Let $n_1,...,n_k \in \mathbb{N}$ and let $a_1,...,a_k \in \mathbb{Z}$. How to prove the following version of the Chinese remainder theorem (see here):
There exists a $x \in \mathbb{Z}$ satisfying system of equations: $$x=a_1 \pmod {n_1}$$ $$x=a_2 \pmod {n_2}$$ $$\ldots$$ $$x=a_k \pmod{n_k}$$ if and only if $a_i=a_j \pmod{\gcd(n_i,n_j)}$ for all $i,j=1,...,k$?
If numbers $n_i$, for $i=1,...,k$, are pairwise coprime, it is a classical version of Chinese remainder theorem.
Thanks.
If we factor $n_k$ into primes, $n_k = p_{1}^{b_{1}}\cdots p_r^{b_{r}}$, then the Chinese Remainder Theorem tells us that $x\equiv a_k\pmod{n_k}$ is equivalent to the system of congruences $$\begin{align*} x&\equiv a_k\pmod{p_1^{b_{1}}}\\ x&\equiv a_k\pmod{p_2^{b_{2}}}\\ &\vdots\\ x&\equiv a_k\pmod{p_r^{b_{r}}} \end{align*}$$ Thus, we can replace the given system of congruences with one in which every modulus is a prime power, $n_i = p_i^{b_i}$.
Note that the assumption that $a_i\equiv a_j\pmod{\gcd(n_i,n_j)}$ "goes through" this replacement (if they were congruend modulo $\gcd(n_i,n_j)$, then they are congruent modulo the gcds of the prime powers as well).
So, we may assume without loss of generality that every modulus is a prime power.
I claim that we can deal with each prime separately, again by the Chinese Remainder Theorem. If we can solve all congruences involving the prime $p_1$ to obtain a solution $x_1$ (which will be determined modulo the highest power of $p_1$ that occurs); and all congruences involving the prime $p_2$ to obtain a solution $x_2$ (which will be determined modulo the highest power of $p_2$ that occurs); and so on until we obtain a solution $x_n$ for all congruences involving the prime $p_n$ (determined modulo the highest power of $p_n$ that occurs), then we can obtain a simultaneous solution by solving the usual Chinese Remainder Theorem system $$\begin{align*} x &\equiv x_1 \pmod{p_1^{m_1}}\\ &\vdots\\ x &\equiv x_n\pmod{p_n^{m_n}} \end{align*}$$ (where $m_i$ is the highest power of $p_i$ that occurs as a modulus).
So we are reduced to solving figuring out whether we can solve the system $$\begin{align*} x &\equiv a_1\pmod{p^{b_1}}\\ x &\equiv a_2\pmod{p^{b_2}}\\ &\vdots\\ x & \equiv a_n\pmod{p^{b_n}} \end{align*}$$ with, without loss of generality, $b_1\leq b_2\leq\cdots\leq b_n$.
When can this be solved? Clearly, this can be solved if and only if $a_i\equiv a_j\pmod{p^{b_{\min(i,j)}}}$: any solution must satisfy this condition, and if this condition is satisfied, then $a_n$ is a solution.
For example: say the original moduli had been $n_1 = 2^3\times 3\times 7^2$, $n_2= 2^2\times 5\times 7$, $n_3=3^2\times 5^3$. First we replace the system with the system of congruences $$\begin{align*} x&\equiv a_1 \pmod{2^3}\\ x&\equiv a_2\pmod{2^2}\\ x&\equiv a_1\pmod{3}\\ x&\equiv a_3\pmod{3^2}\\ x&\equiv a_2\pmod{5}\\ x&\equiv a_3\pmod{5^3}\\ x&\equiv a_1\pmod{7^2}\\ x&\equiv a_2\pmod{7}. \end{align*}$$ Then we separately solve the systems: $$\begin{align*} x_1&\equiv a_1 \pmod{2^3} &x_2&\equiv a_1\pmod{3}\\ x_1&\equiv a_2\pmod{2^2}&x_2&\equiv a_3\pmod{3^2}\\ \strut\\ x_3&\equiv a_2\pmod{5}&x_4&\equiv a_1\pmod{7^2}\\ x_3&\equiv a_3\pmod{5^3}&x_4&\equiv a_2\pmod{7}. \end{align*}$$
Assuming we can solve these, $x_1$ is determined modulo $2^3$, $x_2$ modulo $3^2$, $x_3$ modulo $5^3$, and $x_4$ modulo $7^2$, so we then solve the system $$\begin{align*} x &\equiv x_1\pmod{2^3}\\ x &\equiv x_2\pmod{3^2}\\ x&\equiv x_3 \pmod{5^3}\\ x&\equiv x_4\pmod{7^2} \end{align*}$$ and obtain a solution to the original system.
Hence, if the condition $a_i\equiv a_j\pmod{\gcd(n_i,n_j)}$ holds in the original system, then we obtain a solution for each prime, and from the solution for each prime we obtain a solution to the original system by applying the usual Chinese Remainder Theorem twice.
Let $N = \operatorname{lcm}(n_1, \ldots, n_k)$, $N = n_i \cdot N_i$. Then $\gcd(N_1, \ldots, N_k)= 1$. Take $c_1$, $\ldots$, $c_k$ integers such that $$\sum_{i=1}^k c_i N_i = 1$$ Let's show that $$a= \sum a_i c_i N_i$$ is a solution to the congruences $$a\equiv a_i \mod n_i$$ $i=1,\ldots,k$. Choose an index $i_0 \in \{1, \ldots, k\}$. We have $$a- a_{i_0} = a -a_{i_0} \cdot \sum_{i} c_i N_i = \sum_{i} (a_i -a_{i_0}) c_i N_i$$
For every $i \ne i_0$ we have
$$a_i - a_{i_0} \vdots \gcd(n_i, n_{i_0})$$ and $$N_i = \frac{\operatorname{lcm}(n_1,\ldots n_k) }{n_i} \vdots \frac{\operatorname{lcm}(n_i, n_{i_0})}{n_i} $$ so $$(a_i-a_{i_0})N_i \vdots \gcd(n_i, n_{i_0}) \cdot \frac{\operatorname{lcm}(n_i, n_{i_0})}{n_i} = n_{i_0}$$