Proof: $a^2 - b^2 = (a-b)(a+b)$ holds $\forall a,b \in R$ iff R is commutative We want to show that for some ring $R$, the equality $a^2 - b^2  = (a-b)(a+b)$ holds $\forall a,b \in R$ if and only if $R$ is commutative.
Here's my proof --- I'm not sure if the first part stands up to examination. I'd be grateful if someone could take a look.
Forward: $a^2 -b^2 = (a-b)(a+b) \forall a,b \in R$ implies $R$ is commutative
Let $x = (a-b)$. Then \begin{align}
x(a+b) &= xa+xb\\
&= (a-b)a + (a-b)b\\
&= a^2 -ba + ab - b^2\end{align}
Then we note that $a^2 - ba + ab - b^2 = a^2 - b^2$ iff $-ba + ab = 0$ if and only if $ab=ba$ iff $R$ is commutative.
Backwards: $R$ is commutative implies $a^2 - b^2 = (a-b)(a+b) \forall a,b \in R$.
Let $x = (a+b)$. Then $(a-b)x = ax - bx = a(a+b) - b(a+b) = a^2 + ab - ba - b^2$. $R$ is commutative, so $ab-ba = 0$, so $a^2 + ab - ba - b^2 = a^2 - b^2$.

I think there are not many word necessary.
$$a^2 -b^2 = a^2+ab-ab-b^2=a^2+ab-ba-b^2=a(a+b)-b(a+b)=(a-b)(a+b)$$
