Showing $na \equiv 1 \pmod m$ and $n'a \equiv 1 \pmod m \implies n \equiv n' \pmod m$ for $(a,m) = 1$ We have $(a,m) = 1$ iff exists integer $n$ such that $na \equiv 1 \pmod m$. Prove  $na \equiv 1 \pmod m$ and $n'a \equiv 1 \pmod m$ implies $n \equiv n' \pmod m$.
For this question, I have so far
$na \equiv 1$ and
$n'a \equiv 1$
So, I have the equations 
$na -1 = mk$ and
$n'a - 1 = mk'$
Subtracting, I end up with
$a(n-n') = m(k-k')$ dropping $k$ because its a multiple. Then I just end up with 
$na \equiv n'a \pmod m$ which seems circular. Any help?

So, $m\mid a(n-n')\implies m\mid (n-n')$ as $(a,m)=1$
