Proving that a circle will contain n lattice points?

Question

A lattice point is a point $(x, y)$ in the plane, both of whose coordinates are integers.It is easy to see that every lattice point can be surrounded by a small circle which excludes all other lattice points from its interior. It is not much harder to see that it is possible to draw a circle that has exactly two lattice points in its interior, or exactly 3, or exactly 4. How to prove that it satisfies for n $\forall n\in\mathbb{N}$? I have tried to generalize it. If I take a the centre on the line segment joining 2 latice points ( not the diagonal). Then by increasing the radius I can draw with all the n which are even by keeping it internally within the circle. If I take the centre on the diagonal of 2 lattice points then we can easily draw for all the odd numbers. If i take x as the radius using greatest integer function we can find the no. of points in 1 quadrant. In this way I have tried it but is there a more clear solution to this problem.

Answer 1

For example you can take point $(\frac13, \pi)$ as a centre. Then distances to all lattice points from it are pairwise distinct.

Suppose this is not true. Then there are $x_1, y_1, x_2, y_2 \in \mathbb{Z}$ such that $x_1 \ne x_2 \lor y_1 \ne y_2$ and $$(x_1 - \frac13)^2 + (y_1 - \pi)^2 = (x_2 - \frac13)^2 + (y_2 - \pi)^2.$$

If $y_1 = y_2$ then $(x_1 - \frac13)^2 = (x_2 - \frac13)^2 \land x_1 \ne x_2$. That is either $x_1 - \frac13 = x_2 - \frac13$ or $x_1 - \frac13 = -x_2 + \frac13$. The first equation leads to $x_1 = x_2$ and the second one leads to that at least one of $x_1$ and $x_2$ is not integer.

So $y_1 \ne y_2$. Then $\pi = \frac{x_1^2 + y_1^2 - x_2^2 - y_2^2 - \frac23 x_1 + \frac23 x_2}{2(y_2 - y_1)}$ that is not true because $\pi$ is not rational. This contradiction finishs proof.

Now it is easy to see that increasing radius or circle with centre in $(\frac13, \pi)$ we will cover lattice points one by one.