Difficult ring-theory problem Let $(R,+,\cdot)$ be a ring with at least 2 elements. If we know that $R$ is not a field and $x^2=x$ for any $x \in R$, where $x$ is not invertible, prove that:
a) $a+x$ is not invertible, $\forall a,x\in R$, where $a$ is invertible and $x$ is not invertible, $x \neq0$ 
b)$x^2=x, \forall x\in R$
My solution, which is not correct:

a) Let $U(R)$ be the group of invertible elements from $R$.
Obviously, if $a \in R$ is invertible, then $a^{-1}$ is invertible, too. Also,
 for any $t \in U(R)$ and for any $s \in R-U(R)$, we have that $t \cdot s \in R-U(R)$. $(1)$ 
We suppose $a+x \in U(R)$.
$a+x=a \cdot(1+a^{-1}x)$
Hence $1+a^{-1}x \in U(R)$, which means that $a^{-1}x \in U(R)$, contradiction with $(1)$. So our supposition is false. It follows that $a+x \in R-U(R)$.  (This is wrong, take $\mathbb{Z}_{6}$ for instance)
b) Let $y \in R-U(R), y\neq0$.
From a), we have that $1+y$ is not invertible.
Then $(1+y)^2=1+y\Leftrightarrow 2y=0$.
Let $a$ be an invertible element.
$(a+y)^2=a+y$
$ \Rightarrow a^2+ay+ya+y^2=a+y$ 
$ \Rightarrow a^2+ay+ya+=a$
I thought that $R$ should be commutative, hence $ay+ya=ya+ya=2ya=2y\cdot a=0\cdot a=0$.
So, $a^2=a$, and thus we obtain that $a=1$ is the only invertible element.
Knowing that $x^2=x, \forall x \in R-U(R)$ and that $1^2=1$, we obtained that $x^2=x, \forall x \in R$
 A: EDIT: I hope this is a real answer now:
As Matemáticos Chibchas pointed out, the mere notion of "invertible" means that there is a $1$, $R$ is a unital ring.
Now our ring isn't a field, so there is a not invertible element  $x$. Obviously, $-x$ isn't invertible, either, so $-x=(-x)^2=x^2=x$. Moreover, if $x$ isn't invertible, we have $x(1-x)=0$, i.e. $1+x=1-x$ isn't invertible. This means $1+x=-(1+x)$, i.e. $1=-1$, our ring must have characteristic $2$ (so $\mathbb{Z}_{6}$ doesn't matter, here). If $a$ is invertible, but $x$ isn't, you were right to conclude $a^{-1}x$ is not invertible, so $1+a^{-1}x$ is not invertible, as we have seen. Then, $a+x$ is not invertible , because otherwise, we'd have $(1+a^{-1}x)^{-1}=(a+x)^{-1}a$.
We're done as soon as we show that $1$ is the only invertible element in $R$. So let's still assume $a$ is invertible, and $x$ is not.
First, $ax$ can't be invertible, so $ax\cdot ax=ax$, meaning ($a$ is invertible!)
$$xax=x\tag 1.$$ According to what was shown above, $a+x$ is not invertible, so
$(a+x)^2=a+x$, i.e. $a^2+ax+xa+x^2=a+x$, meaning (remember $R$ has characteristic $2$!)
$$a+a^2+ax+xa=0\tag2.$$ We can rewrite that as $a(1+a+x)=xa$. Multiplying both sides by $x$ from the right and using (1), we get
$$a(x+ax+x^2)=a(x+ax+x)=a^2x=xax=x\tag3.$$
Using the symmetry of (2), we can say as well $(1+a+x)a=ax$, and multiplication by $x$ from the left gives $$xa^2=x\tag4.$$ Now (3) and (4) together mean that $a^2$ and $x$ are commutative. Since $a^2$ must be invertible, and $x$ is not, $a^2+x$ is not invertible, $(a^2+x)^2=a^2+x$, so we must have
$$(a^2+x)^2=a^4+a^2x+xa^2+x^2=a^4+x+x+x=a^4+x=a^2+x.$$ This means $a^4=a^2$, i.e. $a^2(a^2-1)=0$, and that's equivalent (due to the invertibility of $a$) to $a^2-1=(a-1)^2=0$, meaning $$a=1\tag5.$$ (Remember: if $a$ is invertible and $a\neq1$, $a-1$ has to be invertible, because otherwise, $1+(a-1)$ wouldn't be invertible, as was shown above.)
So (5) shows that $1$ is the only invertible element in $R$, and thus we've shown that $x^2=x$ for all $x\in R$.
I was just reminded that I erroneously deleted something from my previous attempt (not yet an answer): a ring with that property has to be commutative. We've seen characteristic $2$ shown above. And as soon as we have shown $x^2=x$ for every element of $R$, we see that $(x+y)^2=x+y$, and that (because of $x^2=x$ and $y^2=y$) implies $xy+yx=0$, i.e. $xy=-yx=yx$. But that's only a conclusion from the claim, we can't use it to prove the claim, that would be circular.
