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Let $\Gamma$ be a hyperbolic group. $g$, $\gamma\in \Gamma$ be fixed non-trivial elements and $g$ and $\gamma$ do not commute. Assume $\gamma$ has infinite order and $g$ and $\gamma$ generate a free semi group. Does the equation $g\gamma^n=h^m$ only have finitely many solutions of triples $(n,h,m)\in \mathbb N\times\Gamma\times \mathbb N$ with $m\neq \pm 1$?

More generally, let $(X,d)$ be a Gromov hyperbolic space, and $\Gamma$ be a subgroup of the group of isometries of X: $Isom(X)$. What is the answer to the same question when $\gamma$ is a hyperbolic isometry of $X$?

Let $\Gamma$ be a hyperbolic group. Let $g$, $\gamma\in \Gamma$ freely generate a non-abelian semigroup (in particular, they don't commute and have infinite order). Does the equation $g\gamma^n=h^m$ only have finitely many solutions of triples $(n,h,m)\in \mathbb N\times\Gamma\times \mathbb N$ with $m\neq \pm 1$?

More generally, let $(X,d)$ be a Gromov hyperbolic space, and $\Gamma$ be a subgroup of the group of isometries of $X$: $\mathrm{Isom}(X)$. What is the answer to the same question when $\gamma$ is a hyperbolic isometry of $X$?

Let $\Gamma$ be a hyperbolic group. $g$, $\gamma\in \Gamma$ be fixed non-trivial elements and $g$ and $\gamma$ do not commute. Assume $\gamma$ has infinite order. Does the equation $g\gamma^n=h^m$ only have finitely many solutions of triples $(n,h,m)\in \mathbb N\times\Gamma\times \mathbb N$ with $m\neq \pm 1$?

More generally, let $(X,d)$ be a Gromov hyperbolic space, and $\Gamma$ be a subgroup of the group of isometries of X: $Isom(X)$. What is the answer to the same question when $\gamma$ is a hyperbolic isometry of $X$?

Let $\Gamma$ be a hyperbolic group. $g$, $\gamma\in \Gamma$ be fixed non-trivial elements and $g$ and $\gamma$ do not commute. Assume $\gamma$ has infinite order and $g$ and $\gamma$ generate a free semi group. Does the equation $g\gamma^n=h^m$ only have finitely many solutions of triples $(n,h,m)\in \mathbb N\times\Gamma\times \mathbb N$ with $m\neq \pm 1$?

More generally, let $(X,d)$ be a Gromov hyperbolic space, and $\Gamma$ be a subgroup of the group of isometries of X: $Isom(X)$. What is the answer to the same question when $\gamma$ is a hyperbolic isometry of $X$?

Let $\Gamma$ be a hyperbolic group. $g$, $\gamma\in \Gamma$ be fixed non-trivial elements and $g$ and $\gamma$ do not commute. Is the equation $g\gamma^n=h^m$ only have finitely many solutions of triples $(n,h,m)\in \mathbb N\times\Gamma\times \mathbb N$ with $m\neq \pm 1$?

More generally, let $(X,d)$ be a Gromov hyperbolic space, and $\Gamma$ be a subgroup of the group of isometries of X: $Isom(X)$. What is the answer to the same question when $\gamma$ is a hyperbolic isometry of $X$?

Let $\Gamma$ be a hyperbolic group. $g$, $\gamma\in \Gamma$ be fixed non-trivial elements and $g$ and $\gamma$ do not commute. Assume $\gamma$ has infinite order. Does the equation $g\gamma^n=h^m$ only have finitely many solutions of triples $(n,h,m)\in \mathbb N\times\Gamma\times \mathbb N$ with $m\neq \pm 1$?

More generally, let $(X,d)$ be a Gromov hyperbolic space, and $\Gamma$ be a subgroup of the group of isometries of X: $Isom(X)$. What is the answer to the same question when $\gamma$ is a hyperbolic isometry of $X$?