I want to show that for $\gcd(m,n) = \gcd(p,q) = 1$, if $qm = pn$ then $m=p $ and $n=q$.
Here is what I am working through:
- If both the gcds are equal to 1, then I can express $\gcd(m,n)$ as a linear combination of m and n, and likewise for $\gcd(p,q)$.
- If I multiply these two linear combinations, I can still get a value of 1 but it will result in complicated algebra.
However, if I try this method:
- I let $qm - pn = 0$. Since the two gcds can divide 0, then naturally they have the same value. I know that this is a linear combination of both q and p, and m and n.
But it seems that this approach also means any gcd can divide qm-pn. So I'm unable to proceed further to show that $m=p$ and $n=q$.
We know $am+bn=1$ for some $a,b$, so $pam+pbn=p$.
But $pbn=bqm$, so $(pa+bq)m=p$, giving $m\mid p$.
Similarly you can show the reverse, and for $q,n$.