There exists a lattice point $(a,b)$ whose distance from every *visible* point is greater than $n$. Question- A lattice point $(x,y)\in\mathbb{Z}^2$ is called visible if $gcd(x,y)=1$. Prove that given a positive integer $n$, there exists a lattice point $(a,b)$ whose distance from every visible point is greater than $n$.
I am totally nowhere near progress on this. I was thinking of trying pigeon hole principle, but cant find any appropriate candidate pigeons. Please give any hints to start.
 A: This follows from Theorem 7.3 in the book “Mathematical Journeys,” by Peter D. Schumer. The proof, which uses the Chinese remainder theorem, can be seen on Google Books here or (summarized) by hovering over the spoiler paragraph below.
Schumer constructs an $n\times n$ square of invisible lattice points as follows. (You can choose $(a,b)$ at the center of such a square of side $2n$.)
The proof goes like this:

 Arrange the first $n^2$ primes in an $n\times n$ array and denote by $m_i$ the product across the $i$-th row and by $M_i$ the product down the $i$-th column. Then consider two systems of congruences: The system $x\equiv -i \mod m_i$ and the system $y\equiv -i\mod M_i$. Modulo $P=\prod m_i=\prod M_i$, these systems have unique solutions $x\equiv_P a$ and $y\equiv_P b$ (by the Chinese Reminder Theorem). Then the square with lower left corner $(a+1,b+1)$ and upper right corner $(a+n,b+n)$ contains no visible lattice points, because for any integers $i,j$ between $1$ and $n$, both $a+i$ and $b+j$ will be divisible by the prime $p$ in row $i$ and column $j$ of the array of primes.

