Mongolian Mathematical Olympiad

By Chinese remainder theorem the system $$\{\begin{array}{l}n\equiv p_0(\mathrm{mod}{p}_0^2)\\ n+1\equiv p_1(\mathrm{mod}{p}_1^2)\\ \text{................}\\ n+k\equiv p_k(\mathrm{mod}{p}_k^2)\end{array}$$ 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 $x^2+y^2\equiv p_t(\mathrm{mod}{p}_t^2)$ and $x^2+y^2\equiv 0(\mathrm{mod}{p}_t)$, $x^2+y^2\not\equiv 0(\mathrm{mod}{p}_t^2)$, $p_t=4l+3$. It implies $x^2\equiv -y^2(\mathrm{mod}{p}_t)\Rightarrow x\equiv y\equiv 0(\mathrm{mod}{p}_t)$ 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.