Equations in free groups satisfying all elements 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?
 A: Yes. Denote by $a,b$ the first two free generators of $F$.
The case $w\in F\cup\langle x\rangle$ is clear. So we can suppose, after conjugation, that $w=u_1x^{n_1}\dots u_kx^{n_k}$ with $k\ge 1$, $u_i\in F\smallsetminus\{1\}$, and $n_i\in\mathbf{Z}\smallsetminus\{0\}$. 
Choose $n$ large enough such that the reduced form of $a^nu_ia^{-n}$ starts with $a$ and finishes with $a^{-1}$ for all $u_i$ that is not a power of $a$. In all cases, the reduced form $v_i$ of $a^nu_ia^{-n}$ starts and finishes with $a^{\pm 1}$. Then $q=\prod v_ib^{n_i}$ is a reduced form and hence $\neq 1$. We have
$$a^{-n}qa^n=a^{-n}\left(\prod a^nu_ia^{-n}b^{n_i}\right)a^n=\prod u_ia^{-n}b^{n_i}a^n=w(a^{-n}ba^n)\neq 1.$$
Added: This actually shows that the canonical map $F\ast\langle x\rangle\to F^F$, $w\mapsto (t\mapsto w(t))$ is injective. Indeed this is a group homomorphism ($F^F$ being a group for the target operation: infinite power of the group $F$ by the index set $F$). Of course this fails when $F$ has rank $r\le 1$: for $r=0$, $x$ is the kernel, while for $r=1$ $axa^{-1}x^{-1}$ is in the kernel. The above shows injectivity of the projection to $F^X$, where $X=\{a^nba^{-n}:n\in\mathbf{Z}\}.$
A: 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.
