Originally Posted by

**Bruno J.** The problem is trivial if $\displaystyle (0)$ is prime, for then it is maximal and $\displaystyle R$ is a field, so that $\displaystyle b=-a^{-1}$ always works for nonzero $\displaystyle a$.

The problem is only hard when $\displaystyle R$ has zero divisors. We can suppose $\displaystyle a$ is neither a unit nor 0, those cases being trivial.

If $\displaystyle a,a^2,...$ are not all distinct then for some $\displaystyle 0<s<t$ we have $\displaystyle a^s=a^{s+t} \Rightarrow a^s(1-a^t)=0$ then $\displaystyle b=-a^{t-1}$ works.

If they are all distinct then I don't know. I suspect this is impossible but I can't see why that would be. I'd love the solution!