TITLE: Equations in free groups satisfying all elements
QUESTION [8 upvotes]: please help me to solve the following problem.
Let $F$ be a non-abelian free group and $w(x)=1$ be an equation in one variable $x$ ($w(x)$ may contain elements of $F$ as constants). Clearly, one can consider $w(x)$ as an element of free product $F\ast \langle x\rangle$. 
Suppose $w(a)=1$ for all $a\in F$. 
Is it true that $w(x)$ equals $1$ in $F\ast \langle x\rangle$?
Do you know a simple proof? Probably, you can remember the papers which can be useful for this propblem?

REPLY [2 votes]: If $F=\langle a_1,\ldots,a_n\rangle=\langle\underline{a}\rangle$ is finitely generated, we can replace $w$ with a coefficient-free word $w'(x,y_1,\ldots,y_n)=w'(x,\underline y)$ such that $w(x)=w(x,\underline{a})$. Now consider the first-order sentence $\phi$:
$$\exists y_1,\ldots,y_n\forall x\ (w'(x,\underline{y})=1)$$
Then the result follows from Sela's proof of Tarski's problem:

Theorem [Sela, Theorem 3]: All finitely generated non-abelian free groups have
  equivalent first-order theories.

In particular, $F$ satisfies $\phi$ (where $\underline{y}=\underline{a}$ works to satisfiy the existential part of $\phi$) if and only if $F*\langle a_{n+1}\rangle$ satisfies $\phi$.
I'm not sure about the case where $F$ is not finitely generated, but because at most finitely many of the generators of $F$ could appear as coefficients in $w(x)$, you might be able to reduce to this case, depending on what you need this result for.