On a ring $R$ where $x^2 = x$, we must have that $2x = 0$ and $R$ is commutative I want to show that on a ring $R$ where $x^2 = x$ for all elements $x$, we must have that $2x = 0$ and $R$ is commutative. Does my answer look sound?
Well, if $x = 0$ we have that $x^2 = x$ is trivially true.
If $x \neq 0$ then $x^2 = x$
$\Longleftrightarrow x \cdot x = x$
$\Longleftrightarrow x \cdot x \cdot x^{-1} = x \cdot x^{-1}$
$\Longleftrightarrow x = 1$
I.e. the only non-zero element that we can possibly find in $R$ is $1$.
So $R = \{0, 1\}$ and as $1 + 0 = 1$ the only way we can satisfy the additive inverse ring axiom is if $1 + 1 = 0$.
I.e. $2(1) = 0 \Longleftrightarrow 2x = 0$
$R$ must be commutative because if we take two elements from $R$, e.g. take $a = 0$ and $b = 0$ such that $ab \neq ba$ we will have that $ab$ is either $0$ or $1$.
$ab = 1$ is not possible as $a = b = 0$. 
If $ab = 0 \implies ba = 1$ and this is not possible as again $a = b = 0$.
 A: You can't use the existence of multiplicative inverses because they don't necessarily exist (consider the inverse of $2$ in the ring $\mathbb{Z}/4\mathbb{Z}$), but there's a different approach that only needs the existence of the multiplicative identity $1$; since $x^2=x$ for all $x$, then likewise $(x+1)^2=x+1$ for all $x$.  Now, expand out the LHS and do some additive cancellations (since addition is a group operation and so has a proper inverse) and you should see the conclusion.  Once you have $x+x=0$ for all $x$ then commutativity can be proven as a separate step.
A: The party line on this problem, since I am such a good party member!:  since for all $x \in R$ we have $x^2 =x$, it follows that
$(x + y)^2 = x + y, \tag{1}$
or
$x^2 + xy + yx + y^2 = x + y, \tag{2}$
or
$x + xy + yx + y = x + y, \tag{3}$
or
$xy + yx = 0, \tag{4}$
whence taking $y = x$,
$2x = x + x = x^2 + x^2 = 0, \tag{5}$
and thus we also have
$x = -x \tag{6}$
holding for all $x \in R$.  Now we see that
$xy = -yx = yx \tag{7}$
follows from (4) and (6).  Equations (4)-(7) express the requisite conclusions.QED
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
