I need to (1) identify the error in this proof, and (2) give a counterexample to show that the conditional is false.
Proposition: Let $u$, $m$, $n$ be three integers. If $u\mid mn$ and $\gcd(u,m) = 1$, then $m = \pm 1$.
Proof: If $\gcd(u,m) = 1$, then $1 = us + mt$ for some integers $s$ and $t$. If $u\mid mn$, then $us = mn$ for some integer $s$. Hence, $1 = mn + mt = m(s + t)$, which implies that $m\mid 1$, and therefore $m = \pm 1$.
For the counterexample, I have:
Let $d = \gcd(u,m)$, then $a = dn$ and $b = dm$ for some integers $m$ and $n$. By Theorem 1.34 (from study guide), for positive integers $a$, $b$ there exists a unique positive integer $c$ such that $c = \gcd(a,b)$ and $c = ax + by$ for some integers $x$ and $y$. Let $c = 1$ and if $1 = ax + by$, then $1 = dnx + dmy = d(nx + my)$. Thus $d \mid 1$, and since $d$ is a positive integer, we conclude $d = 1$. So $d = ax + by$ for some integers $x$ and $y$. Thus $d = dnx + dmy$ and by the cancellation law, $1 = nx + my$, and $1 = \gcd(m,n)$.
Do I have the right idea with this or am I completely off track? If this is not correct for a counterexample any help pointing me in the right direction would be appreciated.
As for the error in the conditional, if $u\mid mn$ and $\gcd(u,m) = 1$, then $m = \pm 1$, I believe it is in "$m = \pm 1$". I believe that $m$ should be positive and be equal to $1$.
