tricky GCD Question I am trying to show that the $ gcd (\frac {n}{gcd(n,d)} , \frac {d}{gcd(n,d)} ) = 1 $. 
My steps let $gcd(n,d) = K = xn+yd$ then I need to find some linear combination of $\frac {n} {K} $ and $\frac {d} {K} $ that gives 1. Any hints would be appreciated.

Let $z=\gcd (n,d)$ and $w=\gcd (n/z,d/z).$ 
We have $n/z= w m$ and $d/z= w e$ with $m,e\in Z.$
So $n=z w m$  and $d=z w e$ . So  $z w$ is a common divisor of $n$ and $d$. So $1\leq z w\leq \gcd(n,d)=z$. So $w=1.$
