$\gcd(m,n) = 1$ and $\gcd (mn,a)=1$ implies $a \cong 1 \pmod{ mn}$ I have $m$ and $n$ which are relatively prime to one another and $a$ is relatively prime to $mn$
and after alot of tinkering with my problem i came to this equality:
$a \cong 1 \pmod m \cong 1 \pmod n$
why is it safe to say that $a \cong 1 \pmod {mn}$?..

$\gcd(m,n)=1$ implies there is an $x$ and $y$ so that
$$
mx+ny=1\tag{1}
$$
$a\equiv1\pmod{m}$ and $a\equiv1\pmod{n}$ imply there is a $j$ and $k$ so that
$$
a-1=jm=kn\tag{2}
$$
multiplying $(1)$ by $j$ and using $(2)$
$$
nkx+jny=j\tag{3}
$$
Plugging $(3)$ back into $(2)$ yields
$$
a-1=jm=n(kx+jy)m\tag{4}
$$
which implies that $a\equiv1\pmod{mn}$.

Idea of the Preceding Argument
The idea of $(3)$ is to show that since $\gcd(m,n)=1$, by $(1)$, and $jm$ is a multiple of $n$, by $(2)$, we have that $j$ is a multiple of $n$. Thus, $a-1=jm$ is a multiple of $mn$, by $(4)$.
