Does the Baumslag Solitar group $B(2,3)$ contain a non-trivial element with arbitrary roots? The Baumslag Solitar groups $B(n,m)$ are defined via the presentation $\langle a,b \mid b a^m b^{-1} = a^n \rangle$. We say that an element $g$ of a group $G$ has an $n$-th root, if the equation $g = x^n$ has a solution in $G$ and an element $g$ has arbitrary roots if for every $n\geq 1$ the equation $g = x^n$ has a solution in $G$.
Does the Baumslag Solitar group $B(2,3)$ contain a non-trivial element with arbitrary roots?
I think one can somehow concludes this from Britton's Lemma/ Normal Form.
 A: The answer is no. Let us show something stronger: for every $u\neq 1$, the set of $n$ such that $u$ has an $n$-root is bounded. Let me show this for $$G=\mathrm{BS}(m,n)=\langle t,x\mid tx^mt^{-1}=x^n\rangle$$ whenever neither $m/n$ nor $m/n$ is an integer.
Indeed, $G$ acts on its Bass-Serre tree $T$, with a vertex $v_0$ with stabilizer $\langle x\rangle$. So every vertex $v$ has cyclic stabilizer, with a unique generator $\xi_v$ conjugate to $x$.
Let $u$ satisfy the previous property. This property forces $u$ to act as an elliptic element (indeed if $u$ were loxodromic of translation length $R$, it could not have $p$-roots for $p>R$).
The Bass-Serre tree has a unique height function $h$ satisfying: $h(tv)=h(v)+1$ and $h(xv)=h(v)$ for all vertices $v$. Then every vertex $v$ has exactly $n+m$ neighbors, consisting of two $G_v$-orbits: one orbit of $n$ neighbors $v'$ with $h(v')=h(v)+1$ (called right neighbors) and one orbit of $m$ neighbors $v'$ with $h(v')=h(v)-1$ (called left neighbors). If $v$ is a vertex and $v'$ is a right neighbor, then we have the formula $\xi_v^n=\xi_{v'}^m$.
Let $s\in G$ be an elliptic element (for the action on $T$). Let $T^s$ be the tree of fixed points of $s$. For every vertex $v\in T^s$, we have $s\in G_v$, so $s=\xi_v^{q_u(v)}$ for some unique $q_s(v)\in\mathbf{Z}$. If $v,v'\in T^s$ with $v'$ right neighbor of $v$, then by the previous formula, we have $q_s(v')=\frac{m}{n}q_s(v)$. Hence by connectedness of $T^s$, for any two vertices $v,v'\in T^s$ we have $q_s(v')=\left(\frac{m}{n}\right)^{h(v')-h(v)}q_s(v)$.
If neither $m/n$ nor $n/m$ is an integer and $s\neq 1$, it follows that $h(T^s)$ is bounded, and in turn, it follows that $q_s$ is bounded on $T^s$.
Now let $u\neq 1$ have roots of unbounded order. As we have seen, $u$ is elliptic. Then for some $R$ we have $|q_u(v)|\le R$ for all $v\in T^u$.  Suppose that $u=s^N$. So $s$ is elliptic as well, so fixes a vertex $v$, and necessarily $v\in T^u$. It follows that $N\le R$.

Comment: if $n/m=\pm 1$, the same conclusion holds by a distinct (easier) argument, while if exactly one of $m/n$ or $n/m$ is an integer $r$ (so $|r|\ge 2$), the conclusion fails as one can then check that $\mathrm{BS}(m,n)$ then contains an isomorphic copy of $\mathbf{Z}[1/r]$: in $\mathrm{BS}(m,mr)$ we have $x^m=(t^{-p}x^mt^p)^{r^p}$ for all positive integers $p$.

Another fact (to address Derek's comments): let $I$ be an infinite set of positive integers. Write $G^n=\{g^n:g\in G\}$. Say that a group $G$ satisfies Property $P_I$ if $\bigcap_{n\in I} G^n=\{1\}$, and $g^n=1$ implies $g=1$ for every $n\in I$. Then Property $P_I$ passes to group extensions.
If $G$ is free, then it satisfies $P_I$ for every infinite $I$.
If $G=\mathbf{Z}[1/k]$, then it satisfies $P_I$ for every infinite $I$ of integers coprime to $k$.
Define $k=mn/\gcd(m,n)^2$. Then there is an obvious homomorphism $\mathrm{BS}(m,n)\to\mathbf{Z}[1/k]\rtimes_{n/m}\mathbf{Z}$. It follows that for every infinite $I$ of elements coprime to $k$, $\mathrm{BS}(m,n)$ has Property $P_I$. In particular, every nontrivial element is divisible by only finitely many primes.
