How to solve this one-variable equation in a free group? Let $F_n$ be a free group, $u, v \in F_n$, where $[u,v] \neq 1 $. Trying to show that the following equation does not have a solution in $F_n$ : 
$$ [v,x] = [u,v] .$$
Any ideas are appreciated. I seem to have a proof but it's ugly (combinatorial, cancellation theory etc.) and not short (so, more likely to have a silly mistake :) )
Immediate thing to notice is that if $x=w$ is a solution, then so is the "line" $x=C(v)w$, where $C(v)$ is a centralizer of $v$. 
Also, not difficult to show that there are no solution in a free group of rank two generated by $u$ and $v$. 
thanks!
 A: This is not a complete solution, but it may help. Suppose the equation has a solution. Then use the Wicks forms (Lyndon, R. C.; Wicks, M. J.
Commutators in free groups. 
Canad. Math. Bull. 24 (1981), no. 1, 101–106.): there exist reduced decompositions $u\equiv a\bar c, x\equiv cb$, $v\equiv p\bar q, u\equiv qr$ such that $[u,x]\equiv abc\bar a\bar b\bar c, [v,u]\equiv pqr\bar p\bar q\bar r$ where $\equiv$ means there are no cancellations in these words, and $\bar z=z^{-1}$. Since $abc\bar a\bar b\bar c=pqr\bar p\bar q\bar r$, we have that either $\bar c$ is a suffix of $\bar r$ or vice versa, etc. The point is that with Wicks forms you can work as in the free semigroup (no cancellations).
A: Hi Mark, just wanted to share the solution to this problem as well as a method used. 
I found quite a powerful tool buried at the end of this paper 
"A Classification of Fully Residually Free Groups of Rank Three or Less",
Benjamin Fine, Anthony M Gaglione, Alexei Myasnikov, Gerhard Rosenberger, Dennis Spellman.
Here is the result that follows from the proofs/discussion of that paper and stated in the Section 8, Theorem 6: 
Let $F_{u,v} = \langle u ,v \rangle $ be a free group of rank two and suppose that $a(u,v)$ and $b(u,v)$ are non-trivial and not proper powers in $F_{u,v}$. If $a(u,v)$ and $b(u,v)$ are conjugate in some overgroup $F_{u,v} < F$ then either $a(u,v)$ and $b(u,v)$ are already conjugate in $F_{u,v}$ or the element $ta(u,v)t^{-1}b(u,v)^{-1}$ is a primitive in a free group of rank three $F_{3} = \langle t,u,v \rangle $.
Let me now demonstrate that the equation $[x,u]=[u,v]$ has no solutions in $F$. 
Suppose the above equation has a solution $x=\omega$. We have then $\omega^{-1}u^{-1}\omega =[u,v]u^{-1}$, so that elements $u^{-1}$ and  $[u,v]u^{-1}$ are conjugated in $F$. First, notice that $u^{-1}$ and  $[u,v]u^{-1}$ are neither proper powers nor conjugated in $F_2= \langle u,v \rangle $. 
Hence, by the above result, element $e=t^{-1}u^{-1}tu[v,u]$ must be primitive in $F_3 = \langle t,u,v \rangle$. However element $e$ belongs to the derived subgroup $F_3 '$ and as such can not be primitive.
