@TobiasKildetoft

$y+1=(y+1)^2=y^2+2y+1=3y+1=y+1$ so the idempotency satisfied with $k=2$?

@TobiasKildetoft The instance "a finite ring R" exist, $x\in R$, is $R=(\mathbb Z_2, 0,1,*,+)$. When $k>0$,

$x^k=x^{2k} = 0 mod 2

Eg.

1^1=1^2=1mod2
2^2=2^4=0mod2
...

this is my candidate solution, is there something wrong here?

@TobiasKildetoft

Consider the polynomial ring $R[x]$. $Z[x]_2$, an instance of $k Z[x]$, is a subring of $R[x]$, an ideal of $R[x]$. The task is to show that an ideal exist with some k>0 of the polynomial ring $R[x]$.

Is this "something in general"?

@TobiasKildetoft I am trying to reformulate the task

"Let R be a finite, but not necessarily commutative ring, and let x\in R. Show that there exists k\in \mathbb N_{>0} such that x^k=x^{2k}$."

in terms of ideals so that I could solve it "in general" by ideal criterion.

Is this wrong "in general" approach here?

@TobiasKildetoft
Consider $n,m\geq 1$ with $x^m=x^{m+n}$.
With $k\geq m$, $x^{k+n}=x^k \forall k\geq m$.
With $n\geq m$, $x^{2n}=x^{k+n}=x^k \forall k\geq m, n \geq m.$
With $k=n$,

$$x^{2k}=x^k \forall k\geq m.$$

@TobiasKildetoft With $k=n$,

$$x^{2k}=x^k \exist k\geq m.$$

@TobiasKildetoft With $k=n$,

$$x^{2k}=x^k \forall n\geq m.$$

@TobiasKildetoft With $k=n$,

$$x^{2k}=x^k$$

@TobiasKildetoft With $k=n$,

$$x^{2n}=x^n$$

@TobiasKildetoft

Consider $n,m\geq 1$ with $x^m=x^{m+n}$.
With $k\geq m$, $\forall k\geq m \quad x^{k+n}=x^k$.
With $n\geq m$, $\forall n \geq m \exists k\geq m \quad x^{2n}=x^{k+n}=x^k.$
With $k=n$, $\forall n \geq m \exists k\geq m \exists k=n \quad x^{2n}=x^n$ so

$$\forall n \geq m \exists k\geq m \exists k=n \quad x^{2n}=x^n$$

which is equivalent to

$$\exists k\in\mathbb N_{>0} \quad x^k=x^{2k}$$

with a finite ring R and $x\in R$.

$\forall n \geq m \exists k\geq m \exists k=n \quad x^{2n}=x^n$ is equivalent to

$$\forall k \geq m \quad x^{2k}=x^k,$$

I cannot yet see where the sufficiency, @TobiasKildetoft.

It is a very fascinating question to me.
Associate to each sequence $a=\{\alpha_n\}$ in which $\alpha_n$ is 0 or 2, the real number $$x(a)=\sum_{n=1}^{\infty}\frac{\alpha_n}{3^n}$$. Prove that the set of all $x(a)$ is precisely the cantor set@BalarkaSen

What is sufficient to prove that $A=\{k+l\sqrt 3 \mid m,n\in\mathbb Z\}$ is closed under multiplication?

Some condition with ideal $\langle A\rangle$?

Or (k+l\sqrt 3) (k+l\sqrt 3) = k^2+kl\sqrt 3+(l\sqrt 3) k +3*l^2 =K+L\sqrt 3 with some K,L\in\mathbb Z and then considering all linear combinations?

@Balarka A kind of observation:
$$\sum_{n=1}^{\infty}\frac{\alpha_n}{3^n}=\frac{\alpha_1}{3}+\frac{\alpha_2}{3^2}+\frac{\alpha_3}{3^3}+...$$ . Now if I set $\alpha_2$=2 and the rest as $0$ I get $\frac{1}{3}$. Now similarly if I set $\alpha_1=2$ and the rest all $0$ I get $\frac{2}{3}$. This is just, when for the first time $(\frac{1}{3},\frac{2}{3})$ is removed from $[0,1]$. Similarly we can go for the other end points of the intervals.