$x+x^2+x^3=x^4+x^5+x^6$ implies $x^4=x$ in a ring

Question

Let $(A,+,.)$ be a ring s.t. $x+x^2+x^3=x^4+x^5+x^6$ for all $x \in A$. Prove that $x^4=x$ for all $x$ in $A$. Can somebody give me some tips, please?

Answer 1

Here is a variant of the solution by @Servaes that seems simpler to me:

Plugging $x=-1$ gives $2=0$ in $A$.

Therefore, $x+x^2+x^3=x^4+x^5+x^6$ is the same as $x+x^2+x^3+x^4+x^5+x^6=0$.

Multiplying that by $x-1$ gives $x^7-x=0$ and so $x^8-x^2=0$.

Now $x^8-x^2=(x^4-x)^2$.

Let $z=x^4-x$. Then $z^2=0$ and $z+z^2+z^3=z^4+z^5+z^6$ imply $z=0$, and so $x^4=x$.

Answer 2

Hint 1: Plugging in $x=-1$ shows that $2=0$ in $A$.

Hint 2: The given equation can be rearranged to get

$$(x+x^2+x^3)(1-x^3)=0,$$ or equivalently $(x^4-x)(1+x+x^2)=0$. Multiplying through by $x-x^2$ shows that $$(x^4-x)^2=0.$$