Show why the proof for the following is false $u\mid mn$ and $\gcd(u,m) = 1$, then $m = \pm1$

Question

Let $u,m,n$ be three integers.

If $u\mid mn$ and $\gcd(u,m) = 1$, then $m = \pm 1.$

If $\gcd(u,m) = 1$, then $1 = us + mt$ for some integers $s, t$.

If $u \mid mn$, then $us = mn$ for some integer s.

Hence, $1 = mn + mt = m(n + t)$, which implies that $m|1$, and therefore $m = 1$.

First dips into discrete math and having trouble with proofs

I've been playing around with various values and found $u=2, m=3, n=4,$ to be a counterexample to the proof but have yet to find the flash of inspiration needed to understand why.

Any help is appreciated.

Answer 1

There exist integers $s ,t$ with $1=us+mt.$ There exists integer $s'$ with $us'=mn.$ IF $s=s'$ THEN $m=\pm 1.$ But if $s\ne s'$ then ... ?