A nonzero unital division ring $D$ A nonzero unital ring  $D$ in which every nonzero element is
invertible is called a division ring.
I think when the ring is divided, the following two are equivalent?Is it right?

1:  For all $a, b \in  D$ with $a \neq 0$, the equations $ax = b$ and $ya = b $ have
  unique solutions in $D$.
2:  For all $a, b \in  D$ with $a \neq 0$, the equation $ax = b$  has a solution in $D$.

Can the below statement  be obtained from The second case above?

3:$D^{2} \neq 0$ and $D$ has no right ideals other than $0$ and $D$.

Does the phrase 3 alone indicate a unital ring??

As I showed you here, (2) implies there are only trivial right ideals.
Then here I showed you that as long as multiplication isn't zero ($D^2\neq \{0\}$) that having only trivial right ideals implies $D$ has an identity.
Then obviously there are no proper left ideals either, and $ya=b$ has a solution when $a$ is nonzero.
Finally, the solutions to $ax=b$ and $ya=b$ are unique since left and right multiplication by nonnegative elements is injective again as previously mentioned to you here. So (1) does indeed follow from (2) as long as you are still talking about $D^2\neq \{0\}$.

I think when the ring is divided,

I don't know for sure what you mean, but I suspect you mean "the ring is divisible" like a divisible module, meaning $D=xD$ for any nonzero $D$, which is just a fancy way of saying $D$ is a division ring when $D^2\neq\{0\}$.

Does the phrase 3 alone indicate a unital ring??

At first glance, perhaps no, but it is true, and you already asked this and got the positive answer.
