A good *guess* would be that for square numbers $n^2$ of white squares with $n \geq 2$, the minimal number of black squares is $4 \cdot n$. All of this is postulate; you'd have to prove that this is the optimal method. Between squares, replace the black border squares with white squares and complete the border in this order: Sample replacement for $4 \text{x} 4$ square 1. Start with replacing the green squares with white squares; this doesn't change the amount of black squares you need to complete the boundary. 2. Replace the orange squares; the first one adds one black square to the boundary (indicated in red), the second one doesn't. 3. Replace the blue squares; both add one black square to the boundary. 4. Replace the red square; this adds one black square to the boundary. So by 'extending' the boundary from a $n \times n$ square to a $n + 1 \times n + 1$ square, you're up net $4$ black squares, which fits with the formula I postulated. You can derive a formula from the four steps.

if you move green cells you increase the number of white cells but fixes the number of black cells, so there can be more than $n^2$ white with 4*$n$ black

Solution(Part 1): Answer(s): For $n=4t\rightarrow 2t^2-2t+1$, $n=4t+1\rightarrow 2t^2-t$, $n=4t+2\rightarrow 2t^2$, $n=4t+3\rightarrow 2t^2+2t$ The first claim is the configuration can not have three consecutive black squares(horizontal or vertical), otherwise we can achieve a configuration with more white squares. It can be proven by the Figure $1$. The second claim is the polygon obtained by the union of the center(s) of the black squares must bound a convex region, otherwise we can achieve a configuration with more white squares. It can be seen by the Figure $2$. The third claim is the sides of the polygon can have four directions(horizontal, vertical, two diagonal(s)), then the polygon will have at most $8$ sides. Let $a,b,c,d$ be the orange sides $AB,CD,EF,GH$ by the Figure $3$, respectively.

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Solution(Part 2:) By the claims above, none of the numbers $a,b,c,d$ can be greater than $2$. If $a=c=2$, it is possible find a configuration with more white squares by the Figure $4$(The blue path has more white squares). So in the best scenario the polygon does not have two opposite sides of size $2$. The fouth claim is we can not have three of the four numbers $a,b,c,d$ be equal to $1$, and the remaining one equal to $2$. The quantity of white squares in each line(horizontal or vertical) have the same parity(two consecutive differ by two or have the same amount). Hence, there are two remaining cases: Case 1: $a=b=c=d=1$. We will have a parallelogram by the Figure $5$. The quantity of black squares is $n=4+2(x+y)$ (I), it is only possible for $n$ even. The quantity of white squares is $xy+(x+1)(y+1)=2xy+x+y+1$ (II), replacing (I) in (II), we have $2x(\frac{n-4}{2}-x)+x+\frac{n-4}{2}-x+1=-2x^2+(n-4)x+\frac{n-4}{2}+1$ this quadratic function has the maximum point in $x=\frac{n-4}{4}$. If $n=4k$, then $\frac{n-4}{4}=k-1$ and the number of white squares is $2k^2-2k+1$. If $n=4k+2$, then $\frac{n-4}{4}=k-\frac{1}{2}$, is not an integer, so looking to the nearest integer(s) $\{k,k+1\}$ the number of white squares is $2k^2$. Case 2: Wlog $a=d=2$ and $b=c=1$. The number of white squares surrounded is $x+1+(x+1)y+(x+2)(y+1)=(2x+3)(y+1)$ and $n=6+2x+1+2y=7+2(x+y)$ so $n$ is odd. Again, looking to the quadratic function $(2x+3)(\frac{n-7}{2}-x+1)=-2(x+\frac{3}{2})(x-\frac{n-5}{2})$ is a quadratic function with maximum point in $x=\frac{n-8}{4}$. if $n=4k+1\rightarrow \frac{n-8}{4}=k-\frac{7}{4}$, so looking to the nearest integers $\{k-2,k-1\}$ the number of white squares is $2k^2-k$. If $n=4k+3\rightarrow \frac{n-8}{4}=k-\frac{5}{4}$, so looking to the nearest integers $\{k-2,k-1\}$ the number of white squares is $2k^2+k$.

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