date: 14/03/2019
Author: LAGRIDA Yassine

# Goldbach's conjecture

Let $\pi(n)$ denote the number of primes less than $n$.

Let $q(n)$ be the greatest prime verify $\displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)} < n$

Let $I_n$ denote the number of numbers less than $n$ and coprime to $\displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$.

Consider Mertens third theoreme :

${\small \prod_{\substack{a \leq q(n) \\ \text{a prime}}} \left({\normalsize 1-\frac{1}{a}}\right)} \sim \frac{e^{-\gamma}}{\log(q(n))}$

Consider the prime number theoreme :

$\log \displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)} \sim q(n)$

Then we have : $I_n \sim \dfrac{n}{\log\log(n)} e^{-\gamma}$

We have $\dfrac{n}{\log\log(n)} e^{-\gamma} = \frac{n}{\log(n)} \frac{\log(n)}{\log\log(n)} e^{-\gamma}$

Then we have this fondamentale equivalence : $I_n \sim \pi(n) \big( \pi(q(n)) e^{-\gamma} \big)$

Now to deduce Goldbach's conjecture, consider $n$ even number.

Consider this function:
$G(n, q) = \#\left\{k \text{ and } n-k \text{ coprime to } \displaystyle{\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)} \text{ and } k \leq \displaystyle{\small \left( \prod_{\substack{a \leq q \\ \text{a prime}}} {\normalsize a} \right)}\right\}$

The objective is to calculate the number of couples $(k,n-k)$ that $k$ and $n-k$ coprime to $\displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$ and $k$ less than $\displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$.

That equivalente to show the number of solutions of this system (unkown $k$), for all $p_i \leq q(n)$ and $b_i \neq \bar{0} \, , \, c_i \neq \bar{0}$ in $\mathbb{Z}/p_i\mathbb{Z}$:

$\left\{ \begin{array}{cl} k & \equiv b_i[p_i] \\ \\ n-k & \equiv c_i[p_i] \end{array} \right.$

Chinese remainder theorem say that the solutions is unique modulo $\displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$.

If $n$ is coprime to $p_i$ then we have $p_i-2$ possibilities that $b_i \neq \bar{0}$ and $c_i \neq \bar{0}$ in $\mathbb{Z}/p_i\mathbb{Z}$.

If $n$ is divisible by $p_i$ then $\bar{b}_i = -\bar{c}_i$ in $\mathbb{Z}/p_i\mathbb{Z}$, then we have $p_i-1$ possibilities.

Then we have the number of couples $(k, n-k)$ that $k$ and $n-k$ coprime to $\displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$ and $k$ less than $\displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}$ :

$\prod_{\substack{3 \leq a \leq q(n) \\ \text{a prime, } a | n}} (a-1) \prod_{\substack{3 \leq a \leq q(n) \\ \text{a prime, } a \nmid n}} {\normalsize (a-2)}$

Then:

${\small \left( \prod_{\substack{a | n \\ \text{a prime} \\ a \leq q(n)}} {\normalsize \frac{a-1}{a-2}} \right)} {\small \left( \prod_{\substack{3 \leq a \leq q(n) \\ \text{a prime}}} {\normalsize (a-2)} \right)}$

Asymptotique Formula:

We denote $G(n) = G(n, q(n))$

If the greatest prime dividing $n$ less than $q(n)$:

Then we have :

$\begin{array}{rcl} \dfrac{G(n)}{\displaystyle{\small \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} }} & = & \displaystyle \dfrac{1}{2} {\small \left( \prod_{\substack{a | n \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} {\small \prod_{\substack{3 \leq a \leq q(n) \\ \text{a prime}}} {\normalsize \left( 1-\dfrac{2}{a}\right)} } \\ \\ & = & \displaystyle \dfrac{1}{2} {\small \left( \prod_{\substack{a | n \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} {\small \prod_{\substack{3 \leq a \leq q(n) \\ \text{a prime}}} \left({\normalsize 1-\frac{1}{(a-1)^2}}\right)} {\small \prod_{\substack{3 \leq a \leq q(n) \\ \text{a prime}}} \left({\normalsize 1-\frac{1}{a}}\right)}^2 \\ \\ & \sim & \displaystyle \dfrac{1}{2} {\small \left( \prod_{\substack{a | n \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} \, C_2 \, 4 \, \dfrac{1}{\log(q(n))^2} e^{-2\gamma} \end{array}$

With $C_2 = {\small \prod_{\substack{3 \leq a \\ \text{a prime}}} \left({\normalsize 1-\dfrac{1}{(a-1)^2}}\right)}$ a constante.

We have $q(n)$ is the greatest prime verify $\displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)} < n$, then :

$G(n) \sim \displaystyle 2 \, C_2 \, {\small \left( \prod_{\substack{a | n \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} \dfrac{n}{\log(\log(n))^2} e^{-2 \gamma}$

If some primes dividing $n$ are larger than $q(n)$:

The expression $G(n) \sim \displaystyle 2 \, C_2 \, {\small \left( \prod_{\substack{a | n \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} \dfrac{n}{\log(\log(n))^2} e^{-\gamma}$ is valide if $a | n \implies a \leq q(n)$, I will show that this equivalent is true also for every even $n$.

The probleme is presente in this terme ${\small \left( \prod_{\substack{a | n \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)}$

Let $q(n) = p_k$, Then by the definition of $q(n)$ we have :

$\displaystyle{\small \left( \prod_{\substack{a \leq p_k \\ \text{a prime}}} {\normalsize a} \right)} < n \leq {\small \left( \prod_{\substack{a \leq p_{k+1} \\ \text{a prime}}} {\normalsize a} \right)}$

The worst case is for $n = 2 {\small \left( \prod_{\substack{p_{k+1} \leq a \leq p_{m} \\ \text{a prime}}} {\normalsize a} \right)}$, with $p_m$ is the greatest prime verify :

$\displaystyle{\small \left( \prod_{\substack{a \leq p_k \\ \text{a prime}}} {\normalsize a} \right)} < 2 {\small \left( \prod_{\substack{p_{k+1} \leq a \leq p_{m} \\ \text{a prime}}} {\normalsize a} \right)} \leq {\small \left( \prod_{\substack{a \leq p_{k+1} \\ \text{a prime}}} {\normalsize a} \right)}$

Then multiplying by $\displaystyle{\small \left( \prod_{\substack{a \leq p_k \\ \text{a prime}}} {\normalsize a} \right)}$ :

$\displaystyle{\small \left( \prod_{\substack{a \leq p_k \\ \text{a prime}}} {\normalsize a} \right)}^2 < 2 {\small \left( \prod_{\substack{a \leq p_{m} \\ \text{a prime}}} {\normalsize a} \right)} \leq p_{k+1} {\small \left( \prod_{\substack{a \leq p_{k} \\ \text{a prime}}} {\normalsize a} \right)}^2$

Then we apply logarithm function:

$\displaystyle 2 \log {\small \left( \prod_{\substack{a \leq p_k \\ \text{a prime}}} {\normalsize a} \right)} < \log {\small \left( \prod_{\substack{a \leq p_{m} \\ \text{a prime}}} {\normalsize a} \right)} + \log(2) \leq \log(p_{k+1}) + 2 \log {\small \left( \prod_{\substack{a \leq p_{k} \\ \text{a prime}}} {\normalsize a} \right)}$

Using Prime number theoreme :

$p_k \sim \log {\small \left( \prod_{\substack{a \leq p_k \\ \text{a prime}}} {\normalsize a} \right)}$ And $p_m \sim \log {\small \left( \prod_{\substack{a \leq p_{m} \\ \text{a prime}}} {\normalsize a} \right)}$

Then from the inequality we have :

$p_m \sim 2 p_k$

Then we have : $p_m = 2 q(n) + o(q(n))$.

For $\delta > 1$ we have :

$\begin{array}{rcl} \displaystyle {\small \left( \prod_{\substack{n \leq a \leq \delta n \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} & = & \displaystyle {\small \left( \prod_{\substack{3 \leq a \leq \delta n \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} {\small \left( \prod_{\substack{3 \leq a < n \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)}^{-1} \\ \\ & = & \displaystyle {\small \left( \prod_{\substack{3 \leq a \leq \delta n \\ \text{a prime}}} {\normalsize \frac{1 - \dfrac{1}{a}}{1 - \dfrac{2}{a}}} \right)} {\small \left( \prod_{\substack{3 \leq a < n \\ \text{a prime}}} {\normalsize \frac{1 - \dfrac{1}{a}}{1 - \dfrac{2}{a}}} \right)}^{-1} \\ \\ & = & \dfrac{\displaystyle{\small \prod_{\substack{3 \leq a \leq \delta n \\ \text{a prime}}} {\normalsize \left( 1 - \dfrac{1}{a} \right)}}}{\displaystyle{\small \prod_{\substack{3 \leq a \leq \delta n \\ \text{a prime}}} \left({\normalsize 1-\frac{1}{(a-1)^2}}\right)} {\small \prod_{\substack{3 \leq a \leq \delta n \\ \text{a prime}}} \left({\normalsize 1-\frac{1}{a}}\right)}^2} \\ \\ & \, & \times \left( \dfrac{\displaystyle{\small \prod_{\substack{3 \leq a < n \\ \text{a prime}}} {\normalsize \left( 1 - \dfrac{1}{a} \right)}}}{\displaystyle{\small \prod_{\substack{3 \leq a < n \\ \text{a prime}}} \left({\normalsize 1-\frac{1}{(a-1)^2}}\right)} {\small \prod_{\substack{3 \leq a < n \\ \text{a prime}}} \left({\normalsize 1-\frac{1}{a}}\right)}^2} \right)^{-1} \\ \\ & \sim & \dfrac{1}{2 C_2} \log(\delta n) e^{\gamma} 2 \, C_2 \, \dfrac{e^{-\gamma}}{\log(n)} \\ \\ & \sim & \dfrac{\log(\delta n)}{\log(n)} \end{array}$

Then we have $\displaystyle {\small \left( \prod_{\substack{n \leq a \leq \delta n \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} \sim 1$ for any fixed $\delta > 1$

We prooved that $p_m = 2 q(n) + o(q(n))$, then we have :

$\displaystyle {\small \left( \prod_{\substack{q(n) < a \leq p_m \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} = {\small \left( \prod_{\substack{q(n) < a \leq (2+o(1)) q(n) \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} \sim 1$

Then for $n = 2 {\small \left( \prod_{\substack{p_{k+1} \leq a \leq p_{m} \\ \text{a prime}}} {\normalsize a} \right)}$ we have :

${\small \left( \prod_{\substack{a | n \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} \sim 1$

And That complete the proof.

Then We have The Weak Goldbach's conjecture form:

$G(n) \sim \displaystyle 2 \, C_2 \, {\small \left( \prod_{\substack{a | n \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} \dfrac{n}{\log(\log(n))^2} e^{-2\gamma}$

And using the fondamentale equivalence:

$I_n \sim \pi(n) \big( \pi(q(n)) e^{-\gamma} \big)$

I conjecture that :

$G(n) \sim G_p(n) \big( \pi(q(n)) e^{-\gamma} \big)^2$

Where $G_p(n) = \#\left\{k \text{ and } n-k \text{ are primes } \text{ and } k \leq \displaystyle{\small \left( \prod_{\substack{a \leq q(n) \\ \text{a prime}}} {\normalsize a} \right)}\right\}$

And that gives immediatly this result:

$G_p(n) \sim \displaystyle 2 \, C_2 \, {\small \left( \prod_{\substack{a | n \\ \text{a prime}}} {\normalsize \frac{a-1}{a-2}} \right)} \dfrac{n}{\log(n)^2}$