prove that for any $(s,t)$ there is a pair $(x,y)$ that is not good 
Let $s$ and $t$ be nonzero integers and let $(x,y)$ be any ordered pair of integers. A move changes $(x,y)$ to $(x-t, y-s)$. The pair $(x,y)$ is good if after some number of moves it becomes a pair of integers that are not relatively prime. Prove that for any $s,t,$ there is a pair $(x,y)$ that is not good.

Working towards a proof by contradiction, suppose there is a pair $(s,t)$ of nonzero integers for which every pair $(x,y)$ of integers is good. Let $g = sx-ty$ for some integers $x,y,$ where $g = \gcd(s,t)$. Then $x$ and $y$ are coprime, since $1 = (s/g) x - (t/g) y$. By assumption, there is some $k$ so that after $k$ moves, $(x -kt, y-ks)$ become not relatively prime. Let $p$ be a prime divisor of the two integers $x-kt$ and $y-ks$. Then $p$ divides $s(x-kt)$ and $t(y-ks)$ so it divides $sx-yt = g.$ Hence $p$ divides $s$ and $t$. But I'm not sure how to get the contradiction from here. For instance, if $s = 4, t = 6, g = 2, x = 2, y= -1,$ then $p=2$ in this case but $p\nmid y$.
 A: Let's start with an example.  Say $(t,s)=(2,3)$.  Then I claim that $(1,1)$ works.
Pf:  We need to show that $\gcd(1-2n,1-3n)=1$ $\forall n\in \mathbb N$. But suppose it was $d$.  Then of course $d$ divides $(1-2n)-(1-3n)=n$.  But if $d$ divides $n$ then $d$ can't divide $1-2n$ so we are done.
Now we seek to generalize this.  So let $(t,s)$ be arbitrary. Suppose first that they are relatively prime.  In that case, we know from Euclid that there exist integers $(a,b)$ with $at+bs=1$.  We claim that the pair $(a, -b)$ is not good.
Pf:  We need to consider $d_n=\gcd (a-tn, -b-sn)$  As before, we remark that $d_n$ divides $b(a-tn)+a(-b-ns)=n(-bt-as)=-n$.  But, as before, if $d_n\,|\,n$ then that would imply that $d_n$ divided both $a$ and $b$, a contradiction (if $d_n>1$).  And we are done.
Now suppose that $\gcd(t,s)=d>1$.  Repeat the argument using integers $(a,b)$ with $bt+as=d$  Again consider the pair $(a,-b)$ noting that at least it is a relatively prime pair.  As before, let $d_n=\gcd (a-tn, -b-sn)$ and deduce that $d_n\,|\,d\times n$.  Now $\gcd (d_n, d)=1$ since, if a prime $p$ divided both $d_n$ and $d$ it would have to divide both $a$ and $b$.  Thus $d_n\,|\,n$ and we are done, as before.
