Combinatorial Problem about putting foxes in a $n\times n$ table Let $n$ be an integer with $n\geq 2$. 
$k$ foxes are put into $n \times n$ table, and each $1 \times 1$ square has at most $1$ fox. They are put in such a way that each $2 \times 2$ table has exactly $2$ foxes. For what values of $k$, in respect to $n$, could that happen?
My intuition told me it is done with induction, but then I realized that for even $n$ we split the table into $n^2/4$ squares that are each $2 \times 2$, so turns out $k=n^2/2$ for even $n$. Right now I'm trying to prove for odd $n$.
 A: You are correct that the minimum and maximum $k$ are the same for even $n$; the only value of $k$ possible is $$k = \frac{n^2}{2}$$

Hint 1: For odd $n$, use the fact that there are four $n-1 \times n-1$ squares in an $n \times n$ square, along with the fact that $n-1$ is even, to show that the minimum $k$ is $$\frac{n(n-1)}{2}$$
This corresponds to an arrangement of the foxes in a "striped" pattern (every other row is all foxes, with the start and end rows being empty).
Similarly, you can show that the maximum $k$ is $$\frac{n(n+1)}{2}$$
This also corresponds to an arrangement of the foxes in a "striped" pattern, but this time the start and end rows are foxes. 

This is only a hint because it remains to show which values of $k$ that are strictly between the minimum and the maximum are possible. Two examples of possible values are 
$$\frac{n^2-1}{2} \qquad \text{and} \qquad \frac{n^2+1}{2}$$
These correspond to "checkerboard" arrangements of foxes. 
In order to find the total number of possible values of $k$ for odd $n$, notice that two foxes next to each other in the same row determine the placement of foxes in both of their columns. Similarly, two foxes next to each other in the same column determine the placement of foxes in both of their rows. 
This will eliminate many arrangements. For example, it is impossible to have an arrangement in which there is a pair of foxes vertically adjacent and another pair of foxes horizontally adjacent. 
