# 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

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?

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 $$x=x^2$$.
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