In the number theory, finding each prime number is a challenging job. However, it is an easy task using the Eratosthenes sieve principle to search composite numbers, after that, those remaining integers should be primes. Now, in the Strong Goldbach problem, we utilize the strategy to explore the optimal Goldbach partitions from all feasible Goldbach partitions.

1) Special cases and effacement even numbers

If m is a prime, then $2m=m+m$, it is an optimal Goldbach partition. For example, $4=2+2$ and $6=3+3$. So, effacing all even integers since even numbers can not be prime except 2. Also, assume that m is not a prime number.

2) Creates a feasible Goldbach partition list

Let 2m be the given integer for the Strong Goldbach Problem, and all feasible Goldbach partitions should be $\left(i,2m-i\right)$, where $2\le i\le m$. Such that, we can obtain all feasible Goldbach partitions saved in a list

$S\left(2m\right)=\left\{\left(2,2m-2\right),\left(3,2m-3\right),\cdots ,\left(m,m\right)\right\}.$(4)

3) Find optimal Goldbach partitions

One should set the feasible Goldbach partitions to false if any object has at least one composite number. So, $S\left(2m\right)$ passing the Eratosthenes sieving in stages by $p_i$, and $p_i\le \sqrt{2m}$, all false objects had been effaced out from $S\left(2m\right)$. The termination condition is $p_i>\sqrt{2m}$. Then, the remaining objects must be optimal Goldbach partitions for the problem.

Since given even number $2m>4$ is composite numbers, we can efface those false feasible Goldbach partitions from $S\left(2m\right)$. To analyze the Goldbach partition problem, let any odd prime p, and $3\le p\le \sqrt{2m}$, also $2m-p=m+m-p=m+d$, where $d=m-p$. Then, any one of Goldbach partitions is an feasible Goldbach partitions containing a pair of integers $\left(m-d,m+d\right)$, or $\left(p,2m-p\right)$, such that they sum to 2m. We compute a Goldbach list modulo p, clearly leading to the following four cases.

Case 1. $p\nmid \left(m-d\right)$ and $p\nmid \left(m+d\right)$.

Exist either $p\nmid m$ or $p|m$.

Case 2. $p\nmid \left(m-d\right)$ and $p|\left(m+d\right)$.

Only $p\nmid m$, then exist $\left(m+d\right)\equiv 0\left(\mathrm{mod}p\right)$, and $\left(m-d\right)\ne 0\left(\mathrm{mod}p\right)$.

Case 3. $p|\left(m-d\right)$ and $p\nmid \left(m+d\right)$.

Only $p\nmid m$, then exist $\left(m-d\right)\equiv 0\left(\mathrm{mod}p\right)$, and $\left(m+d\right)\ne 0\left(\mathrm{mod}p\right)$.

Case 4. $p|\left(m-d\right)$ and $p|\left(m+d\right)$.

Only $p|m$, then exists $\left(m-d\right),\left(m+d\right)\equiv 0\left(\mathrm{mod}p\right)$.

We now analyze the above cases in detail.

1) If $p|m$, there are either $p\nmid d$ or $p|d$, which lead to Case 1 or Case 4, respectively. In each 2p distance region containing p successive feasible Goldbach partitions, Case 4 appears once in which $p|d$, and Case 1 appears $p-1$ times in which $p\nmid d$.

2) If $p\nmid m$, three cases appear, Case 1, Case 2, and Case 3. Because inside an odd numbers arithmetic progression, any 2p distance must have $p-1$ objects that can not be divided by p. So, in the median region, the median number must be one of the numbers, so there are $p-2$ numbers remaining. This situation implies that we can not partition $p-2$ in half equally. Hence, Case 2 must appear once in each 2p distance in which $p|\left(m+d\right)$ and $p\nmid \left(m-d\right)$. Case 3 must appear once in each 2p distance in which $p|\left(m-d\right)$ and $p\nmid \left(m+d\right)$. The remaining events are Case 1 situations in which $p\nmid \left(m-d\right)$ and $p\nmid \left(m+d\right)$.

Following, we take an odd number object list. Each odd prime $p_i$, and $3\le p_i\le \sqrt{2m}$.

$\begin{array}{l}S\left(2m\right)=\{\left(p_i,2m-p_i\right),\left(p_i+2,2m-p_i-2\right),\left(p_i+4,2m-p_i-4\right),\\ \left(p_i+6,2m-p_i-6\right),\cdots ,\left(m,m\right)\}\end{array}$(5)

and consider what $S\left(2m\right)\left(\mathrm{mod}{p}_i\right)$ can be.

First, we partition $S\left(2m\right)$ into three lists by its distinct congruence modulo $p_i$.

1) If any odd prime $p_i|m$, then

$\begin{array}{l}\left(p_i,2m-p_i\right),\left(3p_i,2m-3p_i\right),\cdot \cdot \cdot ,\left(\left(2k+1\right)p_i,2m-\left(2k+1\right)p_i\right),\cdots \left(\mathrm{mod}{p}_i\right)\\ \equiv \left(0,0\right)\left(\mathrm{mod}{p}_i\right)\end{array}$,

where $k\in {\mathbb{Z}}^{+}$.

Otherwise,

$\left(a_1,2m-a_1\right),\left(a_2,2m-a_2\right),\cdots ,\left(a_j,2m-a_j\right),\cdots \left(\mathrm{mod}{p}_i\right)\equiv \left(x,y\right)\left(\mathrm{mod}{p}_i\right)$,

where $p_i\nmid a_j$, and $0<x,y<p_i$.

Hence, each $2p_i$ distance of the list, there is an object with zero modulo $p_i$.

2) If $p_i\nmid m$, then $p_i\nmid 2m$. Hence, there are three cases.

First case,

$\begin{array}{l}\left(p_i,2m-p_i\right),\left(3p_i,2m-3p_i\right),\cdots ,\left(\left(2k+1\right)p_i,2m-\left(2k+1\right)p_i\right),\cdots \left(\mathrm{mod}{p}_i\right)\\ \equiv \left(0,y\right)\left(\mathrm{mod}{p}_i\right)\end{array}$,

where $y\ne 0$ and $0<y<p_i$.

Second case,

$\left(a_1,2m-a_1\right),\left(a_2,2m-a_2\right),\cdots ,\left(a_j,2m-a_j\right),\cdots \left(\mathrm{mod}{p}_i\right)\equiv \left(x,0\right)\left(\mathrm{mod}{p}_i\right)$,

where $x\ne 0$ and $0<x<p_i$. Since, $2m-a_j\equiv 0\left(\mathrm{mod}{p}_i\right)$, and $p_i\nmid a_j$. So, $2m-a_j=k{p}_i$.

The remaining cases are,

$\left(b_1,2m-b_1\right),\left(b_2,2m-b_2\right),\cdots ,\left(b_j,2m-b_j\right),\cdots \left(\mathrm{mod}{p}_i\right)\equiv \left(x,y\right)\left(\mathrm{mod}{p}_i\right)$,

where $p_i\nmid b_j$, and $p_i\nmid \left(2m-b_j\right)$. Hence, $x\ne 0$ and $y\ne 0$, and $0<x,y<p_i$.

Now, we summary the case as below.

Inside each $2p_i$ distance of the list $S\left(2m\right)$, if we ignore $p_i$ itself as a prime, then there are only two objects with zero modulo $p_i$.

Above cases should happen in $p_i$ and $3\le p_i\le \sqrt{2m}$. Continuing in this fashion until $p_i>\sqrt{2m}$ stopped, then we must obtain the optimal solutions. Following, we define the number of the biprimes of the Strong Goldbach partition (BIP).

Definition 4. Let set $A=\left[2,m\right]$ and set $B=\left[m,2m-2\right]$, where, $m\in {\mathbb{Z}}^{+}$ is the median between $\left[2,2m\right]$. Let $q_{m-d_i}=m-d_i$ and $q_{m+d_i}=m+d_i$ be a pair of Goldbach partition, such that $q_{m-d_i}+q_{m+d_i}=m-d_i+m+d_i=2m$, where d is the distance between any integer to m, and $0\le d_i\le m$. Hence, if both $q_{m-d_i}$ and $q_{m+d_i}$ are primes, then $\left(q_{m-d_i},q_{m+d_i}\right)$ is an optimal solution of the Strong Goldbach problem.

Furthermore, let

$BIP\left(m\right)=\left\{d_i:\text{prime}{q}_{m-d_i}=m-d_i,\text{and}\text{prime}{q}_{m+d_i}=m+d_i;\text{with}0\le d_i\le m\right\}$.

Also, let $biu\left(m,d_i\right)=\{\begin{array}{l}\begin{array}{l}1,\hfill \end{array}m-d_i\text{and}m+d_i\text{both}\text{are}\text{primes}\\ 0,\text{otherwise}\end{array}$.

Such that, $BIP\left(m\right)={\displaystyle {\sum }_{0\le d_i\le m}biu\left(m,d_i\right)}$.

We use Eratosthenes’s pair sieve method to find all prime pairs sum to a given natural even number 2m. This sieve method allows us to determine whether any pair numbers are both primes. We now explain how the pair sieve of the Eratosthenes method can be used to find all prime pair numbers that sum to the given even natural number 2m. If a number a is composite, then from 2 to m has at least one prime factor of a less than or equal to $\sqrt{2m}$. Otherwise, a is a prime.

We proceed as follows.

$\left\{\frac{1}{2}\times \frac{1}{3}\times \frac{3}{5}\times \cdot \cdot \cdot \times \frac{p_k-2}{p_k}\times m\right\}$,

where $p_k\le \sqrt{2m}$ and $p_{k+1}>\sqrt{2m}$, where we ignore that $p_k$ is itself prime.

$0,1,2,3,\cdots ,\left(p_i-1\right)\left(\mathrm{mod}{p}_i\right)$

although they may be in a different order. While we consider both the big numbers and the small numbers, there are two pairs included zero number, $\left\{p_{i}c,x\right\}$ or $\left\{y,d{p}_i\right\}$, $p_i\nmid x$ and $p_i\nmid y$ and $c,d$ belong to natural numbers. Hence, we have $\left(\frac{p_i-2}{p_i}\right)$ to remove the two zero included pair composite numbers each time.

For example for number 3, the remainders are 0, 1, and 2. In the worst case, there are one-third list to remain. The three cases are $\left(x,3c\right),\left(3d,y\right),\left(x,y\right)$, where $3\nmid x$ and $3\nmid y$ and c,d belong to natural numbers.

Hence,

$BIP\left(m\right)\ge \left\{\frac{1}{2}\times \frac{1}{3}\times \frac{3}{5}\times \cdot \cdot \cdot \times \frac{p_k-2}{p_k}\times m\right\}$(6)

where $p_k\le \sqrt{2m}$ and $p_{k+1}>\sqrt{2m}$, where we ignore that $p_k$ is itself prime.

Following, let us prove that if $m\ge 2$, then $BIP\left(m\right)\ge 1$.

Theorem 2. Every even integer $2m>2$ can be expressed as the sum of two primes.

Proof of Theorem 3. If m is a prime, then $2m=m+m$, then we are done. So, suppose that m itself is not prime. We have factored m as a product of primes, say

$m=p_1^{k_1}\cdot p_2^{k_2}\cdots p_r^{k_r}$,

where $p_1,p_2,\cdots ,p_r$ are all different primes. By the Eratosthenes sieves principle, any composite numbers $x<2m$ must exist at least a prime factor $p_i|x$ and $p_i\le \sqrt{2m}$, where, $0\le i\le r$. Then the following conditions is true:

1) Efface All Even Numbers

If $2m>4$, then all even numbers can not be prime. So, we efface all of them. The remaining objects equal $\frac{m}{2}$.

2) Search Stages

The search stages about equal the number of primes, $\pi \left(\sqrt{2m}\right)$. Let $p_k\le \sqrt{2m}$, and $p_{k+1}>\sqrt{2m}$. Then we have

$BIP\left(m\right)={\displaystyle {\sum }_{2\le p_i\le \sqrt{2m}}biu\left(m,m-p_i\right)}={\displaystyle {\sum }_{i=1}^{k}biu\left(m,m-p_i\right)}$.

3) Analyze the Case A

If $m=2^n$, $n\ge 2$ and $n\in {\mathbb{Z}}^{+}$, then $p_i\nmid m$ and $i>2$ or $p_i|m$ and $p_i>\sqrt{2m}$, such that, there are two arithmetic progressions of the composite numbers for each $p_i$ and $i>2$ respectively, one is the small side numbers and another is the large side numbers. Hence, we should efface failure pairs

$\frac{m}{2}\times {\displaystyle {\prod }_{i=2}^k\left(\frac{p_i-2}{p_i}\right)}$, where $p_i\in P$, and $3\le p_i\le \sqrt{2m}$.

Ensures at least one optimal solution exists in the case

$\begin{array}{c}BIP\left(m\right)\ge \frac{1}{2}\times \frac{1}{3}\times \frac{3}{5}\times \cdots \times \frac{p_k-2}{p_k}\times m\\ =\frac{1}{2}\times \frac{1}{3}\times \frac{3}{5}\times \frac{5}{7}\times \frac{9}{11}\times \frac{11}{13}\times \frac{15}{17}\times \frac{17}{19}\times \cdots \times \frac{p_k-2}{p_k}\times \frac{2m}{2}\\ \text{reduce the numerator and keep the denominator beyond}\\ \ge \frac{1}{4}\times \frac{1}{3}\times \frac{3}{5}\times \frac{5}{7}\times \frac{7}{11}\times \frac{11}{13}\times \frac{13}{17}\times \frac{17}{19}\times \cdots \times \frac{p_k-2}{p_k}\times \sqrt{2m}\times \sqrt{2m}\\ \ge \frac{1}{4}\times \frac{1}{3}\times \frac{3}{5}\times \frac{5}{7}\times \frac{7}{11}\times \frac{11}{13}\times \frac{13}{17}\times \frac{17}{19}\times \cdots \times \frac{p_k-2}{p_k}\times p_k\times \sqrt{2m}\\ \ge \frac{1}{4}\times \frac{3}{3}\times \frac{5}{5}\times \frac{5}{7}\times \frac{7}{11}\times \frac{11}{13}\times \frac{13}{17}\times \frac{17}{19}\times \cdots \times \frac{p_k-2}{p_k}\times p_k\times \sqrt{2m}\\ \ge \frac{\sqrt{2m}}{4}\end{array}$

4) Analyze the Case B

Assume that any prime $p_i|m$ and $p_i\le \sqrt{2m}$, and for $\frac{p_k-1}{p_k}>\frac{p_k-2}{p_k}$, then

$\begin{array}{c}BIP\left(m\right)\ge \frac{1}{2}\times \frac{2}{3}\times \frac{4}{5}\times \frac{6}{7}\times \frac{10}{11}\times \frac{12}{13}\times \frac{16}{17}\times \frac{18}{19}\times \cdots \times \frac{p_k-1}{p_k}\times m\\ \ge \frac{1}{2}\times \frac{2}{3}\times \frac{4}{5}\times \frac{6}{7}\times \frac{10}{11}\times \frac{12}{13}\times \frac{16}{17}\times \frac{18}{19}\times \cdots \times \frac{p_k-1}{p_k}\times \frac{2m}{2}\\ \ge \frac{1}{4}\times \frac{2}{3}\times \frac{4}{5}\times \frac{6}{7}\times \frac{10}{11}\times \frac{12}{13}\times \frac{16}{17}\times \frac{18}{19}\times \cdots \times \frac{p_k-1}{p_k}\times \sqrt{2m}\times \sqrt{2m}\\ \ge \frac{\sqrt{2m}}{4}\times \frac{2}{3}\times \frac{4}{5}\times \cdots \times \frac{p_k-1}{p_k}\times \sqrt{2m}\\ \ge \frac{\sqrt{2m}}{4}\times \frac{1}{3}\times \frac{3}{5}\times \cdots \times \frac{p_k-2}{p_k}\times p_k\\ \ge \frac{\sqrt{2m}}{4}\end{array}$

Hence, $\frac{\sqrt{2m}}{4}\ge 1$, this means that at least one Strong Goldbach partition to be found. Therefore, $m\ge 8$, the Strong Goldbach Conjecture must be true.

While $m<8$, we list the solutions of the Strong Goldbach partition as below:

$m=2$, then $4=2+2$.

$m=3$, then $6=3+3$.

$m=4$, then $8=3+5$.

$m=5$, then $10=3+7$, or $10=5+5$.

$m=6$, then $12=5+7$.

$m=7$, then $14=3+11$, or $14=7+7$.

Therefore, every even integer $2m\ge 4$ can be expressed as the sum of two primes. The Strong Goldbach Conjecture is true.

Given $2m=100$. Then the odd numbers feasible solutions listed below,

$\begin{array}{l}S\left(100\right)=\{\left(3,97\right),\left(5,95\right),\left(7,93\right),\left(9,91\right),\left(11,89\right),\left(13,87\right),\left(15,85\right),\left(17,83\right),\\ \left(19,81\right),\left(21,79\right),\left(23,77\right),\left(25,75\right),\left(27,73\right),\left(29,71\right),\left(31,69\right),\left(33,67\right),\\ \left(35,65\right),\left(37,63\right),\left(39,61\right),\left(41,59\right),\left(43,57\right),\left(45,55\right),\left(47,53\right),\left(49,51\right)\}.\end{array}$

Does exist at least one optimal solution to the Strong Goldbach problem?

Answer: The median $m=50$ and $\lceil \sqrt{100}\rceil =10$. So $p_i=\left\{3,5,7\right\}$.

1) First Stage, S(100) modulo 3

Since $3\nmid 50$, we must efface one arithmetic progressions from S(100). For details, see Table 1 .

2) The Second Stage, (x,y) list modulo 5

Since $5|50$, choose the middle row list modulo 5 from Table 1 . For details, see Table 2 .

3) The Third Stage, (x,y) list modulo 7

Since $7\nmid 50$, choose the middle row list modulo 7 from Table 2 . For details, see Table 3 .

4) The Fourth Stage

Since ${11}^2=121>100$, stop.

Now, we obtain four optimal solutions: $S\left(100\right)=\left\{\left(11,89\right),\left(17,83\right),\left(29,71\right),\left(41,59\right),\left(47,53\right)\right\}$.

$BIP\left(47\right)=5>1$, the Strong Goldbach Conjecture is correct.

Note: Ignore $p_i=\left\{3,5,7\right\}$ as primes.