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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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    Some solution is given as Theorem 4 [here](http://math.ucsd.edu/~jwavrik/web00/ISSAC_99.pdf). It was among the first google hits for ["ab^2=ba^2" commutative](http://www.google.com/search?hl=en&q=%22ab%5E2%3Dba%5E2%22+commutative). [By this I do not suggest that googling should be considered the general approach for such problems ;-)]2011-11-18
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    @MartinSleziak is there any other way to solve this?2017-12-12

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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 $xy=yx.$