<s>[INST] A ring with $8$ elements 
Let $(A,+,\cdot)$ be a unitary ring with 8 elements. Prove that :
$a)8=0$  and $k\neq 0$, for any odd $k$.
b)If $\exists a\in A$ such that $a^3+a+1=0$, then $a\neq 0$, $a\neq 1$, $2=0$, $a^7=1$ and $A$ is a field.

a) is pretty straightforward, it is obvious that $8=0$ from Lagrange's theorem in the additive group $(A,+)$, and then if we had some odd $k$ such that $k=0$ then we would have that $1=0$, which is a contradiction to $|A|=8$. 
For b) the most difficult part is showing that $\operatorname{char}A=2$.
The fact that $a\neq 0$ and $a\neq 1$ is really trivial.
Proving that $a^7=1$ is easy as well if we know that $2=0$. From here we easily get that $A$ is a field and we are done.
The only progress I made towards showing that $2=0$ was that $a$ is invertible(I guess it might help) because we have that $a(-a^2-1)=(-a^2-1)a=1$.
Obviously, we also have that $\operatorname{char}A\in \{2,4,8\}$, but I didn't get any further.
 [/INST] Suppose that $a = n \cdot 1$ for some $n \in \mathbb{Z}$. Then $\left( n^{3} +n +1 \right) \cdot 1 = 0$, and hence $n^{3} +n +1$ is even since it is a multiple of $\text{car}(A) \in \lbrace 2, 4, 8 \rbrace$. As this is not possible, $a \notin \mathbb{Z} \cdot 1$. In particular, $\mathbb{Z} \cdot 1 \neq A$, and so $\text{car}(A) \in \lbrace 2, 4 \rbrace$.
Now, assume that $\text{car}(A) = 4$. Then $A = \lbrace 0, 1, 2 \cdot 1, 3 \cdot 1, a, a +1, a +2 \cdot 1, a +3 \cdot 1 \rbrace$ as these elements are pairwise distinct. It follows that $2 \cdot a \in \mathbb{Z} \cdot 1$ since $a \notin \mathbb{Z} \cdot 1$, and hence $2 \cdot a \in \lbrace 0, 2 \cdot 1 \rbrace$ because $4. a = 0$. Therefore, we have either $2 \cdot a = 0$ or $2 \cdot (a +1) = 0$. This is impossible since both $a$ and $a +1 = -a^{3}$ are invertible and $\text{car}(A) = 4$.
Thus, we have proved that $\text{car}(A) = 2$.
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