What are all pairs $(a,b)$ such that if $Ax+By \equiv 0 \pmod n$ then we can conclude $ax+by = 0 \pmod n$? All these are good pairs:
$$(0, 0), (A, B), (2A, 2B), (3A, 3B), \ldots \pmod{n}$$
But are there any other pairs?
actually it was a programming problem with $A,B,n \leq 10000$ but it seems to have a pure solution.
 A: If $\rm\:c\ |\ A,B,n\:$  cancel $\rm\:c\:$ from $\rm\:Ax + By = nk.\:$ So w.l.o.g. $\rm\:(A,B,n) = 1,\:$ i.e. $\rm\:(A,B)\equiv 1$.
Similarly, restricting to "regular" $\rm\:x,y,\:$ those such that $\rm\:(x,y,n) = 1,\:$ i.e. $\rm\:(x,y)\equiv 1,\:$ yields 
Theorem $\rm\:\ If\:\ (A,B)\equiv 1\equiv (x,y)\:\ and\:\ Ax+By\equiv 0,\ then\:\ ax+by\equiv 0\iff aB\equiv bA$
Proof $\ $ One easily verifies
$$\rm\:\ \ B(ax+by)\: =\: (aB-bA)x + b(Ax+By) $$
$$\rm -A(ax+by)\: =\: (aB-bA)y - a(Ax+By)$$
$(\Rightarrow)\ $ Let $\rm\:z = aB-bA.\:$ By above $\rm\:ax+by\equiv 0\ \:\Rightarrow\ xz,\:yz\equiv 0 \ \Rightarrow\ z \equiv (x,y)z\equiv 0$.
$(\Leftarrow)\ $ Let $\rm\:z = ax+by.\:$ By above $\rm\:aB-bA\equiv 0\ \Rightarrow\ Az,Bz\equiv 0\ \Rightarrow\ z \equiv (A,B)z\equiv 0.\ \ $ QED
Note $\rm\ (x,y)\equiv 1\pmod n\:$ means $\rm\:ix+jy = 1 + kn\:$ for some $\rm\:i,j,k\in \mathbb Z$ 
Thus we infer $\rm\:xz,yz\equiv 0\ \Rightarrow z \equiv (ix+jy)z\equiv i(xz)+j(yz)\equiv 0\pmod n$
i.e. $\rm\ \ ord(z)\ |\ x,y\ \Rightarrow\ ord(z)\ |\ (x,y) = 1\ $ in the additive group  $\rm\:(\mathbb Z/n,+)$
