Slightly weaker than twin prime conjecture, using $\Omega: \Bbb{Q}^{\times} \to \Bbb{Z}$

Let $$\Bbb{Q}^{\times} \xrightarrow{\Omega} \Bbb{Z}$$ be the group hom given by $$\Omega$$ from the study of arithmetic functions (you can extend it uniquely (and homomorphically) to all of $$\Bbb{Q}^{\times}$$.)

It's kernel is generated by $$\{\dfrac{p}{q} : p, q$$ are prime numbers $$\} = \ker \Omega$$. Does the weaker-than twin prime conjecture:

$$\ker \Omega \cap A = \ker \Omega \cap \{\dfrac{n}{n+2} : n \in \Bbb{Z}\}$$ is an infinite set hold?

Assume the intersection is finite, then $$\ker \Omega \cap A' = \ker \Omega \cap \{\dfrac{n}{n-2} : n \in \Bbb{Z}\}$$ is also finite. The set $$B = \{ \dfrac{n^2}{n^2 - 4} : n \in \Bbb{Z}\} \subset AA'$$ elemenwise. So that $$\ker \Omega \cap B \subset \ker \Omega \cap AA'$$. That's as far as I've gotten.

What happens elementwise with respect to $$\cap$$ is:

$$A(B\cap C) \subset AB \cap AC$$

but not the other way around in general. So that:

$$(\ker \Omega \cap A')(\ker \Omega \cap A) \subset (\ker \Omega)^2 \cap AA' \cap (\ker \Omega)A \cap (\ker \Omega)A' = \\ \ker \Omega \cap A\ker \Omega \cap A'\ker \Omega \cap AA'$$

So that in particular, the left side is a subset of $$\ker \Omega \cap AA'$$. So maybe this route is ruled out?

Note that $$A^{-1} = A'$$ by construction.

Other facts: $$\Bbb{Q}^{\times} = \biguplus_{k \in \Bbb{Z}} p^k G$$, where $$G = \ker \Omega$$, for any fixed prime $$p \in \Bbb{P}$$.

Take $$A = \{\dfrac{n -1}{n+1} : n \in \Bbb{Z}, n \neq -1\}$$, then similarly $$G \cap A$$ would be finite.

$$\dfrac{r_1\cdots r_i}{q_1\cdots q_i} = x \in G \cap A \iff \\ (n-1)q_1 \cdots q_i = (n+1) r_1 \cdots r_i, \\ q_j, r_j \in \Bbb{P} \\ \\ \iff \\ \Omega(\dfrac{\text{lcm}(n-1, n+1)}{n-1}) = \Omega(\dfrac{\text{lcm}(n-1, n+1)}{n+1}) \\ \iff \\ d := \gcd(n-1, n+1), \\ \Omega(\dfrac{n+1}{d}) = \Omega(\dfrac{n-1}{d})$$

• Can you explain in more detail what in the world $\Omega$ is supposed to be
– D_S
Jan 2, 2021 at 4:55
• Doesn't ker $\Omega$ contain $\frac{pq}{rs}$ for distinct $p,q,r,s$? Jan 2, 2021 at 4:58
• I'm not convinced this is weaker; I haven't sketched a proof, but I suspect that this can be easily shown to imply the twin prime conjecture. Jan 2, 2021 at 5:42
• @MorganRodgers There could be many numbers in $\ker \Omega \cap \{ \frac{n}{n+2} : n \in \Bbb{Z} \}$ that aren't twin primes. For instance, $\frac{33}{35}$ and $\frac{42}{44}$ are both elements of $\ker \Omega$, but it's hard to see how either or both could be used to get twin primes. Jan 2, 2021 at 6:19
• @RiversMcForge Thank you, I hadn't written a proof or done any computation, but had a hunch (incorrect, as you show) that two composites with the same number of prime factors wouldn't differ by 2. Jan 2, 2021 at 6:24

For any $$B\geq 5$$ there are infinitely many integers $$x$$ with $$\Omega(x) = \Omega(x + 2) = B$$.
This gives a much stronger statement than you ask for; not only are there infinitely many rationals $$\frac{x}{x+2}$$ for which $$\Omega\left(\frac{x}{x+2}\right)= \Omega(x) - \Omega(x+2) = 0,$$ but there are infinitely many if you restrict the pair $$(\Omega(x),\Omega(x+2))$$ to be any $$(B,B)$$ with $$B$$ sufficiently large. This isn't quite enough to guarantee "twin semiprimes," but it's rather close.
Their results seem to be more general; in particular, it seems that they have also proven this where $$2$$ is replaced with any other positive integer, or where $$\Omega$$ is replaced with any sufficiently well-behaved number-theoretic function.
• @StudySmarterNotHarder I think if you replace $2$ with $1$, it holds where $\Omega$ is replaced with any function $f$ for which $$f(p_1^{a_1}\cdots p_s^{a_s})=g(\{a_1,\dots,a_s\})$$ for some function $g$; that is, $f$ depends only on the exponents in the prime factorization of its input. I'm not sure about other "distances" besides $1$, but the specific results given seem to suggest that they should hold in general. I haven't read the paper in depth. Jan 3, 2021 at 4:32