Let the number of interior $k$-points be $I={\mathrm{Ak}}^2-\mathrm{Bk}+C$, and the number of $k$-points in the closed rectangle be $R={\mathrm{Ak}}^2+\mathrm{Bk}+C$.

When $k=1$

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{I=A-B+C}& =2\times 1=2\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(1)\\ \multicolumn{1}{c}{R=A+B+C}& =2+10=12\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(2)\end{array}}$

Subtracting (2)-(1) gives $2B=10,B=5$.
Substituting into the original equations

${\displaystyle A+C=7\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(1\text{'}).}$

When $k=2$

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{I=4A-2B+C}& =5\times 3=15\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(3)\\ \multicolumn{1}{c}{R=4A+2B+C}& =15+20=35\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(4)\end{array}}$

Subtracting (4)-(3) gives $4B=20,B=5$, (agreeing with the above).
Substituting into the original equations

${\displaystyle 4A+C=25\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(3\text{'}).}$

Subtracting (3') - (1') gives $3A=18,A=6$ and $C=7-A=1$.

To verify these values for $A$, $B$ and $C$, we will look at the case for $k=3$.

When $k=3$

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{I=9A-3B+C}& =8\times 5=40\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(5)\\ \multicolumn{1}{c}{R=9A+3B+C}& =40+30=70\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(6)\end{array}}$

Subtracting (6)-(5) gives 6B=30, B=5, (agreeing with the above). Substituting $A=6$, $B=5$, $C=1$ into equation (5):

${\displaystyle 9\times 6-3\times 5+1=54-15+1=40}$

and therefore these values of $A$, $B$ and $C$ hold: $A=6,\hspace{0.5em}B=5,\hspace{0.5em}C=1$

In order to find a connection between $A$, $B$, $C$, the area of the rectangle and the number of $k$-points we can investigate for a rectangle of dimension $p\times q$. The number of interior $k$-points $I={\mathrm{Ak}}^2-\mathrm{Bk}+C=(\mathrm{pk}-1)(\mathrm{qk}-1)$. This is because there are $\mathrm{pk}-1$ points on each row of interior $k$-points and $\mathrm{qk}-1$ points in each column of interior $k$-points. Therefore the total number of interior $k$-points would be $(\mathrm{pk}-1)(\mathrm{qk}-1)$. So ${\displaystyle {\mathrm{Ak}}^2-\mathrm{Bk}+C={\mathrm{pqk}}^2-(p+q)k+1.}$

Comparing coefficients we find that

$A=\mathrm{pq}$ which is the area of the rectangle,

$B=p+q$ which is half the perimeter of the rectangle

$C=1$ (a constant).

To explain why, for large $k$, the area of any polygon with ${\mathrm{Ak}}^2-\mathrm{Bk}+C$ interior $k$-points, is given by

${\displaystyle \underset{k\to \infty }{lim}\frac{\mathrm{number}\mathrm{of}\mathrm{interior}k-\mathrm{points}}{k^2}}$

consider the interior of the polygon divided into a very large number of tiny squares of area $\frac{1}{k}\times \frac{1}{k}.$ The number of these small squares is roughly the same as the number of interior $k$-points, that is ${\mathrm{Ak}}^2-\mathrm{Bk}+C.$ (This will not be exact, some of these squares on the boundary of the polygon will be partly inside and partly outside the polygon but as $k$ increases so the approximation gets better and better.)

${\displaystyle \mathrm{Area}\mathrm{of}\mathrm{polygon}=\underset{k\to \infty }{lim}[\mathrm{number}\mathrm{of}\mathrm{interior}k-\mathrm{points}\times \frac{1}{k^2}]=\underset{k\to \infty }{lim}\left(\frac{{\mathrm{Ak}}^2+\mathrm{Bk}+C}{k^2}\right)=\underset{k\to \infty }{lim}(A+\frac{B}{k}+\frac{C}{k^2})=A.}$

(This is because as $k$ approaches infinity, $\frac{B}{k}$ and $\frac{C}{k^2}$ both approach zero.)

So $A$ is equal to the area of the polygon as we found for the rectangle in this example.