How to prove or disprove matrices $A$ and $B$ commute iff $(A-B)(A+B)=A^2-B^2$?

Question

Prove or show the statement is false with counterexample:
Square matrices $A$ and $B$ commute if and only if $(A-B)(A+B)=A^2-B^2$

Now, this is a biconditional statement, so we must prove $X=>Y$ $\land$ $Y=>X$

I managed to prove $X=>Y$: if the statement $(A-B)(A+B)=A^2-B^2$ holds, the two matices must commute but how do you prove, or disprove $Y=>X$: if matrices commute the statement $(A-B)(A+B)=A^2-B^2$ holds?
Here is the part of the proof:
$(A-B)_{ik}=a_{ik}b_{jk}$
$(A+B)_{kj}=a_{kj}b_{kj}$

$(A-B)_{ik}(A+B)_{kj}$ =$\sum\limits_{k=1}^\mathbb{n}(A-B)_{ik}(A+B)_{kj}$

=$\sum\limits_{k=1}^\mathbb{n}(a_{ik}-b_{ik})(a_{kj}+b_{kj})$

=$\sum\limits_{k=1}^\mathbb{n}a_{ik}a_{kj}+a_{ik}b_{kj}-n_{ik}a_{kj}-b_{ik}b_{kj}$

=$\sum\limits_{k=1}^\mathbb{n}a_{ik}a_{kj}+\sum\limits_{k=1}^\mathbb{n}a_{ik}b_{kj}+\sum\limits_{k=1}^\mathbb{n}-b_{ik}a_{kj}+\sum\limits_{k=1}^\mathbb{n}-b_{ik}b_{kj}$

=$A^2_{ij}-B^2_{ij}+AB_{ij}-BA_{ij}$
From this we can see if the statement $(A-B)(A+B)=A^2-B^2$ holds, $AB_{ij}-BA_{ij}=0$ which is true only if matrices $A$ and $B$ commute.

How would you now prove or disproove that if matrices $A$ and $B$ commute $(A-B)(A+B)=A^2-B^2$ must be the case?

Answer 1

You're thinking way too overcomplicated. We have $$(A+B)(A-B)=A^2-AB+BA-B^2.$$ Both directions follow at once.