<s>[INST] Prove that there is a unit $u \in R$ such that $ub = bu = a$ 
Let $R$ be a ring with identity containing elements $a$ and $b$ with $ab = b$ and $b^2 = a$. Prove that there is a unit $u \in R$ such that $ub = bu = a$.
Source: Problem $15.3.6$, Algebra in Action: A Course in Groups, Rings, and Fields by Shahriar Shahriari.

My work:
From $ab = b$ and $b^2 = a$, we have $ab^2 = a^2 =  b^2$. So, $a^2 = a$, i.e. $a$ is idempotent. $u = b$ satisfies $ub = bu = a$, but we don't know if $b$ is a unit. I believe the main reason I'm stuck is that $a$ and $b$ may not be units, but from the given information, we likely want to find $u$ in the form $u = a^\alpha b^\beta$ for some $\alpha,\beta\in \mathbb N$. How do I proceed?
Thank you!
 [/INST] We know that $b^3 = b$ and so $b^4 = b^2$. We may then compute that
$$
(b^2+b-1)^2 = b^4 + 2 b^3 - b^2 - 2 b + 1 = b^2 + 2b - b^2 - 2b + 1 = 1.
$$
Hence $b^2+b-1$ is a unit. Furthermore, we have
$$
(b^2+b-1)b = b(b^2+b-1) = b^3 + b^2 - b = b + b^2 - b = b^2,
$$
so it is the unit we are after.
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