$a_{mn}=\prod_{k=1}^{n-1} a_{mk}=\prod_{k=1}^{m-1} a_{kn}.$$
Since both thes products are equal to the product of entries of the matrix $A$, the condition uniquely determines $a_{mn}$.
A: This $4k$-claim is not true:
$$
\pmatrix{ -1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & -1 }
$$
which has six entries equal to $-1$.
