If $\forall x \in R, x^2-x \in Z(G)$, than $R$ is commutative

Question

Let $R$ be a ring such that for every $x\in R$ we have $x^2-x \in Z(G)$. Show that $R$ is a commutative ring.

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My thoughts

What should I do? I could show that every $y \in R$ could be written in the shape $x^2-x$ but I don't know if that is true. We know that $y \cdot(x^2-x) \ = \ (x^2-x)\cdot y$ for any $x,y \in R$, but that doesn't mean that $xy = yx$. Please give me a hint

Answer 1

One possible hint: Working on $$(x+y)^2-(x+y).$$ to show that $xy+yx\in Z(R)$ could pave our way to solution. Indeed, when $xy+yx\in Z(R)$ then $$(xy+yx)y=y(xy+yx)\to xy^2=y^2x.$$ Now since $y^2-y\in Z(R)$ so $$y^2x-yx=xy^2-xy$$

Answer 2

Suppose that $R$ has $1$.

Then, for any $x \in R$,

$(1+x)^2-(1+x) \in Z(R)$, which shows: $(1+x^2+2x)-(1+x) \in Z(R)$, which implies that $x^2+x$ is in $Z(R)$ for all $x \in R$. So, $(x^2+x)-(x^2-x) \in Z(R)$ which implies $2x \in R$ for all $x \in R$. This easily implies $R$ is commutative.