Let $(A,+,.)$ be a ring such that, if $x \in A$ with $6x = 0$, then $x=0$. Let $a,b,c \in A$ such that $a-b$ , $b-c$ , $c-a$ are idempotent. Prove that $a=b=c$.
Unfortunately, I haven't made any big progress on this one. I noticed that $(a+b+c)^2 = 3(a^2+b^2+c^2)$ and I tried finding an expression $E$ with $6E =0$, but with no success. I haven't encountered many problems with rings, so I am quite a beginner in this area.
Can you help me on this?
Note that
$\begin{align*} (a-c)^2=(a-c)&=((a-b)+(b-c))^2\\ &=(a-b)^2+(b-c)^2+(a-b)(b-c)+(b-c)(a-b)\\ &=a-b+b-c+ab-ac-b^2+bc+ba-b^2-ca+cb \end{align*}$
It follows that
$\begin{align*} 0=&ab-ac-b^2+bc+ba-b^2-ca+cb\\ =&\color{blue}{(ab+ba)-b^2} +\color{red}{(bc+cb)-b^2}-(ac+ca)\\ =&\color{blue}{a^2-a+b}+\color{red}{c^2-c+b}-(ac+ca)\\ =&(a^2+c^2-ac-ca)-(a+c)+2b\\ =&(a-c)^2-a-c+2b\\ =&2b-2c \end{align*}$
It follows that $6(b-c)=0$, whence $b=c$.
Similarly, show that $a=b$
Explanation for blue/red colored parts:
$(a-b)^2=a-b\implies a^2+b^2-ab-ba=a-b\implies ab+ba-b^2=a^2-a+b$