I don't see how this holds, or maybe I'm understanding modular arithmetic incorrectly. If $n = 12$ then $a ⊙ b = 0$ can be satisfied by $a = 3$ and $b = 4$ no?
I see how this holds when I write the proof for it.
If $(a * b)$ mod $n = 0$, $a * b = n$.
We can rewrite $a ⊙ b = 0$ as: $ab + kn = 0$, now as $ab + ab(k) = 0$.
Factoring out an $ab$ we get $ab( 1 + k) = 0$.
Since $k$ is $a \geq 0$, $a*b = 0$, meaning $a ∨ b = 0$.
But I still don't see how this makes sense!
This hold for prime numbers. Because if $a \times b$ is the choosen prime n, then $a$ or $b$ is $n$ and the other is 1