If haven't done the calculations, but can tell you how they can be done for the general problem of placing $k$ kings on a $p\times q$ board.
Let $I=\{1,\ldots,p\}$ represent a column of the board. For simplicity, assume that $p\le q$. Let $\cal{C}$ be the set of subsets of $I$ so that no pair $x,x+1\in C$ for any $C\in\cal{C}$: these sets represent permitted placements of kings in a column.
Now we define a matrix $A$ with elements $A_{ij}$ for $i,j\in\cal{C}$: you may enumerate the sets in $\cal{C}$ any way you like. We define the elements
$$
A_{ij}=\left\{\begin{array}{l l}
0&\text{if there are $x\in i$, $y\in j$ with $|x-y|\le1$}\\
x^{|j|}&\text{otherwise, where $|j|$ is the number of elements in $j$}
\end{array}\right.
$$
so $A_{ij}$ represents the addition of a column with kings in the $j$ positions after having had a column with kings in the $i$ positions.
If we let $e=\{\}\in\cal{C}$ represent the initial state, i.e. no kings to the left of the board, and successively add $q$ columns, we get
$$
e\cdot A^q\cdot 1=\sum_{k} a_kx^k
$$
where the $1$ represents the vector $(1,\ldots,1)$ and $a_k$ is the number of ways to place $k$ kings on the $p\times q$ board.
This is fairly ok to compute for arbitrary $k$ and $q$, but quickly gets tough is $p$ increases. That's why selecting $p\le q$ is an advantage if the board is rectangular but not square.
You are on the right track.
- If the white king is in a corner ($4$ possibilities), the black king can be at any of $64-4$ positions.
- If the white king is at the boundary, but not in a corner ($24$ possibilities), the black king can be at any of $64-6$ positions
- In all other cases for the white king ($36$ possibilities), the black king can be at any of $64-9$ positions.
In total:
$$4\cdot 60+24\cdot 58+36\cdot 55. $$
Alternatively:
There are $64\cdot 63$ ways to place the kings without respecting the rules (except that they are on different squares).
Subtract invalid positions: There are $7\cdot 8$ positions with the black king directly to the east of the white king; the same holds for west, north, and south.
And there are $7\cdot 7$ positins with the black king south-east of the white king; the same for south-west, north-west, north-est.
Hence:
$${64\cdot 63}-4\cdot 56-4\cdot 49. $$
Best Answer
HINT: A corner square has $3$ neighbors; an edge square that is not in a corner has $5$ neighbors; and every non-edge square has $8$ neighbors. That means that if one king is in a corner, there are $mn-4$ possible locations for the other, and if one king is on an edge but not in a corner, there are $mn-6$ possible locations for the other. If one king is on any non-edge square, how many possible locations are there for the other king?
Now imagine a graph with a vertex for each square of the board and an edge between two vertices if and only if they are not adjacent. The number of edges in this graph is the number of ways to place two non-adjacent kings on an $m\times n$ board. The $4$ vertices corresponding to the corner squares will each have degree $mn-4$: each of them will be connected by an edge to $mn-4$ other vertices. What is the sum of the degrees of the vertices? How many times does that sum count each edge? How many edges are there?
If you’re not familiar with graphs, you don’t need them, but if you are, they make the argument exceptionally easy. Without them you have to realize that if you assign to each square the number of acceptable placements with one king in that square and then sum these numbers, you’re counting every acceptable placement twice.
You can instead use the same basic approach to count pairs of adjacent squares and subtract that number from the total number of ways to put two kings on an $m\times n$ board.