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Let's look some definitions before we start.

Let $n \in \mathbb{Z}^+$ and $G_n := \{\text{primes } p : p \nmid n \, \wedge \, p<n\}$. Let $U(n)$ be group of units of the cyclic group $\mathbb{Z}/n\mathbb{Z}$ where $(n) := \{nm : m \in \mathbb{Z}\}$. More specifically, $U(n) := \{m \in \mathbb{Z}^+ : 1 \leq m \leq n \, \wedge \, \gcd(m,n) = 1\}$ and the binary opertion is multiplication modulo $n$.

I noticed that $U(n)$ contains some nice properties. Notice that if $m \in U(n)$ (hence, $\gcd(m,n)=1$ and $1 \leq m \leq n$), then $\gcd(m,n-m) = 1$. This property is true because one can use the Extended Euclidean Algorithm. This implies, that one can always find a pair of numbers $(p,q) \in U(n) \times U(n)$ such that $p+q = n$. In particular, $|U(n)|$ is always even for all $n \geq 3$. This is not a surprise since by definition the Euler's totient function $\varphi(n) = |U(n)|$, for which it has a lot of properties found here.

I was wondering that if $G_{2n}$ is nonempty for all $n \geq 2$, then Goldbach's Conjecture is true? In particular, $|G_{2n}| \geq 2$. I was thinking about this because for all positive integers $n$, we have $G_{n} \subsetneq U(n)$.

To prove that $G_n \subsetneq U(n)$, we notice that if $n = 1$ or $2$, $G_n = \varnothing \subsetneq U(n)$ since there no primes that are less than $n$. All is left to show is that $$\forall n \forall p \,\, (n \geq 3 \,\, \& \,\, p\in G_n \Longrightarrow \gcd(n,p) = 1 \,\, \& \,\, 1\leq p \leq n)$$ By definition of $G_n$, if $p \in G_n$, then $p$ is prime. Hence, $1< 2 \leq p < n$. So, $1\leq p \leq n$. Also, since $p$ is prime and $p \in G_n$, we have the $$p \nmid n \iff \gcd(p,n) =1$$

Summary: I was wondering that if $|G_{2n}| \geq 2$ for all $n \geq 4$, then Goldbach's Conjecture is true just by using the fact that $G_n \subsetneq U(n)$ by searching primes $(p,q) \in G_{2n} \times G_{2n}$ such that $p+q = 2n$?

Edit: My hypothesis is that my conjecture is nothing compared to the actual Goldbach's Conjecure. That is, we need more than just verifying if one set is a subset of another set.

Yoyos Tutoring
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    Not following. $G_{2n}$ will be quite large, in general. After all, it surely contains all the primes between $n$ and $2n$, so it has more that $\pi(2n)-\pi(n)$ elements. Not seeing the connection to Goldbach. Or have I misread the definitions? – lulu Jan 27 '24 at 00:00
  • @lulu, let me explain. I was thinking that since $U(n)$ has the property that there exists a pair $(r,s) \in U(n)$ s.t. $n = r+s$, maybe one can find a pair of primes $(p, q) \in G_{2n} \times G_{2n}$, such that $p + q = 2n$? Let me edit my post again. – Yoyos Tutoring Jan 27 '24 at 00:05
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    Well, I don't see it. As I say, $G_{2n}$ will be quite large. this question together with the prime number theorem let you estimate $|G_{2n}|$. But, again, I don't think this will help with Goldbach. – lulu Jan 27 '24 at 00:10
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    @lulu I agree. Maybe I was thinking so randomly and just thought of this conjecture as if it would be so simple to prove the Goldbachs's Conjecture, you know? Lol. Regardless, thanks for your comment. – Yoyos Tutoring Jan 27 '24 at 00:14

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It seems to me that you are solving $p+q=n$ for $p,q\in U(n)$ and hoping to translate this into a solution with $p,q\in G_n$ which, if this worked for all even $n$, would prove Goldbach's conjecture. Now it is true that the equation has solutions with $p,q\in U(n)$, for instance $p=1$ and $q=n-1$, but there is no obvious way to turn this into a solution for $p,q\in G_n$.

Joshua Tilley
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