It is well known that $\exists x \in \mathbb{N}$ such that
$$x \equiv a\_1 \mod b\_1$$ $$x \equiv a\_2 \mod b\_2$$
if and only if
$a\_1 \equiv a\_2 \mod \text{gcd}(b\_1, b\_2)$.


Is there such a simple condition for the following system?
$$x \equiv a\_1 \mod b\_1$$ $$x \not\equiv a\_2 \mod b\_2$$



Conditions for congruences
If $b\_2$ divides $b\_1$ then an obvious necessary and sufficient condition is that $a\_1\not\equiv a\_2\ \mathrm{mod}\ b\_2$. 


If $b\_2$ does not divide $b\_1$ then $\gcd(b\_1,b\_2)$ is a proper divisor of $b\_2$, so there are at least two residue classes $y\ \mathrm{mod}\ b\_2$ such that $y\equiv a\_1\ \mathrm{mod}\ \gcd(b\_1,b\_2)$. Pick one that satisfies $y\not\equiv a\_2\ \mathrm{mod}\ b\_2$, then by the criterion you quote there is an $x$ satisfying $x\equiv a\_1\ \mathrm{mod}\ b\_1$ and $x\equiv y\ \mathrm{mod}\ b\_2$. This $x$ solves the second system.


To summarize, the second system has a solution unless $b\_2\mid b\_1$ and $a\_1\equiv a\_2\ \mathrm{mod}\ b\_2$.


