Circles in an infinite chessboard Is it true that for every natural number we can draw a circle, in an infinite chessboard, that intersect exactly that number of little squares in their interior?
 A: The answer is "yes". We can pick a single origin $O$ in $\mathbb{R}^2$ and define $n(r)$ to be the number of squares covered by the circle with centre $O$ and radius $r > 0$. Then $n(r)$ is an increasing function of $r$ and it takes all positive integer values if there is never an $r$ where the expanding circle crosses two new grid-line segments at once. But this is the case if $O$ is not equidistant from any two grid lines, or any two vertices, or any vertex and any grid line.  
The locus equidistant from two parallel lines is a line (parallel to them both). The locus equidistant from two lines at right-angles is the pair of lines at 45 degrees to them both. The locus equidistant from any two vertices is also a line, the perpendicular bisector. The locus equidistant from a point and a line is a parabola.
There are only countably many pairs of grid lines, so the set we must avoid is the union of countably many lines and parabolas. But this is a set of Lebesgue measure $0$ in $\mathbb{R}^2$, so its complement is dense everywhere. Hence almost any choice of origin $O$ is the centre of a family of circles with the desired property.
