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Let $R = (A,+,\cdot)$ be a ring. If there exists a positive integer $n$ such that any element $x$ of a ring $R$ satisfies $x^{4^n+2} = x$, then every element $x$ in $R$ is idempotent.

I have studied some group theory but I don't think it's that helpful here. Basically by multiplying both sides a bunch of times with $x^{4^n + 1}$ you can show that $$x^{k(4^n + 1) + 1} = x \hspace{5px} \forall k \in \mathbb{N}.$$

My intuition goes along the lines of "if $x^a = x$ and $x^b = x$ then $x^{(a,b)} = x$" but in the absence of multiplicative inverses that wouldn't work (the reason I thought this might have been helpful is because the exponent of $2$ in $4^n + 2$ is always $1$, so by finding an appropriate $k$ we could make the gcd $2$).

How should I proceed?

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1 Answer 1

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Let $N$ be $4^n+2$ for short. It is an even number. Then from $$ x = x^N=(-x)^N=-x $$ we obtain $2x=0$ for all $x\in R$. So we work "in the characteristic two". In particular, $(x+y)^{4^n}=x^{4^n}+y^{4^n}$ for all $x,y$ in the ring. This implies: $$ \begin{aligned} x+1&=(x+1)^N \\ &=(x+1)^{4^n}(x+1)^2 \\ &=\left(x^{4^n}+1\right)(x^2+1) \\ &=x^N +x^{4^n}+x^2+1\ . \end{aligned} $$ This implies $x^{4^n}=x^2$. Then the given equation reads simpler, $$ x = x^N = x^{4^n}\cdot x^2=x^2\cdot x^2=x^4\ . $$ From here $x=x^4=(x^4)^4=x^{4^2}$, and inductively we get $x=x^{4^n}$. This implies $x=x^N=x\cdot x^2=x^3$. So $(x+1)=(x+1)^3$ and after cancellations finally $x=x^2$.

$\square$

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