Suppose that $n(r)$ denotes the number of integer points on a circle of radius$>1$...

Question

Suppose that $n(r)$ denotes the number of points with integer co-ordinates on a circle of radius $r>1$.

Prove that, $n(r)<2\pi r^{2/3}$

I could not get much help from a similar question, though a much weaker inequality $n(r)<6{\pi^{1/3} r^{2/3}}$ from Iranian Mathematical Olympiad 1999.

Answer 1

Because $r>1$ and $2\pi>6$, we may assume $n\ge7$

Label the $n$ lattice points $P_1,P_2,\ldots,P_n$.The (counterclockwise)arcs $P_1P_3,P_2P_4,\ldots,P_nP_2$ cover the circle twice so they sum upto $4\pi$. Therefore, one of them, say arc $P_1P_3$ measures at most $\frac{4\pi}{n}$.

Consider the triangle $P_1P_2P_3$ which is inscribed in an arc of measure $\frac{4\pi}{n}$. Because $n\ge7$,the arc is less than a quarter of the circle. The area of $P_1P_2P_3$ will be maximized if $P_1$ and $P_3$ are the endpoints and $P_2$ is the midpoint of the arc.In that case,

$\Delta(P_1P_2P_3)=\dfrac{abc}{4r}=\dfrac{2r\sin\frac{\pi}{n}2r\sin\frac{\pi}{n}2r\sin\frac{2\pi}{n}}{4r}\le\dfrac{2r\frac{\pi}{n}2r\frac{\pi}{n}2r\frac{2\pi}{n}}{4r}=\dfrac{4r^2\pi^3}{n^3}$

And in general, the area of $P_1P_2P_3$ cannot exceed $\dfrac{4r^2\pi^3}{n^3}$.On the other hand, if the co-ordinates of the point $P_1,P_2,P_3$ are respectively $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ then $\Delta(P_1P_2P_3)=\dfrac12|x_1y_2-x_2y_1+x_2y_3-x_3y_2+x_3y_1-x_1y_3|$
Because the co-ordinates of the points are integers, the area cannot be less than $\dfrac12$.

We obtain the inequality $\dfrac12\le\dfrac{4r^2\pi^3}{n^3}$
Which proves $2\pi r^{2/3}\ge n$

Answer 2

There is a well-known result (more-or-less equivalent to the fact that $\mathbb{Z}[i]$ is an euclidean domain) that states that the number of $(x,y)\in\mathbb{Z}^2$ such that $x^2+y^2=n$, say $r_2(n)$, is equal to four times a multiplicative function of $n$, for instance: $$ r_2(n) = 4\cdot(\chi_4*1)(n) = 4\cdot(d_1(n)-d_3(n)), $$ where $$d_1(n)=|\{d\in\mathbb{N}: d|n, d\equiv 1\!\!\!\pmod{4}\}|,$$ $$d_3(n)=|\{d\in\mathbb{N}: d|n, d\equiv 3\!\!\!\pmod{4}\}|.$$ So the number of integer points on the circle $x^2+y^2=n^2$ is bounded by: $$ r_2(n) \leq 4\cdot\!\!\!\!\!\!\!\! \prod_{\substack{p|n\\p\equiv 1\!\!\pmod{\!\!4}}}\!\!\!\!\!(1+2\,\nu_p(n))\leq 4\cdot d(n^2),$$ where $\nu_p(n)$ is the maximum $h\in\mathbb{N}$ such that $p^h|n$, and equality is attained when all the prime divisors of $n$ belong to the arithmetic progression $4k+1$. Following the spirit of [https://terrytao.wordpress.com/2008/09/23/the-divisor-bound/1, consider now that for every prime $p\geq 23>e^3$ and every $\alpha\geq 0$ $$\frac{2\alpha+1}{p^{2\alpha/3}}\leq 1$$ holds in virtue of the Bernoulli inequality, while: $$\frac{2\alpha+1}{5^{2\alpha/3}}\leq\frac{3}{5^{2/3}},\qquad \frac{2\alpha+1}{13^{2\alpha/3}}\leq 1,\qquad \frac{2\alpha+1}{17^{2\alpha/3}}\leq 1$$ hold for any $\alpha\in\mathbb{N}$. This trivially gives: $$ r_{2}(n)\leq \frac{12}{5^{2/3}} n^{2/3} < \frac{37}{9} n^{2/3}, $$ or $$ n(r) < \left(4+\frac{1}{9}\right) r^{2/3} $$ in your notation.