Mongolian Mathematical Olympiad

By Chinese remainder theorem the system
$$\left\{ \begin{array}{l} n \equiv p_0 \pmod{p_0^2} \\ n + 1 \equiv p_1 \pmod{p_1^2} \\ \text{................} \\ n + k \equiv p_k \pmod{p_k^2} \end{array} \right.$$
has $n$ positive integer solutions for $\forall k \in \mathbb{N}$, where $p_0, p_1, \dots, p_k$
are different primes of the form $4t + 3$. From this follows that there is no
lattice point in the ring $\{(x, y) \in \mathbb{R}^2 \mid n \le x^2 + y^2 \le n + k\}$. Proceeding
by contradiction, if suppose that $x^2 + y^2 = n + t$, $0 \le t \le k$, $x, y \in \mathbb{Z}$ then
$x^2 + y^2 \equiv p_t \pmod{p_t^2}$ and $x^2 + y^2 \equiv 0 \pmod{p_t}$, $x^2 + y^2 \not\equiv 0 \pmod{p_t^2}$, $p_t = 4l + 3$. It
implies $x^2 \equiv -y^2 \pmod{p_t} \Rightarrow x \equiv y \equiv 0 \pmod{p_t}$ and we get $p_t^2 \mid x^2 + y^2$. This is
a contradiction. If we take $k > \frac{l^2}{2}$ then one can locate the needle in the ring
$\{(x, y) \in \mathbb{R}^2 \mid n \le x^2 + y^2 \le n + k\}$ and it is possible to rotate the needle
inside the ring not passing a lattice point.