I have a result that holds on $\mathbb{Z}_n$ and I like to hold in any euclidean domain $A$ with unique quotient and remainder (for instance $K[X]$ with $K$ a field). The first part of the result remains true:
Given $A$ any euclidean domain with unique quotient and remainder. Let $n \in A - \{0\}$ then
- Given $a,b \in A$ with $a \neq 0$, $ax = b$ in $A_n$ has solution $\iff d = (a,n) | b$.
I'm having difficulties to extend the proof of the second part (perhaps it doesn't hold):
- In this case, the equation has exactly $d$ solutions that from the initial one $x_0$ with $x_0 = 0 \lor \phi(x_0) < \phi(\frac{n}{d})$ where $\phi$ es the euclidean function of the domain are $\{x_0 + k \frac{n}{d}:k = 0,\cdots,d-1\}$
I think it is clear that the number of solutions should be less than $d$ just by using euclidean division in the set of generic division that the equation has, $\{x_0+k\frac{n}{d}:k \in \mathbb{Z}\}$, i.e., we write $k = qd+r$ and see that the only possible solutions are of the form $x_0+k\frac{n}{d}$ with $k = 0, \cdots,d-1$. But I don't know how to prove these are all different.
The idea in the case of $\mathbb{Z}_n$ was to take $0 \le k' < k < d$ and then realize that $0 < (x_0+k\frac{n}{d}) - (x_0 + k'\frac{n}{d}) = (k-k')\frac{n}{d} < d \frac{n}{d} = n$ but in the general case I don't know how to get this nice order.
Does the general result for 2. hold in general? Do you have a counterexample?