$A$ is a unitary ring with $xy=1$ implies $yx=1$, prove that if $a^2=3b^2$ and $3ab=1+4ba$, then $ab=ba$ Let $(A,+, \cdot)$ be a unitary ring with the property that if $xy=1$ for $x,y \in A$, then $yx=1$. Let $a,b \in A$ with $a^2=3b^2$ and $3ab=1+4ba$. Prove that $ab=ba$.
We easily get that $(a-2b)(2a+3b)=1$, so, by hypothesis, we have that $(2a+3b)(a-2b)=1$. From here we get that $3ba=1+4ab$.
Then, $3ab-4ba=1=3ba-4ab$, and so $7ab=7ba$.
Now, $9ab=3+12ba$ and $12ba=4+16ab$, so $9ab=3+4+16ab$, thus $7ab=-7$.
Of course, we have $7ba=-7$ too.
Next, from $3ab=1+4ba$ and $3ba=1+4ab$ we get that $1+ba=3ab-3ba=-1-ab$, so $ab+ba=-2$.
We can go further and write that $aba+ba^2=-2a$ and $a^2b+aba=-2a$.
We get that $a^2b=ba^2$ and, analogously $ab^2=b^2a$.
Now, we can easily prove that $ab=6ba+5$, $ba=6ab+5$, $2ab=5ba+3$ and $2ba=5ab+3$. Here I'm stuck.

Counterexample: $A = \mathbb{F}_7^{2\times 2}$ (the $2\times 2$ matrix ring over $\mathbb{F}_7$), $a = \left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right)$, $b = \left(\begin{matrix} 0 & 0 \\ 5 & 0 \end{matrix}\right)$.
