Prove that a circle exists for which at least $n \in \mathbb{N}$ points have integer coordinates This question is based on the fourth question in the 2003 edition of the Flemish Mathematics Olympiad.

Consider a grid of points with integer coordinates. If one chooses the number $R$ appropriately, the circle with center $(0, 0)$ and radius $R$ crosses a number of grid points. A circle with radius $1$ crosses 4 grid points, a circle with radius $2\sqrt{2}$ crosses 4 grid points and a circle with radius 5 crosses 12 grid points. Prove that for any $n \in \mathbb{N}$, a number $R$ exists for which the circle with center $(0, 0)$ and radius $R$ crosses at least $n$ grid points.

                        
I have tried to solve this question by induction, considering a given point $(i, j)$, $i \gt j$, on the circle with radius $R$ and attempting to extract multiple points from this on a larger circle. In this case, the coordinates $(i+j,i-j)$ and $(i-j, i+j)$ are both on a circle with radius $\sqrt{2}R$. However, since $(j, i)$ is also a point on the circle, the number of crossed grid points remains the same. What is a correct way to prove the above statement?
 A: Following arguments are not mathematically rigorous, but I think it will explain the main idea.
The solution hinges on the fact that there are infinitely many values of $\phi \in [0, 2\pi)$ such that $\cos{\phi}$ and $\sin{\phi}$ are both rational. (This  in turn depends on the fact that there are infinitely many Primitive Pythagorean Triples. A Pythagorean Triple $(a,b,c)$ is primitive if all three numbers are pairwise coprime).
For every Pythagorean Triple $(a,b,c)$, the point $(\frac{a}{c}, \frac{b}{c})$ lies on the unit circle and both co-ordinates are rational.
Given any $n \in \mathbb{N}$, choose at least $n$ primitive Pythagorean Triples $(a_1, b_1, c_1), (a_2, b_2, c_2), \dots, (a_n, b_n, c_n)$ such that the hypotenuse lengths $c_1, c_2, \dots c_n$ are all pairwise coprime. (You will need to prove that such a choice is possible). Then let $R = lcm(c_1, c_2, \dots c_n)$. This circle will contain at least $n$ grid points.
(In fact this circle will contain at least $4n$ grid points, so this is a huge overestimate. But the question asks for at least $n$ grid points).
Edit 1
It's not necessary to choose $n$ primitive triples such that $gcd(c_i,c_j) = 1$. Any $n$ primitive triples will suffice. 
A: Let $a_1 + b_1 i, \ldots, a_n + b_n i$ be distinct irreducibles of the Gaussian integers $\mathbb{Z}[i]$, with $a_k > b_k > 0$ for each $k$.  Then $a_1 + b_1 i, a_1 - b_1 i, \ldots, a_n + b_n i, a_n - b_n i$ are also distinct irreducibles such that no pair has ratio equal to a unit of $\mathbb{Z}[i]$.  Therefore, since $\mathbb{Z}[i]$ is a UFD, the $4 \cdot 2^n$ numbers $(\pm 1, \pm i) (a_1 \pm b_1 i) \cdots (a_n \pm b_n i)$ are distinct.  Each of them has $|z|^2 = (a_1^2 + b_1^2) \cdots (a_n^2 + b_n^2)$; therefore, the circle with radius $R = \sqrt{(a_1^2 + b_1^2) \cdots (a_n^2 + b_n^2)}$ has at least $4 \cdot 2^n$ integer points.
(This could easily be modified; for example, by taking $(\pm 1, \pm i) \cdot (2+i)^k (2-i)^{n-k}$, you get that the circle with radius $R = 5^{n/2}$ has at least $4(n+1)$ integer points.)
