TITLE: If $x^{2}-x\in Z(R)$ for all $x\in R$, then $R$ is commutative.
QUESTION [6 upvotes]: If $x^{2}-x\in Z(R)$ for all $x\in R$, then $R$ is commutative.
I need to proof this theorem and I have something like this below. However, I do not know how to continue this proof. 
$(x+y)^{2}-(x+y)=x^{2}+xy+yx+y^{2}-x-y=(x^{2}-x)+(y^{2}-y)+xy+yx\in Z(R)$
so $xy+yx\in Z(R)$

REPLY [13 votes]: You’re on the right track. From $xy+yx\in Z(R)$ we get $x(xy+yx)=(xy+yx)x$, or, after expanding, $x^2y+xyx=xyx+yx^2$, and hence $x^2y=yx^2$. This shows that every square is in $Z(R)$. Can you see how to finish it from there?

REPLY [4 votes]: The answer is Theorem 2 in the paper
http://archive.maths.nuim.ie/staff/sbuckley/Papers/bm_variations.pdf