For every fixed integer a we consider the following system of congruences:
$$ax \equiv 1 \pmod{12}$$
$$x \equiv a \pmod{13}$$
If for example $a=0$, it's easy to see that the system does not have any solutions. The question is, for what integer values $a$ does the system have at least one solution?
Here comes my answer to the question:
The first equation has a solution if and only if $gcd(a,12)|1$. The only divisors that divide $1$ are ${-1,1}$. Therefore $a$ and $12$ must be co-prime due to the fact that $gcd(a,12)$ must be $1$ or $-1$ in order to divide $1$.
Knowing that $a$ is co-prime with $12$, we can use Euler's Phi Function
$$\phi(N) = | \{1 \leq a < n : gcd(a,N)=1\}|$$
to find the possible values of $a$.
$$\phi(12) = |\{1,5,7,11\}|=4$$
With this new information regarding what possible values $a$ can attain. It is clear that the second equation has at least one solution for every allowed $a$.
$a=1 \rightarrow x=1$ solves the system.
$a=5 \rightarrow x=5$ solves the system.
$a=7 \rightarrow x=7$ solves the system.
$a=11 \rightarrow x=11$ solves the system.
So the answer to the question is:
The system has at least one solution when $gcd(a,12)=1$.
Is my answer/reasoning correct?
EDIT: Changed The system has at least one solution when $a \in \phi(12)$. to The system has at least one solution when $gcd(a,12)=1$.
