I'm working on this problem, but I'm missing some manipulation.
Suppose $R$ is a rng without zero divisors and has elements $a$ and $b\neq 0$ such that $ab+kb=0$ for some $k\in\mathbb{N}$ (that is, $ab+\underbrace{b+b+\cdots+b}_{k\ \textrm{times}}=0$.) I'm trying to show that $ca+kc=0=ac+kc$ for all $c\in R$.
I observe that $ (ca+kc)b=cab+kcb=c(ab+kb)=c\cdot 0=0 $ so $ca+kc=0$ since $b\neq 0$ and $R$ has no zero divisors. I can't get the other equation. I note it's equivalent to showing $ac=ca$ for all $c\in R$, so I was trying to multiply $ac-ca$ by something nonzero to get $0$ to conclude but with no luck. Does anyone see what to do to get the other equality? Thanks.