I want to prove the following:
Theorem: For integers $a,b,x$ and integers $m,n>0$, if \begin{align*} x&\equiv a \pmod{m} \newline x&\equiv a \pmod{n} \newline \gcd(m,n)&=1 \newline \end{align*} Then $x\equiv a \pmod{m\cdot n}$
I tried using the fact that for some integers $l_1, l_2$ we have $x-a=ml_1, x-a=nl_2$. I wasn't able to manipulate this in any way that gives me $mn|x-a$.
For simplicity, let $y = x-a = m\ell_1 = n\ell_2$.
Since $m$ and $n$ are relatively prime, there are integers $s$ and $t$ such that $sm + tn = 1$. Then $$ y = (sm+tn)y = smy + tny = sm \cdot n\ell2 + tn \cdot m\ell_1 = (s\ell_2 + t\ell_1)mn $$ and so $mn$ divides $y$, i.e. $y \equiv 0 \bmod mn$.
I tried using the fact that for some integers l1,l2 we have x−a=ml1,x−a=nl2. I wasn't able to manipulate this in any way that gives me mn|x−a.
But that would mean $ml_1 = nl_2$.
That means $m|nl_2$. But $m,n$ are relatively prime so have no factors in common.
So $m|l_2$. Write $l_2$ as $l_2 =ml_3$ and we have.... $x - a= nl_2 = nml_3$
So $x\equiv a \pmod nm$.
You have $ml_1 = nl_2$ and $\gcd(m,n) = 1$. This gives $m \mid l_2$, so there is an integer $l_3$ such that $x - a = mnl_3$. Therefore, $mn$ divides $x-a$.
