Characteristic of a Ring Problem Let $A$ be a ring such that $\underbrace{1+1+\dots+1}_{n}=0$ for natural number $n\geq 2$. If $a\in A$ and there exists natural number $k$ such that $a^k=a+1$, prove that there exists natural number $m$ such that $a^m=1$. My handwork so far: The problem implicitly tell that $A$ has multiplicative identity $1$. Moreover, if $n$ is minimum number such that $\underbrace{1+1+\dots+1}_{n}=0$, then $char(A)=n$. It imply $Order(A)=nc$ for natural number $c$. Could I try this idea ? \begin{align*} a^k&=a+1\\\ a^k-a&=1\\\ a(a^{k-1}-1)&=1 \end{align*} Hence, $a^{k-1}-1$ is multiplicative inverse of $a$. By $n.1=0$, we have $n.a=0$, $na^{k-1}=0$ and $n(a^{k-1}-1)=0 \iff na^{k-1}=n$. But it means $n=0$. Do you have any idea ?

Let $f \in Z_n[x]$ be defined by $f(x) = x^k - x - 1$.

Let $B$ be the subring of $A$, generated by $a$.

Then every element $b \in B$ can be expressed in the form $b=g(a)$, where $g \in Z_n[x]$, and moreover, we can take $g$ such that $\text{deg}(g) < k$, else, since $f$ is monic of degree $k$, we can replace $g$ by its remainder when divided by $f$.

There are exactly $n^k$ polynomials in $Z_n[x]$ of degree less than $k$, hence the ring $B$ is finite.

Thus, the elements $1,a,a^2,a^3,...$ can't be distinct, so we must have $a^j = a^i$, for some $i,j$ with $0 \le i < j$.

As you noted, $a$ is invertible, hence, from $a^j = a^i$, we get $a^{j-i}=1$, which proves the claim.

With regard to your question about $n=0$, in the context where you obtained that result, the symbol $n$ represents the element $n$ in the ring $A$, and you already know it's zero, so it yields no new information.