This question is about a ring for some chosen $n \in \mathbb{N}$
I wanted to find the number $M_n$ of combinations $(a,b) \in \mathbb{Z}_n \times \mathbb{Z}_n$ can be found such that $a \cdot b = 0$ and $a\neq 0 \neq b$. I hoped that I could find one formula that works for any $n \in \mathbb{N}$, in a manner that we can maximize the quotient $\frac{M_n}{n}$
What I did
First of all, we can find a bound in a very rough way, namely. $$ (n - \phi(n))^2 \quad = \quad n^2 \left(1 - \prod_{p |n}(1-\frac 1p) \right) $$ Another thing I thought about is that if $(a,b)$ is such a combination, then we also have: $$ (a,-b) \ , \ (-a,b) \ , \ (-a,-b) \ , \ (b,a) \ , \ (b,-a) \ , \ (-b,a) \ , \ (-b,-a) $$ So each pair we would find would give a set of at most seven other pairs.
If we want to count in the most straight-forward way, I believe that a zero-divisor $a$, and look for a value $b$ such that $a\cdot b = 0$. But if we want to find those values $b$, we have to make use of factorisation in prime numbers of $n$.
I could use your help here. (If you want me to try to make I big formula of what could be the number of combination I seek, please tell me. I thought it would be useless here.)
For each $a \in \mathbb Z_n$, $a \cdot b = 0 \iff ab \equiv 0 \pmod n$. This will occur exactly when $$\frac n {(a,n)} \big\vert\ b$$which can happen exactly $\big((a,n)-1\big)$ times: since we must have $0< b<n$, possible values for $b$ are $$\frac{n}{(a,n)},\frac{2n}{(a,n)},\frac{3n}{(a,n)},\ldots,\frac{\big((a,n)-1\big)n}{(a,n)}$$
Hence, you are looking to calculate $$\sum_{a=1}^{n-1}\big((a,n)-1\big)=\operatorname P(n)-(2n-1)$$ where $\operatorname P(n)=\displaystyle\sum_{a=1}^n\operatorname{gcd}(a,n)$ is Pillai's arithmetical function.