Proving the commutativity of a ring with identity Let $\left(A,\ +,\ \times \right)$ be a ring such that $1_A$ is the identity in $\left(A,\ \times \right)$.
Let $a$ and $b$ be two elements in $A$ such that:
$$\begin{cases}
ab+ba=1_A \\
a^2b+ba^2=a
\end{cases}$$
I need to prove that ab=ba. Since the elements of a ring are per def. not invertible I did not know how I can approach it. Any hints how to get started
I'm also interested knowing under which assumptions are the elements of $A$ $\times-$invertible
 A: If the equations hold for any $a,b$, then we can take $b=1$ to get $a+a=1$. That holds for any $a$, so we have $ab+ab=1$ for any $a,b$.
But $ab+ba=1$, so $ab=ba$. That proves commutativity.
But multiplying $a+a=1$ by $b$ we get $ab+ab=b$. We also have $ab+ba=1$, and $ba=ab$, so $b=1$ for all $b$. 
In other words the ring is only commutative because it is trivial.
A: From $ab+ba=1_A$, we get $a^2b+aba=a$ and subtracting the second equation gives $(ab-ba)a=0$. Now, consider $a$ and $(ab-ba)$. Then, this pair satisfies your first equation as well. So $$(ab-ba)a+a(ab-ba)=1.$$ Since $(ab-ba)a=0$, we have $a(ab-ba)=1$. Now, multiply $(ab-ba)a=0$ by $(ab-ba)$ from the right. So, $$0=(ab-ba)a(ab-ba)=(ab-ba)1=ab-ba.$$ Hence, $ab=ba$.
A: This proves it for any particular $a$ and $b$ that satisfy these equations regardless of whether the equations are satisfied by all elements of the ring. Multiplying the first equation by $a$ once on the left and another on the right and solving the equations each time yields $$ a(ab-ba)=(ab-ba)a=0 $$
Multiplying the first equation on the left by $a$ and on the right by $b$ yields $$aabb+abab=ab \tag{1}$$And multiplying it on the left by $b$ and on the right by $a$ gives $$baba+bbaa=ba \tag{2} $$ Multiplying $(ab-ba)a=0$ on the right by $b$ gives $abab=baab$ and multiplying $a(ab-ba)=0$ on the left by $b$ gives $baab=baba$ so $baba=abab$ And what remains is to show that $aabb=bbaa$. To do this, we multiply $ a(ab-ba)=0$ on the right by $b$ to get $aabb=abab$. Now multiply the equation $(ab-ba)a=0$ on the left by $b$ to get $bbaa=baba$. So $bbaa=aabb$.
