Rings in which binomial theorem holds for at least one integer $n>2$

Let $R$ be a ring ; if $(a+b)^2=a^2+2ab+b^2 , \forall a,b \in R$ , then we know that $R$ is commutative ; also if $R$ is commutative then we know that $(a+b)^n=\sum _{r=0}^n {n \choose r} a^{n-r}b^r , \forall a,b \in R , \forall n \in \mathbb Z^+ $ ; so I would like to ask , under what additional conditions on $R$ , does for a fixed given $n \in \mathbb Z^+$ , $(a+b)^n=\sum _{r=0}^n {n \choose r} a^{n-r}b^r , \forall a,b \in R$ implies $R$ is commutative ? what if we also include the condition for some positive integer (or all ) $m>1$ , $(ab)^m=a^mb^m , \forall a,b \in R$ ?

Is it true that for every integer $n>2$ , there exist a non-commutative ring $R$ such that $(a+b)^n=\sum _{r=0}^n {n \choose r} a^{n-r}b^r , \forall a,b \in R$?

Solution 1:

Let $R$ be the ring of $3\times3$ real matrices of the form $\begin{bmatrix} 0&0&0\\ x&0&0 \\ y&z&0\end{bmatrix}$.

This ring is not commutative; for example $\begin{bmatrix} 0&0&0\\ 1&0&0\\ 0&0&0\\ \end{bmatrix}$ and $\begin{bmatrix} 0&0&0\\ 0&0&0\\ 0&1&0\\ \end{bmatrix}$ do not commute.

Every triple product is $0$, so, for $n\ge3$, the binomial theorem reduces to $0=0$.