When a congruence system can be solved? How to prove that a congruence system with $n$ equations can be solved if and only if all the equations can be solved two by two?
\begin{cases} x \equiv a_1 \phantom ((mod\phantom mm_1) \\ x \equiv a_2 \phantom ((mod\phantom mm_2) \\ ... \\ x \equiv a_n \phantom ((mod\phantom mm_n) \end{cases}
I know that if $n=2$, the system can be solved if and only if $\gcd(m_1,m_2)\mid a_1-a_2$.
The thesis is: the system is solvable $\Longleftrightarrow$ $\gcd(m_i,m_j)|\mid a_i-a_j$ for all $i\neq j$.

Hint $\ $ Suppose for induction there is a solution $\,x = a\,$ to the first $\,n\!-\!1\,$ congruences, i.e.
$$a\equiv a_i\!\!\pmod{m_i},\,\ i=1,\ldots n-1\tag{#}$$
Combining the solution with the final congruence we obtain the system
$$\begin{align} &x \equiv a\! \pmod m,\ \ m  = {\rm lcm}(m_1,\ldots m_{n-1})\\
 &x\equiv a_n\!\!\!\! \pmod{m_n}\end{align}$$
We know this is solvable iff $\,(m_n,m)\mid a-a_n.\ $  Since gcd distributes over lcm we have
$$(m_n,m) = (m_n,{\rm lcm}(m_1,\ldots m_{n-1}) = {\rm lcm}((m_n,m_1),\ldots,(m_n,m_{n-1})) =: \ell $$
Hence $\ (m_n,m)\mid a-a_n \iff \ell\mid a-a_n\iff (m_n,m_i)\mid a-a_n,\ $ for all $\,i \le  n-1$
$\ \ (m_n,m_i)\mid m_i\mid a-a_i\,$ since $\,a\,$ is a solution of the first $\,n-1\,$ congruences $\,({\#})$
$\qquad\ \  (m_n,m_i)\mid a_n\!-\!a_i\,$ by hypothesis. Therefore it divides their difference, i.e.
$\qquad\ \ (m_n,m_i)\mid a-a_n = a-a_i -(a_n - a_i).\ $  Therefore a solution exists.
Remark $\ $ Domains satsifying CRT = Chinese Remainder Theorem are known as Prüfer domains. They are non-Noetherian generalizations of PIDs and Dedekind domains. Their ubiquity stems from a remarkable confluence of interesting characterizations. See this answer for a couple dozen characterizatons.
