Proving $a+a =0$ for Boolean ring 
Suppose $R$ is a Boolean ring. Prove that $a+a=0$ for all $a\in R$.
  Also prove that $R$ is commutative. Give an example (with explanation)
  of a Boolean ring.

From what I know, a Boolean ring is a ring for which $a^2=a$ for all $a\in R$.
Under addition a ring is a commutative group. 
$a + b = b + a$ (commutative)
$(a + b) + c = a + (b + c)$ (associative)
$a + (-a) = 0$ (inverse exists for every element)
$a + 0 = a$ (identity exists)
Where $a,b,c \in R$
But I'm not really sure how to proceed with the proof from here. Any idea?
 A: Hint Applying the definition of Boolean gives $$a + a = (a + a)^2 = a^2 + a^2 + a^2 + a^2 .$$ What does applying it again give?
A: Hint. Compute $(a+1)^2$ in two ways -- once by the Boolean ring property, another time by using the distributive law that works in all rings to multiply out $(a+1)(a+1)$. Then apply $a^2=a$ once more and cancel terms that appear in both results.
A: $a+a=(a+a)^2=a^2+a^2+a^2+a^2=a+a+a+a\Rightarrow 0=2a$
A: Take $x \in R$, with $R$ boolean ring, in particular is a ring, therefore $(x+x) \in R$ and $(x+x)^{2}=(x+x)$. Thus
\begin{align*}
        (x+x)^{2} &= (x+x)(x+x) \\
               &=(x+x)x + (x+x)x \\
               &= x^2 + x^2 + x^2 + x^2  
    \end{align*}
since $x^2 = x$ and $(x+x)^2 = (x+x)$ because $R$ is boolean, then       \begin{align*}
        (x+x)^{2} &= x+x+x +x\\
               &= (x+x) + (x+x)\\
               &= (x+x)  
    \end{align*}
and finally we add both sides the right inverse of $-(x+x)$.
\begin{align*}
        (x+x) + (x+x) -(x+x) &=  (x+x) - (x+x)
    \end{align*}
we conclude
\begin{align*}
       (x+x) = 2x = 0
    \end{align*}
