# Proof strategy based on factorization

Prerequisites:

The final aim of the proof strategies that we are carrying out, as already indicated, is to prove that, given an even number $2n$, there exists at least one pair $(p, q)$ where $p + q = 2n$, and $p$ and $q$ are both primes. Since $q = 2n - p$, it suffices to prove that there exists at least one prime number $p$ smaller than $2n$ such that also $2n - p$ is prime; the proof strategy we are dealing with here is based precisely on this principle.

The starting point for finding the possible values of $p$ is a well-known property of prime numbers, Bertrand’s postulate, according to which there is at least one prime number $p$ between $n + 1$ and $2n$. This prime number, by the Property L.F.3 (Prime spaces on the right side of the factorization dashed line of an even number), is also a space in the factorization dashed line; so, on the opposite side, in a symmetrical position (i.e. in correspondence with $2n - p$), there will in turn be another space, due to Property L.F.2 (Symmetry of factorization dashed lines).

So, we have obtained a pair of spaces, $p$ and $2n - p$, one of which is surely a prime number. On the basis of this we prove the following result, which we have called Goldbach-Bertrand Theorem, since it has a similar formulation to Goldbach’s conjecture, and the proof employs Bertrand’s postulate. However, it is a temporary name, because it is possible that, in the future, this Theorem will be replaced by some better result, which would more deserve such an important name.

Goldbach-Bertrand Theorem

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

In order to prove the Theorem, first of all we have to 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.

From now on, given a sum that satisfies the Theorem for a given $2n$, we will always call $p$ the prime addend and $q$ the coprime one. The condition that $p \gt n$ (hence $q \lt n$) allows us to represent each possible sum $2n = p + q$ in a unique way with the pair $(p, q)$. Indeed, even when $q$ is also prime, as in $10 = 7 + 3$, the addend $p$ is always distinguishable by the fact to be the largest, so in this case the matching pair would be $(7, 3)$. Also, the pair would be the same even if we wrote the sum equivalently as $10 = 3 + 7$, because $7$ is the largest addend and hence we have $7 = p$. So the condition $p \gt n$ allows us to treat the sums $2n = p + q$ and $2n = q + p$ as a single sum, represented by the pair $(p, q)$.

The condition that $p \gt n$ is also important for other reasons: first of all, it reminds us of the proof of the Theorem, because it comes directly from Bertrand’s Postulate; moreover, only thanks to this assumption we can be sure that $q$ is coprime to $2n$. For example, we have $18 = 3 + 15$, $3$ is prime but $15$ is not coprime to $18$; instead for all prime numbers $p > 9$ we have that, if $18 = p + q$, then $q$ is coprime with $18$: in fact we have the sums $18 = 11 + 7$, $18 = 13 + 5$, $18 = 17 + 1$.

The Goldbach-Bertrand Theorem, contrary to Goldbach Conjecture, does not take into consideration the possibility of expressing an even number $2n$ as the sum of two identical primes, i.e. $2n = n + n = p + p$, with $p$ prime. In fact, by Goldbach-Bertrand Theorem, one of the two addends (the first one, denoted by $p$) must be greater than $n$. Furthermore, even if we remove this assumption, it still could not occur that $p = q = n$, because in this case $p$ in addition to being prime should also be coprime with $2n$, and this is not possible. In fact, if $2n = p + p$, then $2n = 2p$, then $\mathrm{HCF}(2n, p) = \mathrm{HCF}(2p, p) = p \gt 1$, whereas it should be $\mathrm{HCF}(2n, p) = 1$ if $2n$ and $p$ were coprime.

At this point we might ask ourselves: since decompositions of the type $2n = p + p$ are not taken into consideration by Goldbach-Bertrand Theorem, what happens for even numbers that by Goldbach’s conjecture admit only this type of decomposition, i.e. $4 = 2 + 2$ and $6 = 3 + 3$? 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 $6 = 5 + 1$ which fall within the cases compliant with Goldbach-Bertrand Theorem, even if they are not acceptable for the Goldbach’s Conjecture. In fact, in both decompositions, the first addend is a prime number greater than half the initial number and the second, equal to 1, is coprime with both 4 and 6 (in general 1 is coprime with 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 can 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.

The fundamental condition is that the first addend must be a prime number greater than $n = 7$ (and obviously less than 14, otherwise $q$ would be negative), so the compliant decompositions are as follows (the prime number that makes up the first addend is in bold):

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

Both the second addends, respectively 3 and 1, are coprime to 14, therefore both decompositions are compliant with the statement of the Goldbach-Bertrand Theorem.

We observe that, if we removed the constraint $p \gt n$, we would obtain more sums formed by a prime number and a coprime number:

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

However, one of the additional sums, $14 = \mathbf{3} + 11$, is equivalent to an already found one (it only changes the order of the addends), while one is actually new, $14 = \mathbf{5} + 9$. However, even if the condition $p \gt n$ causes us to lose this sum, the Goldbach-Bertrand Theorem assures us that, despite this, we will find at least one sum (in this case we have found two). Furthermore, as we have observed previously, having set $p \gt n$ we have the certainty of considering non-equivalent sums, i.e. with addends that differ numerically and not only due to the order.

The connections between Bertrand’s postulate and Goldbach’s conjecture do not end with Goldbach-Bertrand 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’s conjecture, it will be necessary to find a condition that allows to choose $p$ so that also $2n - p$, that is the column number which is in a symmetrical position with respect to the column $p$, is prime.

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:

• $q \lt n$;
• $q$ is a space of the factorization dashed line of $2n$;
• $q$ 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, the sum of which 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 $p \gt n$ is prime and $q$ is coprime with $2n$. So, in order to arrive to the Goldbach’s 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.

The first point, however, never occurs, because it is possible to prove that, for $2n > 8$, the minimum number of pairs is 2. In fact:

1. In 1952, in the article On the interval containing at least one prime number, Jitsuro Nagura proved that for $n \ge 25$, i.e. $2n \ge 50$, a prime number $p$ always exists such that $n \lt p \lt \frac{6}{5}n$;
2. Applying the same theorem, but setting $m = \frac{6}{5}n$, we have that, in turn, a prime number $p'$ exists such that $m \lt p' \lt \frac{6}{5}m$, i.e. $\frac{6}{5}n \lt p' \lt \frac{6}{5}(\frac{6}{5}n) = \frac{36}{25}n$;
3. Being $\frac{36}{25}n \lt 2n$ for each $n$, we have $\frac{6}{5}n \lt p' \lt \frac{36}{25}n \lt 2n$;
4. Surely $p \ne p'$, because $p \lt \frac{6}{5}n \lt p'$, due to points 1 and 2;
5. Thus, for $n \ge 25$, i.e. $2n \ge 50$, two distinct prime numbers $p$ and $p'$ exist between $n$ and $2n$, excluding bounds; so, due to Goldbach-Bertrand Theorem, two distinct pairs $(p, q)$ and $(p', q')$ exist, such that $p + q = p' + q' = 2n$, with $q$ and $q'$ coprime with $2n$;
6. For each $2n$ in the remaining interval $8 \lt 2n \lt 50$, it is easy to verify that at least two pairs exist; for example for $2n = 20$ there are three pairs: $(3, 17)$, $(7, 13)$ and $(11, 9)$;
7. So, in total, for $8 \lt 2n \le 50$ and $2n \ge 50$, i.e. for $2n \gt 8$, there are at least two pairs.
Besides being at least 2, the number of pairs also tends to increase as $n$ increases; this has not yet been proved for the pairs related to the Goldbach Conjecture, but it can be easily proved for the ones of Goldbach-Bertrand Theorem. In fact, the number of prime numbers $p$ between $n + 1$ and $2n$ tends to increase as $n$ increases; since each such prime number $p$ is related to a pair $(p, q)$ and therefore a sum $2n = p + q$ (where $q = 2n - p$ is coprime with $2n$, as we saw in the proof of the Goldbach-Bertrand Theorem), the number of pairs (and corresponding sums) tends to increase as $2n$.

Why does the number of primes between $n$ and $2n$ tend to increase as $n$ increases?

This is a direct consequence of the Prime Number Theorem, because:

• Due to the Prime Number Theorem, we have an approximation of the number of primes up to $n$: $\pi(n) \sim \frac{n}{\log(n)}$;
• Then $\pi(2n) - \pi(n) \sim \frac{2n}{\log(2n)} - \frac{n}{\log(n)} = \frac{2n}{\log 2 + \log(n)} - \frac{n}{\log(n)} \sim \frac{2n}{\log(n)} - \frac{n}{\log(n)} = \frac{n}{\log(n)}$;
• We know that $\frac{n}{\log(n)}$ increases as $n$ increases;
• Thus, $\pi(2n) - \pi(n)$ tends to increase as $n$ increases.

However, there is a substantial difference between the expression “tends to increase” and the verb “increases”: in the first case there is no monotonous increase, but only an increase to infinity, therefore it is possible that the number of pairs (equal to the number of primes between $n + 1$ and $2n$) has fluctuations, although the general trend is increasing (for example, there are 4 prime numbers between 11 and 20 ($n = 10$) but only 3 between 12 and 22 ($n = 11$)

A questo punto, per dimostrare la Congettura di Goldbach utilizzando il Teorema di Goldbach-Bertrand, resta da dimostrare che, per $n > 8$, tra le (almeno due) coppie $(p, q)$ con $p \gt n$ primo e $q$ coprimo, ce n’è almeno una in cui $q$ è primo. Le indagini in tal senso sono tuttora in corso, e non sono ancora giunte a una conclusione.
Da questo punto in poi, la dimostrazione potrà tralasciare il caso $2n \le 8$, dato che, per quei valori di $2n$, sappiamo già che essi sono somma di due numeri primi, com’è facile verificare.