# Proof strategy based on factorization

Prerequisites:

Properties of the spaces of the factorization dashed lines can be used to set up a particular proof strategy, which can use not only the dashed line theory, but also the known theorems about prime numbers.
For example, as a starting point for the proof, it is possible to take advantage of the Bertrand’s postulate, according to which there is a prime number $p$ between $n + 1$ and $2n$. In fact, this prime number, due to the Property L.F.3 (Prime spaces on the right side of the factorization dashed line of an even number), is also a space; so on the opposite side of the dashed line, in a symmetrical position, there will in turn be another space, for the Property L.F.2 (Symmetry of factorization dashed lines).

We have thus obtained a pair of spaces, one of which is a prime number. Based on this the following result is proved, which we have called Goldbach-Bertrand Theorem, given that it has a formulation similar to Goldbach’s conjecture, and its proof uses the Bertrand’s postulate. However, this is a temporary name, because it is possible that in the future this Theorem will be replaced by some better result, which would deserve more of such an important name.

Goldbach-Bertrand Theorem

Every even number $2n \gt 2$ can be written as the sum of two integers, one of which is prime and the other is coprime with $2n$.

In order to prove the Theorem, first of all we build the factorization dashed line $T$ of $2n$.

Due to Bertrand’s postulate, there is certainly a prime $p$, between $n + 1$ and $2n$.

Since $p$ is a prime number, due to the Property L.F.3 (Prime spaces on the right side of the factorization dashed line of an even number) it is also a space of $T$;

Since the dashed line $T$ is symmetrical due to the Property L.F.2 (Symmetry of factorization dashed lines), also $q := 2n - p$ is in turn another space;

Since $q$ is a space, by definition it is not divisible by any of the components of the dashed line;

Hence, $q$ is not divisible by any of the prime factors of $2n$, i.e., by definition, $q$ is coprime with $2n$;

Furthermore, we have $p + q = p + (2n - p) = 2n$, that is $p + q = 2n$.

In summary, we have found two numbers $p$ and $q$, of which $p$ is prime, $q$ is coprime with $2n$, and their sum is $2n$, which was the thesis to be proved.

Goldbach-Bertrand Theorem, contrary to Goldbach’s conjecture, does not take into consideration the possibility of expressing an even number $2n$ as the sum of two identical prime numbers, i.e. $2n = n + n = p + p$, with $p$ prime. In fact, by Goldbach-Bertrand Theorem, one of the two addends must be prime, and the other one must be coprime with $2n$. But if both addends were equal to the same prime $p$, then that $p$ must also be coprime with $2n$, and this is not possible. In fact, if $2n = p + p$, then $2n = 2p$, so $\mathrm{LCM}(2n, p) = \mathrm{LCM}(2p, p) = p \gt 1$, whereas it must be $\mathrm{LCM}(2n, p) = 1$ if $2n$ and $p$ are coprime.
Now you can ask yourself: since decompositions of the kind $2n = p + p$ are not taken into considerations by Goldbach-Bertrand Theorem, what happens for the even numbers that by Goldbach’s conjecture admit only this kind of decomposition, namely $4 = 2 + 2$ or $6 = 3 + 3$? This question brings us to another difference between Goldbach-Bertrand Theorem and Goldbach’s conjecture: the coprime number may also be 1. In fact in the case of 4 and 6 there are the decompositions $4 = 3 + 1$ and $6 = 5 + 1$ that are allowed by Goldbach-Bertrand Theorem, though they are not acceptable by Goldbach’s conjecture. In fact, in both decompositions, the first addend is a prime number and the second one, equal to 1, is coprime to both 4 and 6 (generally 1 is coprime to any integer number).
So, recapitulating, in Goldbach-Bertrand Theorem:

• The prime number and the coprime number that constitute the sum cannot be identical
• The coprime number may be 1

We can also note that the coprime number must be odd, because it’s coprime with $2n$ and so it cannot have 2 as a prime factor.

We thank our reader Ultima for the starting points that let us write this remark and the following example.

As an example, let’s see what decompositions of the number 14 are compliant with the Goldbach-Bertrand Theorem.

Since one of the conditions is that the first addend must be prime, first of all let’s see what decompositions of 14 respect this condition (the prime number that constitutes the first addend is written in bold):

\begin{aligned} 14 &= \mathbf{2} + 12 \\ &= \mathbf{3} + 11 \\ &= \mathbf{5} + 9 \\ &= \mathbf{7} + 7 \\ &= \mathbf{11} + 3 \\ &= \mathbf{13} + 1 \end{aligned}

Now let’s scan these decompositions, in order to check which ones satisfy also the other condition, i.e. which ones have a second addend that is coprime with 14. The second addends are respectively 12, 11, 9, 7, 3 and 1, but among them only 11, 9, 3 and 1 are coprime with 14; the other two numbers are not because $\mathrm{LCM}(12, 14) = 2 \gt 1$ and $\mathrm{LCM}(7, 14) = 7 \gt 1$. So the following decompositions are left, that satisfy the statement of Goldbach-Bertrand Theorem:

\begin{aligned} 14 &= \mathbf{3} + 11 \\ &= \mathbf{5} + 9 \\ &= \mathbf{11} + 3 \\ &= \mathbf{13} + 1 \end{aligned}

As we noted earlier, the decomposition 14 = 7 + 7 does not appear, though it’s compliant with Goldbach’s conjecture, whereas there are the decompositions 14 = 5 + 9 and 14 = 13 + 1 that are not compliant with the conjecture.
Of course, according to Goldbach’s conjecture as well as Goldbach-Bertrand Theorem, one of the two decompositions 14 = 3 + 11 e 14 = 11 + 3 can be removed, because they only differ by the order of the addends.

The connections between Bertrand’s postulate and Goldbach’s conjecture do not end with Goldbach-Bertrand’s theorem. Indeed, it can also be shown that Goldbach’s conjecture implies Bertrand’s postulate; that is, if Goldbach’s conjecture were true, it could be used to prove Bertrand’s postulate. More details can be found in the article Goldbach’s Conjecture Implies Bertrand’s Postulate by H. J. Ricardo. The proof is also shown on this page: https://proofwiki.org/wiki/Goldbach_implies_Bertrand.

The Goldbach-Bertrand Theorem is a starting point, but it is still far from the final goal. In order to arrive to the proper proof of the Goldbach conjecture, it will be necessary to find a condition that allows to choose $p$ so that also $2n - p$ is prime, that is the column number which is in a symmetrical position with respect to the column $p$.

A possible starting point is asking ourselves if there is a rule on how our potential values of $2n - p$ are made, which we will call $q$, of which we know so far some characteristics:

• $q \leq n$;
• $q$ is a space of the factorization dashed line of $2n$;
• $q$ it is certainly odd, because, being 2 a component of the dashed line, if it were even it would not be a space.

This proof strategy is represented in the following image: Figure 1: Proof strategy based on factorization. In the factorization dashed line of 2n=14, we must look for two spaces p and 2n-p (in green), both primes, whose sum is 2n. Existence of p and its primality are guaranteed by Bertrand’s postulate.

The next step is to tighten the circle: Goldbach-Bertrand’s theorem assures us that there is at least one pair $(p, q)$ in which $p$ is prime and $q$ is coprime with $2n$. So, in order to arrive to the Goldbach conjecture, we must prove that:

• When the pair $(p, q)$ is unique, $q$ is prime;
• When many pairs $(p, q)$, exist, at least one of them is such that $q$ is prime.