Prove that $n(r) < 2\pi \sqrt[3]{r^{2}}$ 
Suppose that $n(r)$ denotes the numbers of points with integer coordinates on a circle of radius $r > 1$. Prove that

$$
n(r) < 2\pi \sqrt[3]{r^{2}}
$$
What process would you use to resolve the last one?
 A: We have:
$$ n(r) = 4\left(\chi_4* 1\right)(r^2) = 4\sum_{d\mid r^2}\chi_4(d)\leq 4 \,d(r^2) $$
where $\chi_4(d)$ equals $1$ if $d\equiv 1\pmod{4}$, $-1$ if $d\equiv -1\pmod{4}$ and zero otherwise.
The claim hence follows from the divisor bound (thanks to Terence Tao) for $\varepsilon=\frac{1}{3}$.
A: Label the $n(r)$ points $P_1, P_2, \ldots, P_n$ counterclockwise. A general strategy with problems regarding lattice points is to somehow use the area (using the shoelace formula with integer coordinates gives you good bounds). In this case, looking at any triangle the area will be at least $1/2.$ Now we can in particular pick a triangle with "small" area. Restricting our attention to triangles $P_i P_{i+1} P_{i+2}$ and realizing that this area effectively depends on the angle $P_i P_{i+1} P_{i+2},$ and hence the arc $P_i P_{i+2},$ we relabel so that $P_1OP_3$ has angle at most $\frac{4\pi}{n}.$
Now $P_1 P_2 P_3$ has area at most $\frac{1}{2} P_1P_2\cdot P_2P_3\sin \frac{2\pi}{n} \le \frac{\pi}{n} P_1P_2 \cdot P_2P_3.$ But $P_1 P_2 = 2r\sin \alpha,$ where $\alpha$ is half the measure of the arc $P_1 P_2$ (divide by $r$ of course) and $P_2 P_3 = 2r \sin \beta,$ with $\beta$ defined similarly. Notice that $\alpha + \beta$ is at most $\frac{2\pi}{n}$ and thus $\sin \alpha \cdot \sin\beta = \frac{1}{2} \cos(\alpha-\beta) - \frac{1}{2}\cos(\alpha + \beta)\le \frac{1}{2} - \cos\frac{2\pi}{n},$ with equality when $\alpha = \beta = \frac{\pi}{n}.$
Consequently $\frac{1}{2}\le [\triangle P_1 P_2 P_3] \le \frac{4\pi r^2}{n}\sin \alpha \sin\beta \le \frac{4\pi^3 r^2}{n^3},$ whence $n \le 2\pi \sqrt[3]{r^2}.$ 
Strict inequality isn't difficult, if you really want it. Question for you: where was the assumption $r>1$ used?
