A group $G$ is finitely generated divisible iff $G=(1) $
Attempt: One implication is trivial.
Let's try to prove the forward implication.
I want to prove some particular cases first to get some insight for the bigger picture.
$\color{red}{\text{$G$ is finite}}$ :
Since $G$ is divisible, $\forall a\in G$ and $\forall n\in\Bbb{N}$ , $\exists x\in G$ such that $x^n=a$
For $n=|G|$ , $a=x^n=1$ and identity have $n$ th roots for every $n\in\Bbb{N}$ .
Hence $G=(1) $
$\color{red}{\text{$G$ is cyclic}}$ :
Let $G=<a>$
If $|a|<\infty$ then it boils down to the first case. Hence assume $|a|=\infty$
Since $G$ is divisible, for all $k\in\Bbb{Z} $ ,$a^k$ has $n$ th root for all $n\in \Bbb{N}$ i.e
$x^n=a^k$
Implies $a^{mn}=a^k$ (where $x=a^m$ for some $m\in\Bbb{Z}$)
Implies $a^{mn-k}=1$
Implies $a=(e) $ ( since $|a|$ divide $mn-k$)
$\color{red}{\text{$G$ is finitely generated}}$ :
Let $G=<a_1, a_2,..., a_k>$ is divisible.
For $k=1 $ , it boils down to the above case.
Suppose it is true for $k$.
To show that it is also true for $k+1$.
For $a=\Pi_{i}^{k+1}a_i^{n_i}$ , $\exists x=\Pi_{i}^{k+1}a_i^{m_i}$ such that $$(\Pi_{i}^{k+1}a_i^{m_i})^n=\Pi_{i}^{k+1}a_i^{n_i}$$
Implies $\Pi_{i}^{k}a_i^{nm_i-n_i}a_{k+1}^{{nm_{k+1}}-n_{k+1}}=1$
$\Pi_{i}^{k}a_i^{nm_i-n_i}=a_{k+1}^{-({nm_{k+1}}-n_{k+1})}$
Since $a_{k+1}^{-({nm_{k+1}}-n_{k+1})}\in <a_1,a_2,...,a_k>$ and $\Pi_{i}^{k}a_i^{nm_i-n_i}$ has $n$ th root for all $n\in\Bbb{N}$ implies $a_i=1$ for all $i\in\Bbb{N}_k$ and that implies $a_{k+1}=1$.
Hence $G=(1) $
Let $G$ be a finitely generated divisible group.
Since $G$ is finitely generated abelian group,$G = \mathbb{Z}^r \oplus \mathbb{Z}_{n_1} \oplus ... \oplus \mathbb{Z}_{n_k}$
$r=0$ implies $G$ is finite and finite divisible group is trivial.
For $r\ge 1$ , an element $(a_1, a_2,..., a_k) $ is divisible by every $n\in\Bbb{N}$ implies $a_1=0$ and $a_i\cong 0 \mod (n_i) $
Since $\Bbb{Z}$ is nt divisible, for $r\ge 1$ , $G$ can't be divisible as well.
Guba, V. S., A finitely generated complete group, Math. USSR, Izv. 29, 233-277 (1987); translation from Izv. Akad. Nauk SSSR, Ser. Mat. 50, No. 5, 883-924 (1986). ZBL0631.20025.
In the paper Guba constructs infinite 2-generated divisible groups. These groups, of course, are non-abelian. Proofs are very difficult.