Congratulations Tiffany from Island School
on your very clear and well explained solution.
Let the number of interior $k$points be $I=Ak^2Bk+C$, and the
number of $k$points in the closed rectangle be
$R=Ak^2+Bk+C$.
When $k=1$
\begin{eqnarray} I=AB+C &= 2\times 1 = 2 \quad (1)
\\ R=A+B+C &= 2 + 10 =12 \quad (2) \end{eqnarray} Subtracting
(2)(1) gives $2B=10, B=5$.
Substituting into the original equations $$A+C=7 \quad (1').$$
When $k=2$
\begin{eqnarray} I=4A2B+C &= 5\times 3 = 15 \quad (3) \\
R=4A+2B+C &= 15 + 20 = 35 \quad (4) \end{eqnarray} Subtracting
(4)(3) gives $4B=20, B=5$, (agreeing with the above).
Substituting into the original equations $$4A+C=25 \quad (3').$$
Subtracting (3')  (1') gives $3A=18, A=6$ and $C= 7A = 1$.
To verify these values for $A$, $B$ and $C$, we will look at the
case for $k=3$.
When $k=3$
\begin{eqnarray} I=9A3B+C &= 8\times 5 = 40 \quad (5) \\
R=9A+3B+C &= 40 + 30 = 70 \quad (6) \end{eqnarray} Subtracting
(6)(5) gives 6B=30, B=5, (agreeing with the above). Substituting
$A=6$, $B=5$, $C=1$ into equation (5): $$9\times 6 3\times 5 + 1 =
5415+1 = 40$$ and therefore these values of $A$, $B$ and $C$ hold:
$A=6,\ B=5,\
C=1$

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 $I=Ak^2Bk+C =(pk1)(qk1)$.
This is because there are $pk1$ points on each row of interior
$k$points and $qk1$ points in each column of interior $k$points.
Therefore the total number of interior $k$points would be
$(pk1)(qk1)$. So $$Ak^2Bk+C = pqk^2  (p+q)k +1.$$

Comparing coefficients we find that
$A=pq$ 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
$Ak^2Bk+C$ interior $k$points, is given by $$\lim_{k\to \infty}
{{\rm number of interior kpoints}\over k^2}$$ consider the
interior of the polygon divided into a very large number of tiny
squares of area ${1\over k}\times {1\over k}.$ The number of these
small squares is roughly the same as the number of interior
$k$points, that is $Ak^2Bk+C.$ (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.) $${\rm Area of polygon} =
\lim_{k\to \infty}[{\rm number of interior kpoints} \times {1\over
k^2}]= \lim_{k\to \infty}\left({Ak^2+Bk+C\over k^2}\right) =
\lim_{k\to \infty}\left(A + {B\over k} + {C\over k^2}\right) = A.$$
(This is because as $k$ approaches infinity, ${B\over k}$ and
${C\over k^2}$ both approach zero.)
So $A$ is equal to the area of the polygon as we found for the
rectangle in this example.