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Let $R$ be a ring with unit element $1$ such that $(ab)^2=(ba)^2$ for all $a,b$ in $R$. If in $R$, $2x=0$ implies $x=0$, how do I show that R is commutative?

Is there any general approach to attack this kind of problem?

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  • $\begingroup$ Some solution is given as Theorem 4 here. It was among the first google hits for "ab^2=ba^2" commutative. [By this I do not suggest that googling should be considered the general approach for such problems ;-)] $\endgroup$ Nov 18 '11 at 11:44
  • $\begingroup$ @MartinSleziak is there any other way to solve this? $\endgroup$
    – Vincent
    Dec 12 '17 at 15:53
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Here is the proof in Martin's link (so this question doesn't go unanswered and we don't have to link out):

Let $F(a,b) = (ab)^2 - (ba)^2 = abab - baba = 0 $. Then routine verification shows $ F(1+x, 1+y) - F(1+x,y) - F(x,1+y) + F(x,y) = -2yx+ 2xy $ so $ 2(xy-yx) = 0 $, which implies $xy=yx.$

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