In English: Given a ring $\left( A,+,\cdot \right)$ that meets both of the following conditions: (1) $A$ is not a field, and (2) For every non-invertible element $x$ of $ A$, there is an integer $m>1$ (depending on $x$) such that $x=x^2+x^3+\ldots+x^{2^m}$. Show that (a) $x+x=0$ for every $x \in A$, and (b) $x^2=x$ for every non-invertible $x\in A$.

Actually, $(1)$ should be "$A$ is not a skew-field". This condition gives us at least one non-invertible element.

Having proven (a), (b) is not so difficult.

(a) As usual if $n\in \mathbb{N}$ we denote by $n_{A}$ or simply by $n$ , $n:=\underbrace{1+1+\ldots+1}_{n}$. Let $x\in A$ be non-invertible and $x\neq 0$. It holds: $x=x^2+x^3+\ldots+x^{2^{m}} \, \Rightarrow 1+2x=1+x+x^2+\ldots + x^{2^m} \Rightarrow$ $ \text{(3)} x(1-2x+x^{2^m})=0$ Case 1. $2_{A}$ is invertible. Using (3) it follows that $2x-1-x^{2^m}$ is non-invertible since otherwise we will get $x=0$. It means that we can plug in (2) , $2x-1-x^{2^m}$ instead of $x$. Thus we have: $\text{(4)} 2 = x\cdot Q(x)$. (here $Q$ is a polynomial with integer coefficients) Because $2$ commutes with $x$, $2^{-1}$ will also commute with $x$ and we get: $x\cdot Q(x)\cdot 2^{-1}=Q(x)\cdot 2^{-1}\cdot x =1$ , showing that $x$ is invertible, a contradiction. Case 2. $2_A$ is non-invertible and not equals $0$. Then pluging $2$ instead $x$ in (3) it follows: $2(2^{2^m}-3)=0$ Let $N$ is the least integer for which $N\cdot 1=\underbrace{1+1+\cdots+1}_{N}= 0$. Then $N \mid 2(2^{2^m}-3)$. If $N =2$ it follows $2=0$, a contradiction. If $N$ is odd, $2$ will be invertible, a contradiction. Let $N=2k$ , where $k>1$ is odd. Thus $k+k=2\cdot k=0$ , therefore $k\cdot(k-1)=0$, because $k-1$ is even. Hence $k-1$ is non-invertible and there exists some $n\in \mathbb{N}^*$ such that: $(k-1)=(k-1)^2+(k-1)^3+\ldots +(k-1)^{2^n} \, \Rightarrow $ $k\cdot \ell = 2$ , for some $\ell \in \mathbb{Z}$. If $\ell$ is even, we get $2=0$, a contradiction. If $\ell$ is odd it follows $k=2$, a contadiction. Thus, the only remaining case is $2=0$, i.e. $1+1=0$, which means $x+x = x\cdot(1+1)=0\,,\, \forall x\in A$. (b) For every non-invertible $x\in A$ , (3) implies $x\cdot(x^{2^m}-1)=0 \,\Rightarrow x=x^{2^m+1} $ , for some $m\in \mathbb{N^*}$. We have: $x\cdot(x-1) \cdot (x\cdot Q(x)+1)=0$ , where $Q$ is a polynomial with integer coefficients. It is easy to see that $x^{2}-x$ can not be invertible. Getting to the power $2^m$ we have: $\text{(5)} (x^2-x)^{2^m}= x^{2^{m+1}}- x^{2^m}=0 $ . (here we use $x=x^{2^{m}+1}$). On the other hand, because $x^2-x $ is non-invertible, there will exist $n$ such that $(x^2-x)^{2^{n}+1}=(x^2-x)$ , and combining with (5) we get $x^2-x=0$. Comment. Yesterday I posted a solution (instead of post #4), but infortunately, when I did a second glance, it had a flaw in excerpt (a). The point was in case 2, where we took the min $N$ for which $N\cdot 1=0$. Then $N \mid 2(2^{2^m}-3)$ and I automatically presumed that $N$ was a prime, which is OK for fields (or skew-fields), but could not be so when $A$ is just a ring. So, the case $N=2k$ , where $k>1$ is odd, remained untouched. It was fixed in this post.