ways to color black a $8*8$ square that all of the $1*1$ square that are in the same line or column with a $1*1$ black should be colored? ways  to color black  a $8*8$ square that  all of the $1*1$ square that are in the same line or column with a $1*1$ black should be colored?
My attempt:Except the all $8$ lines colored and non of the lines colored we can color $1,2,3,\dots ,7$ columns which gives us the answer:
$(\binom{8}{1}+\binom{8}{2}+\dots+\binom{8}{7})(\binom{8}{1}+\binom{8}{2}+\dots+\binom{8}{7})+2=(2^8-2)(2^8-2)+2=64518$
But the book gives the answer $65026$ which is equal to $(2^8-1)(2^8-1)+1$.
Where did I make a mistake?
 A: I'm going to make an educated guess that the question is the following: How many ways are there to blacken some squares on an $8\times8$ board, in such a way that if a square has a black square in its row and a black square in its column then it too is black? We quickly see that this is equivalent to: How many subsets of $\{1,\dots,8\}\times\{1,\dots,8\}$ are of the form $A\times B$ for some $A,B\subseteq \{1,\dots,8\}$?
The intended solution would then be: if either $A$ or $B$ is empty, then $A\times B=\emptyset$, which should only be counted once. Otherwise, distinct $A$ or distinct $B$ yields distinct $A\times B$. Since there are $2^8-1$ distinct nonempty $A$ and the same for $B$, the total number of nonempty subsets of the form $A\times B$ is $(2^8-1)(2^8-1)$, hence the total number of (possibly empty) subsets is $(2^8-1)(2^8-1)+1$.
My guess is that you mistakenly thought that the cases $A=\{1,\dots,8\}$ or $B=\{1,\dots,8\}$ were also special cases that should be combined like the empty-set case (that would yield $(2^8-2)(2^8-2)+2$). But $A=\{1,\dots,8\}$ doesn't mean that $A\times B$ is the whole board; indeed, all the $\{1,\dots,8\}\times B$ are distinct for distinct $B$.
