About rings having finite number of non-invertible elements 
Let $(A, +, \cdot)$ a ring and $D$ the set of non-invertible elements
  of $A$. We know $a^2=0, \forall a \in D$. 
Prove:
  
  
*
  
*$axa=0, \forall x \in A, \forall a \in D$
  
*If $D$ is finite having at least two elements, then there is $c \in D, c \ne 0$ such that $ac=ca, \forall a \in D$
  


I was able to prove 1) as follows:
First, let's notice $ax \in D, \forall x \in A$ therefore $ax + a \in D$. 
Then $0 = (ax + a)^2= (ax)^2 + aax + axa + a^2=axa$
I can't prove the second one. Any help in appreciated.
UPDATE
To prove $ax \in D$ suppose $ax=y$ where $y$ is invertible. Then, $aax=ay$ therefore $ay=0$ so $a=0$ and $y=0$ contradiction
UPDATE 2
Example of such ring: $\mathbb{Z}_4$
 A: Define $[x,y]=xy-yx$ in $R$.
Claim I: For all $a\in D$ and $x\in R$ we have $a[a,x]=[a,x]a=0$. In particular $[a,[a,x]]=0$. Proof: Immediate consequence of part (a).
Claim II: If $a,b\in D$, then $[a,b]=ab-ba\in D$.
Proof by contradiction: Suppose $x$ is a two-sided inverse to $[a,b]$. Then $$1=[a,b]x\Longrightarrow a=a[a,b]x=0$$
This means $0=[a,b]$ is invertible which is absurd. Also $[a,b]^2=0$ is again an immediate consequence of claim I. Therefore $[a,b]\in D$.
Suppose $D$ has $n\geq 3$ elements. Take $a,b\in D$ such that $[a,b]\neq 0$ (if we can't do this there is nothing to prove). Then $[a,[a,b]]=[b,[a,b]]=0$ meaning $[a,b]$ commutes with $a,b$. Choose $c\in D$ such that $[c,[a,b]]\neq 0$ (again if no such $c$ exists we are done), then $[c,[c,[a,b]]]=0$. Also, by Jacobi identity (this bracket is obviously a Lie bracket)
$$0=[a,[c,[a,b]]]+ [c,[[a,b],a]]+[[a,b],[c,a]]=
[a,[c,[a,b]]]
$$
again because of claim I. This shows that $[c,[a,b]]$ commutes with all $a,b,c$. Continue in this fashion until you run out of elements in $D$ (or you have gotten lucky somewhere in the process).
A: Just want to know how are you saying that $ax\in D$ ,$a\in D$  gives you $ax+a\in D$. If you can do so then i guess you can proceed in the following manner to prove B.
Take a non-zero $c$ and set $s=ac-ca $ for any $a$ in $D$. so $as=a^2c-aca=0-0=0$(by the first you have $axa=0$. Now by hypothesis you have $1$ in $D$. so $1+a\in D$ so $1+a+a=1+a+a+a^2=(1+a)^2 \in D$ so $1+a+a=0$.$s+as+as=0$. But you have $as=0$ so $s=0$. Hence $s=ac-ca=0$ .Thus proven 
