Goldbach-type problem: the valuation of irreducible elements of a subgroup of $\mathbb{Q}_{> 0}$

Question

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 ?

Answer 1

Assuming the Hardy-Littlewood conjecture for pairs of primes, we have the bound $v(r) \leq 5$. If $v(r) \geq 6$, we can split the prime factors of $r$ into two piles, each of size at least $3$, such that the signed sum of the primed factors in each pile is even. Take that even number and find two primes whose difference is that number. Add those primes appropriately to the numerator and denominator of both piles. They're in $\mathcal K$, and smaller. So it's not irreducible.

Answer 2

Yes, here is a proof by induction on an integer $n$, assuming the Goldbach conjecture, or alternatively, just using the Goldbach's weak conjecture (see at the end), proved in 2013, by 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.

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Edit: We can also just using the Goldbach's 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.