My assignment is the following:  Prove that the system  $$x\equiv a \pmod m$$ $$x\equiv b \pmod n$$ has a solution if and only if $\gcd(m,n) \mid (a-b)$  My approach is the following:  [$\longleftarrow$] Let $x=a +r_1m = b + r_2n$, where $r_1$ y $r_2$ are whole numbers. So, $a-b=r_2n-r_1m\ (*)$. If we find  $d=$ GCD$(n,m)$, then $$n=N\cdot d$$ $$m=M\cdot d$$ Replacing in $(*)$ $$a-b=d(r_2N-r_1M)$$  [$\longrightarrow$] If $d=\gcd(n,m)$ and $d \mid a-b$, so $$d=\dfrac{n}{N}$$ $$d=\dfrac{m}{M}$$ and $\dfrac{n}{N}\Big |a-b$ y $\dfrac{m}{M}\Big |a-b$  How can I finish both sides of this proof?
We want to find the solution $x$ to the congruence system  $$\begin{align} x &amp;\equiv r_1 \!\!\!\pmod{\!m_1}\\ x &amp;\equiv r_2 \!\!\!\pmod{\!m_2}\end{align},\ \ {\rm where}\ \  \gcd(m_1, m_2) = 1$$  These can be rewritten as  $$\begin{align} x &amp;= r_1 + m_1 j\\ x &amp;= r_2 - m_2k\end{align}$$  for unknown integers $j, k$. So we set those two equations as equal and rearrange them to  $$m_1j + m_2k = r_2 - r_1$$  Now assume we perform $\text{egcd}(m_1, m_2)$, which gives us the solution to  $$m_1u + m_2v = \gcd(m_1, m_2)$$  Let the integer $\,h\,$ make $\,h \gcd(m_1, m_2) = r_2 - r_1$.  Then $\qquad hm_1u + hm_2v = h\gcd(m_1, m_2) =  r_2 - r_1 = m_1j + m_2k$  At this point I am lost because I don't think we can just assume  $hm_1u = m_1j$ and extract $j = hu$.   How do I get the values of $j$ or $k$ so I can get the value of $x$?
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