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](http://en.wikipedia.org/wiki/Valuation_(algebra)), 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](http://en.wikipedia.org/wiki/Twin_prime), 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 ?



Goldbach-type problem: the valuation of irreducible elements of a subgroup of $\mathbb{Q}_{> 0}$
**Yes**, here is a proof by induction on an integer $n$, assuming the [**Goldbach conjecture**](http://en.wikipedia.org/wiki/Goldbach%27s_conjecture), or alternatively, just using the *[Goldbach's weak conjecture](http://en.wikipedia.org/wiki/Goldbach%27s_weak_conjecture)* (see at the end), *proved* in 2013, by [Harald Helfgott](http://en.wikipedia.org/wiki/Harald_Helfgott).


Note that an element $a \in \mathcal{K}$, $a \neq 1$, is given by an integer $n$ and two different prime partitions of $n$, $p\_1 + \dots + p\_r = q\_1 + \dots + q\_s = n$, with $p\_1 \le \dots \le p\_r$, $q\_1 \le \dots \le q\_s$, and $p\_i$,$q\_j$ prime numbers,
such that $a= \frac{p\_1 \dots p\_r}{q\_1 \dots q\_s}$. 


**Base case**: The smallest $n$ having two different prime partitions is $n\_0=6=2+2+2=3+3$.  

But $a=\frac{2.2.2}{3.3} = \frac{2.11}{13}.\frac{13}{3.3.7}.\frac{2.2.7}{11}$, so it's true for $n\_0 = 6$. 


**Inductive step**: Now, suppose it's true up to $n$, we will prove it's also true for $n+1$.  

Let two different prime partitions of $(n+1)$, $p\_1 + \dots + p\_r = q\_1 + \dots + q\_s = n+1$, and $a= \frac{p\_1 \dots p\_r}{q\_1 \dots q\_s}$ such that $v(a) = r+s \ge 5$. We will show that $a$ is not an irreducible element.  

Without loose of generality, we can suppose that $\forall i, j$ , $p\_i \neq q\_j $. 


**Case 1**: If $p\_r$ and $q\_s$ are odd, for example $p\_r > q\_s$, then $p\_r = q\_s + 2k$ and $s>1$.  

Now $a = \frac{2^k . p\_1 \dots p\_{r-1}}{q\_1 \dots q\_{s-1}}.\frac{ p\_{r}}{2^k.q\_{s}}$.  

**Case 1.1**: If $r>1$ or $k \le 2$, then we can apply the induction.  

**Case 1.2**: If $r=1$ and $k > 2$, then $s \ge 4$, because $v(a) \ge 5$.  

By the **Goldbach's conjecture**, $2k = p+q$ with $p$, $q$ odd prime numbers, then $a = \frac{2^k}{q\_1 \dots q\_{s-1}}. \frac{ p.q}{2^k}. \frac{ p\_{1}}{p.q.q\_{s}}$, and we can apply the induction.


**Case 2**: If $p\_r$ or $q\_s = 2$, for example $p\_r=2$.  

Then $p\_1= \dots = p\_r = 2$ (because $p\_1 \le \dots \le p\_r$), 
and $\forall j$ , $q\_j \neq 2$  

**Case 2.1**: If $s > 2$, then $s \ge 4$. Now, $q\_1 + q\_2 = 2k < 2r$ and  

$a= \frac{2^r}{q\_1 \dots q\_s}= \frac{2^k}{q\_1 . q\_2} . \frac{2^{r-k}}{q\_3 \dots q\_s}$, so we can apply the induction.  

**Case 2.2**: If $s = 2$ and $q\_2 \neq 3$, then $q\_2 = 2k+ 3$ with $k \ge 1$ and  

$a= \frac{2^{r-k}}{3.q\_1}. \frac{2^k.3}{q\_2}$, so we can apply the induction.  

**Case 2.3**: If $s = 2$ and $q\_2 = 3$, then $q\_1 = 3$ and $n=6= n\_0$, it's the base case. 




---


**Edit**: We can also just using the **[Goldbach's weak conjecture](http://en.wikipedia.org/wiki/Goldbach%27s_weak_conjecture)**: 


*Case 1.2*: if $r=1$.  

By Goldbach's weak conjecture, $p\_1 = d\_1+d\_2+d\_3$ with $(d\_i)$ odd prime numbers,  

then $a = \frac{p}{d\_1.d\_2.d\_3}.b$ with $b = \frac{d\_1.d\_2.d\_3}{q\_1 \dots q\_s}$. If $\forall i, j$, $d\_i \neq q\_j$, then we can apply the case 1.1 to $b$, else we can simplify the fraction and apply the induction.


