Let $\phi : \mathbb{Q}_{>0} \to \mathbb{Z}$ be the group morphism defined by $\phi(p) = p$ for $p$ a prime number.

It follows that $\phi(\prod_i p_i^{n_i}) = \sum_i n_i p_i$, with $p_i$ a prime number and $n_i \in \mathbb{Z}$.

Let $v: \mathbb{Q}_{>0} \to \mathbb{N}$ be the map defined by $v(\prod_i p_i^{n_i}) = \sum_i \vert n_i \vert$ (with $i \neq j \Rightarrow p_i \neq p_j$).

The map $v$ is not a really a valuation, nevertheless $v = \sum_p \vert v_p \vert$, with $v_p$ the $p$-adic valuation.

Let $\mathcal{K} = \{ r \in \mathbb{Q}_{>0} \ \vert \ r=\prod_i p_i^{n_i} \text{ and } \sum_i n_i p_i = 0 \} = ker (\phi) $, a subgroup of $\mathbb{Q}_{>0}$.

**Definition:** An element $r \in \mathcal{K} $ is called **irreducible** if $r \neq 1$ and if : $$ r = \prod_i r_i \text{ (with } r_i \in \mathcal{K}) \Rightarrow \exists i \text{ such that } v(r_i) \ge v(r) $$
**Warning**: The notion of irreducible defined above is **different** with the notion of "irreducible fraction".

**Example**: Let $(p,p+2)$ be twin primes, then $r=\frac{2p}{p+2} $ is an irreducible element of $\mathcal{K}$ and $v(r) = 3$.

Question: Let $r \in \mathcal{K} $ be an irreducible element. Is it true that $v(r) \in \{ 3,4 \}$ ?

**Application**: The group $\mathcal{K}$ is generated by its irreducible elements, so (modulo a positive anwser), it's generated by the elements $v$ with $v(r) \in \{ 3,4 \}$.

**Edit** (14/07/14): It's a **Goldbach-type problem**: two days after having posted my question, I've found a proof using the Goldbach conjecture (see my answer below). So an alternative question could be:

Is it possible to answer my question without using the Goldbach conjecture or is it equivalent to it ?

Warning: The notion of irreducible defined above isdifferentwith the notion of "irreducible fraction". For example $8/9$ is not irreducible in the sense above because $\frac{2.2.2}{3.3} = \frac{2.11}{13}.\frac{13}{3.3.7}.\frac{2.2.7}{11}$. $\endgroup$ – Sebastien Palcoux Jul 14 '14 at 15:44