If $x ^ 6 = x$, prove that $x ^ 2 = x$, in a ring

Consider the case where the ring has a unit (if not, then one could consider $R$ as a $\mathbb{Z}$-algebra, but the details would change in that case).

Observe $2^6=2$ and $3^6=3$. In other words, $64=2$ and $729=3$. So $62=0$ and $726=0$. Since $\gcd(62,726)=2$, it follows that $2=0$ (by repeated subtraction).

Therefore, we have a ring where $2=0$. Now, consider $(x+1)^6=(x+1)$. The LHS expands as $$ x^6+6x^5+15x^4+20x^3+15x^2+6x+1=x+1. $$ Simplifying the even coefficients, it follows that $$ x^6+x^4+x^2+1=x+1. $$ Since $x^6=x$, we know that $x^4+x^2=0$ or that $x^4=x^2$. Since $x^4=x^2$, by multiplying by $x^2$, we have $x^6=x^4$ so $x^6=x^2$, but since $x^6=x$, $x=x^2$.

(There are a few places in this calculation, where one should be careful to make sure that I'm not cheating, but the ideas should work, at least in the case where $R$ has a unit.)

See Theorem $2.6$ in the lecture notes here. If $x^6=x$ for all $x$, then it is easy to show that $x^3+x^5=0$ for all $x$. Also $-x=(-x)^6=x^6=x$, so that $x=x^6=x\cdot x^5=x\cdot x^3=x^4$. A similar step shows that $x^2=x$ and that $R$ is commutative.

Hint $\ j := 2^6\!-2 = 0 = 3^6\!-3=:k \,$ so by Bezout their gcd $\,\color{#c00}2 = (j,k) = nj\!+\!mk = 0$.

Similarly $\ f(x) = x^6\!-x = 0 = f(x\!+\!1)\,$ so over $\,\color{#c00}{\Bbb F_2}\,$ their gcd $= x^2-x = 0.\ \ \ $ QED