Prove that there exists a $100 \times 100$ square with no integer point visible from the origin. 
If $x$ and $y$ are integers, we say that the point $P=(x,y)$ in $\Bbb R^2$ can be seen from the origin $O=(0,0)$, if there is no other point on the line segment $OP$ whose coordinates are both integers. Prove that there exists a $100 \times 100$ square with no integer point visible from the origin.

I have a hypothesis that the only way line $\ell$ to a point $(x,y)$ does not contain any other points with integer coordinates is if $\gcd(x,y)=1$. I have not yet proven this, but I think it should help me with the problem.
I think I need to also use the Chinese remainder theorem somehow here, but I don't know in what way. I the hypothesis holds I would get pairs of integers with $\gcd$'s equal to $1$ which I can use in CRT?
 A: I'll leave the gcd claim to you because it's significantly easier than the rest of the problem. For the construction, I will abuse CRT.
Choose large enough primes $p_1, p_2, \cdots$. Pick $x$ such that $p_i|x$ for all $1\le i\le 100$. Pick $y$ such that $y\equiv 0\pmod{p_1}, y+1\equiv 0\pmod {p_2}, \cdots,y+99\equiv 0\pmod{p_{100}}$. Then we have that $\gcd(x,y+k)\ne 1$ for $0\le k\le 99$. Great. The trick is that we can do the exact same process again, pick the $x$ such that $p_i|x+1$ for all $101\le i\le 200$. Pick the $y$ such that $y\equiv 0\pmod{p_{101}}, \cdots, y+99\equiv 0\pmod{p_{200}}$. Choosing such $x,y$ now additionally guarantees that $\gcd(x+1, y+k)\ne 1$ for $0\le k\le 99$. Now just repeat this over and over, and you'll get the following:
Let $p_1, p_2, \cdots$ be sufficiently large primes. Pick $x$ such that $p_i|x+k$ for all $100k+1\le i\le 100k+100$ for all $0\le k\le 99$. Pick $y$ such that $y\equiv -t\pmod{p_{100r+1+t}}$ for all $0\le r\le 99$. Consider the square with vertices $(x,y), (x,y+99), (x+99, y), (x+99, y+99)$.
