I was asked to solve a system of modular equations. Without knowing the system, at first I thought about using the Chinese Remainder Theorem. When I saw the question, I knew I couldn't use it.
The system I was asked to solve for $n$ was: \begin{cases} 132\equiv 5\mod n\\ 342\equiv 5\mod n \\ 450\equiv 5\mod n \end{cases} Since using CRT wasn't an option because there was only one moduli, I proceeded to use the definition. I knew that this was equivalent to: $$\begin{cases} n\mid(132-5)\\ n\mid(342-5)\\ n\mid(450-5) \end{cases} \Rightarrow \begin{cases} n\mid(127)\\ n\mid(337)\\ n\mid(445) \end{cases} \Rightarrow n\mid\gcd(127,337,445)$$ Since $127$ is prime, that $\gcd$ is $1$. Then $n$ must be $1$.
Everything turned out fine because I was hoping that would be the case. In another scenario, $n\in\lbrace d\in\mathbb{N}:d\mid\gcd(127,337,445)\rbrace$ and I wouldn't be able to answer the question completely.
My problem
The question I want to ask is basically, for any number of equations of that sort, is $n$ always $1$?
Also if it isn't always $1$, what are sufficient conditions to guarantee this? Is there a generalization of some sort? By which I mean, is there a way to guarantee that $n$ can be any number I want?
My thoughts
I noticed that the first condition is equivalent to the following:
Given $\{a_i\}\subseteq\mathbb{N} ,i =1,\cdots,n$, and $k\in \mathbb{N}$ such that $a_i\equiv k\mod m$ for all $i$, then is $m =1$?
This would be equivalent to $\gcd(\{a_i-k\}_{i\in[n]})=1$. Which means a sufficient condition is that there exists an $i$ such that $a_i-k$ is prime and all the other terms are not a multiple of said prime.
I haven't thought of a counterexample to see that $n$ might not be $1$.
The solution $n = 1$ always works. There may be other solutions as well, however, like with $$ \cases{2\equiv 7\mod n\\1\equiv 16\mod n} $$ where $n = 1$ and $n = 5$ both work.
As long as all moduli are equal, your approach to the original problem is fine in general: rewrite each equation to the form $a_i\equiv 0\mod n$, and then see that it is sufficient and necessary that $n$ divides all of the $a_i$, which is to say that $n$ divides the $\gcd$ of the $a_i$. If that $\gcd$ is not $1$, then you have multiple solutions.