Let $F$ be a free group of finite rank, and $p, b \in F$, where $b$ is a root element (i.e. not a proper power). I have a case where $p^{n_k} = V_{n_k}^{-1}b^{-1} V_{n_k} \cdot U_{n_k}^{-1}b U_{n_k}$, for some $n_k \in \mathbb{Z}$ ...i.e. some powers of $p$ are products of two conjugates of $b$ and $b^{-1}$.

What can be said about $p$ and $b$ ?

Some immediate implications: By Karras-Magnus-Solitair, since $b$ is root, one-relator group $ < F \ | \ b >$ is torsion-free, so if a proper power of $p$ is in $ncl(b)$ then $p\in ncl(b)$.

Also, by going to an abelianization of $F$, it is clear that $p\in \[ F, F \]$.

I was hoping that $p$ is conjugate to $b$...by Magnus, if we can show that $b$ is also in the the normal closure of $p$ then that would be the case.

I also have a somewhat related general question. Say, if we have $b$ as normal root of $p$ i.e. $p \in ncl(b)$, so that $p = \displaystyle{\prod_{i=1}^{n}} T_i^{-1} b^{\epsilon _i} T_i$, where $\epsilon _i = \pm 1$. Clearly the above product is not unique. Is there a notion of associating to $p$ a minimal integer $n_p \geq 1$ so that $p$ can be decomposed to the product $p = \displaystyle{\prod_{i=1}^{n_p}} T_i^{-1} b^{\epsilon _i} T_i$? For example if $n_p=1$ then $p$ and $b$ are conjugate.

Apologies if I missed something obvious or the question(s) doesn't make sense.

Thanks!