Is $a+b=ab=a^{'}+b^{'}=a^{'}b^{'}$ possible for different elements of $\mathbb Z / n \mathbb Z$?

Question

Let $a, b, a^{'}, b^{'} \in \mathbb Z / n \mathbb Z\ $such that $a+b=ab=a^{'}+b^{'}=a^{'}b^{'}$. Is it possible if $a^{'} \ne a\ $ and $a^{'} \ne b$? I try to solve this problem for a couple of hours, but all my attempts fail. I figured out when $a+b=ab$ is true, but I can't solve if the mapping is injective.

Answer 1

Consider in $\mathbb{Z}/4$ the couples $(\bar 2,\bar 2)$ and $(\bar 0,\bar 0)$.

Answer 2

In complex algebra we have $(1+i)+(1-i)=(1+i)×(1-i)=2$. Now suppose you select a modulus where there are two (or more) different additive inverse pairs of square roots of $-1$. We may then substitute either pair for $\pm i$ from the complex relation and get things like this:

$48+19\equiv 48×19\equiv 2 \bmod 65$

$58+9\equiv 58×9\equiv 2 \bmod 65$

This is the smallest example (by modulus) I can find with a square free modulus, if there is a smaller one I would have to use brute force to see it.