# Our proof strategies: an overview

Prerequisites:

As already stated, our ultimate goal is proving Goldbach’s conjecture by using dashed line theory. It provides many ways to try coming towards the solution of the riddle; in this page and in related ones, we expose the proof strategies which we are following. Investigations are in progress yet, so their content is constantly evolving.
In them, assertions which have not yet verified will be indicated with the wording Hypothesis and will be numbered with the initial “H” (from “hypothesis”), in order to distinguish them clearly from other properties which have already been proved; instead, definitions, properties and so on, seen in these sections, will be indicated by letter “L” (from Italian “Lavori in corso”, i.e. “Work in progres”). The following naming will also be used:

Goldbach’s equation and pairs

• Equation $p + q = 2n$, corresponding to the problem which is the basis of Goldbach’s conjecture, in which $2n$ is the given even number and $p$ and $q$ are the two prime numbers which must be found, will be named Goldbach’s equation.
• Pairs $(p, q)$, ehich are solutions of Goldbach’s equation, will be named Goldbach’s pairs.

## Strategy based on the analysis of dashes and spaces

As already exposed in the article From prime numbers to dashed lines, it is evident that between the spaces of a dashed line and the prime numbers there’s a fairly strong bond. In particular, Property T.1 (Spaces and prime numbers), sets some well precise conditions about when a space is a prime number and when it’s not, referring to the linear dashed line $(2, \ldots, n)$ having, as components, all integer numbers between $2$ and some $n \geq 2$. Moreover, the Property T.2 (Dashed lines having the same spaces) states that the same conditions are valid also for dashed line $(p_1, \ldots, p_k)$, obtained from the previous one removing all components which aren’t prime numbers. Thus, if we were to reason in terms of spaces, the original Goldbach’s conjecture could be reformulated in this way:

Every even number greather than some fixed number $m \geq 2$ can be expressed as the sum of two spaces of a dashed line of type $(p_1, \ldots, p_k)$, with an appropriate $k$ (or of a dashed line of type $(2, \ldots, n)$, with an appropriate $n$), such that they are both prime numbers.

We have deliberately generalized the conjecture, replacing 2 with a generic number $m$ because, as we will see later, for this number some particular cases can occur, which prevent from finding Goldbach’s pairs. However, this does not affect the generality of a possible demonstration because, if it were possible to prove that each even number greater than $m$ is the sum of two prime numbers, only a finite number of cases would remain to be verified, that is the even numbers between 2 and $m$, but this verification has already been performed by Tomás Oliveira e Silva until $4 \cdot 10^{18}$; then, if $m$ does not exceed this limit, Goldbach’s conjecture would be proved anyway.

Regarding the way to search Goldbach’s pairs, the purpose of our study is formulating an argument which, starting from a given even number, would allow us to assert that certainly a dashed line exists, containing at least a pair of spaces which are both prime, whose sum is the starting number. The basic issue, however, is that there are many dashed lines which can be defined, then, looking for the right one is like looking for a needle in a haystack.
For example, let’s suppose having to find two prime numbers which have 10 as their sum. We can try with a first dashed line $T = (2)$, whose spaces are highlighted in yellow:

 0 1 2 3 4 5 6 7 8 9 10 2 – – – – – –

In this dashed line, we have three pairs of spaces which have 10 as their sum:

• 1 and 9;
• 3 and 7;
• 5 and 5 (adding a space to itself is allowed, because Goldbach’s conjecture does not require that the two addends must be different).

It would lead one to think that this is the right dashed line, because the last two pairs are Goldbach’s pairs. But there’s also another pair of columns, the first, which, instead, isn’t formed by two prime numbers; so, this dashed line cannot be considered completely resolutive, also because, actually, it allows us only to conclude that an even number can be written as the sum of two odd numbers. Then, we can try with a dashed line which is more “restrictive” (i.e. with the same columns but less spaces), for example $U = (2, 3, 5)$:

 0 1 2 3 4 5 6 7 8 9 10 2 – – – – – – 3 – – – – 5 – – –

This dashed line, differently from previous one, does not contains pairs of spaces with 10 as their sum: the only pair, 1 and 7, has not 10 as sum, neither is formed by two prime numbers, since only 7 is prime. Then, neither this dashed line is the proper one, because is too restrictive, so that it does not give Goldbach’s pairs.
Therefore, let’s try with a middle ground, the dashed line $V = (2, 3)$:

 0 1 2 3 4 5 6 7 8 9 10 2 – – – – – – 3 – – – –

In this dashed line, differently from the previous two, all pairs of spaces with 10 as their sum are Goldbach’s pairs: column 5, in fact, is a space and is a prime number too, 5 plus 5 is 10, and there aren’t other pairs of spaces with 10 as their sum.
We are looking for a universal solution, so we cannot proceed in this way, i.e. taking an even number $2n$ and proceeding by attempts until finding the right dashed line: otherwise, we would be forced to build a different proof for every even number, which is impossible, since they are infinite; instead, our goal is finding a single solution valid for all of them.
Trying to find some rules based on this first example, we can already make a first set of observations about the dashed line which has been revealed as correct for finding Goldbach’s pairs for munber 10:

• Its components are the first two items of the sequence of prime numbers 2, 3, 5, 7, 11 and so on;
• If it had less than two components, we would obtain some pairs of spaces whose sum is 10 but whose spaces atr not both prime numbers;
• If it had more than two components, we would get no pairs of spaces whose sum is 10.

So, choosoing two components seems an important bound, then we could look for a criterion which states, given an even number $2n$, which must be the order $k$ of a dashed line, having the first prime numbers as components, such that almost a Goldbach’s pair is obtained.

### Choosing the right order

In order to make some hypotheses about why 2 is the right $k$ for our example, we can observe that, for Property T.1 (Spaces and prime numbers), all the spaces of the dashed line $(p_{1}, p_{2}, ... p_{k})$, which will be named $T_k$ from now, surely are prime numbers if they are within the interval from $p_{k} + 1$ to $p_{k+1} ^ 2 - 1$. Now, if we set that the starting number $2n$ belongs to this interval, we have good probability that also the two spaces $p$ and $q$ belong to it. In fact:

• If $2n$ belongs to this interval, then we have $2n \leq p_{k+1} ^ 2 - 1$;
• Then, we have also $p \leq p_{k+1} ^ 2 - 1$, because $p \lt 2n \leq p_{k+1} ^ 2 - 1$, and the same is true for $q$;
• Then, $p$ and $q$ are less than or equal to the upper bound of the interval, so they cannot belong to it only if they were less than the lower bound, i.e. less than $p_{k} + 1$;
• Then, in order to have more probably $p$ and $q$ within the interval, the numbers less than $p_{k} + 1$ would have to be few, and those between $p_{k} + 1$ and $2n$ would have to be many;
• But, if we want to have many numbers between $p_{k} + 1$ and $2n$, since $2n$ cannot be more than the upper bound of the interval $p_{k+1} ^ 2 - 1$, our $2n$ should be as near as possible to this bound. This is equivalent to stating that $k$ must be the smallest integer such that $p_{k+1} ^ 2 - 1 \geq 2n$, i.e. $p_{k+1} ^ 2 \gt 2n$.

With this reasoning based on intuition, we have estabilished a criterion for choosing $k$: it has been set as the smallest integer such that $p_{k+1} ^ 2 \gt 2n$. In our example, we have started from $2n = 10$, and indeed, the smallest integer $k$ such as $p_{k+1} ^ 2 \gt 10$ is just 2. In fact, for $k = 2$, the inequality is satisfied, because $p_{k+1} ^ 2 = p_{3}^2 = 5^2 = 25 \gt 10$, while for $k = 1$ it wouldn’t be true, because $p_{2} ^ 2 = 3^2 = 9 \lt 10$.

To recap, we have started from the even number $2n$ of Goldbach’s equation, and we have found a criterion for calculating $k$, so that the dashed line $T_k = (p_{1}, p_{2}, ... p_{k})$ allows to obtain the greatest possible number of Goldbach’s pairs formed by spaces. In fact, all the spaces of the dashed line $T_k$ between $p_{k} + 1$ and $p_{k+1} ^ 2 - 1$, interval containing also $2n$, are prime numbers, so they can be replace $p$ and $q$ in Goldbach’s equation. Let’s formalize these concepts in the following definition:

Validity order, dashed line and interval

For every integer $h \gt 0$, we denote with the symbol $T_h$ the dashed line $(p_1, p_2, \ldots p_h)$, having the first $h$ prime numbers as components.
Given an even number $2n \gt 2$, let $k$ be the smallest integer such as $p_{k+1}^2 \gt 2n$. Then, the integer $k$ and the related dashed line $T_k$ are called respectively validity order and validity dashed line related to the integer $2n$. Moreover, the interval of integers between $p_{k} + 1$ and $p_{k+1} ^ 2 - 1$ (bounds included) is called validity interval of the dashed line $T_k$.

For example, let’s calculate the validity interval of our dashed line $V = T_2 = (2, 3)$, for which:

• $p_{1} = 2, p_{2} = 3$;
• $k = 2$;
• $p_{k} = p_{2} = 3$;
• $p_{k} + 1 = p_{2} + 1 = 3 + 1 = 4$;
• $p_{k+1}$ is the prime number following $p_{k} = p_{2} = 3$, i.e. 5;
• $p_{k+1} ^ 2 - 1 = 5 ^ 2 - 1 = 25 - 1 = 24$;
• The validity interval is made by the integers between $p_{k} + 1 = 4$ and $p_{k+1} ^ 2 - 1 = 24$.

Property T.1, applied to the dashed line $V$, states that all spaces between 4 and 24 are also prime numbers, and vice versa. It’s true, because 5, 7, 11, 13, 17, 19 and 23 are the spaces within this interval, but are also the prime numbers within the same interval, so the Property is verified.

At this point, the main problem is still there: now, we know what’s the right order for finding Goldbach’s pairs formed by spaces, but we cannot still be sure of finding them, it must be proved. We can translate this problem into a hypothesis:

Hypothesis of existence of Goldbach pairs based on dashed lines

Let $2n \gt 4$ be an even number, and let $k$ be the smallest integer number such as $(p_{k+1})^2 > 2n$. Then, the dashed line $T_k$ contains at least two spaces $p$ and $q$, such as $(p, q)$ is a Goldbach’s pair for $2n$.

In this statement, we have set $2n \gt 4$, and not $2n \gt 2$ like in Goldbach’s conjecture, for a precise reason. If we set $2n = 4$, the corresponding $k$ would be 1, because $p_{1}^2 = 2^2 = 4$ and $p_{2}^2 = 3^2 = 9 \gt 4$, so we would have to use the dashed line $T_1 = (2)$; but, in this one, the only pair of spaces having 4 as their sum is $(1, 3)$, but 1 is not prime.
But, starting from $2n = 6$, until now we have found no even number violating the hypothesis. Then, even if it’s a hypothesis, i.e. it has not been proved universally, it hasn’t neither been refuted. It can be verified for any number using the Dashed line viewer, which has a specific option for components, “Prime numbers up to smallest $p_{k}$​ where $(p_{k+1})^2 > n$“.

Still using the viewer, one can easily note that there are many even numbers corresponding to the same $k$ (and then to the same dashed line $T_k$). Particularly, given a $k \geq 1$, the first even number corresponding to it is the smallest $2n \geq p_k^2 + 1$ and the last is $2n = (p_{k+1})^2 - 1$. For example, the smallest even number corresponding to $k = 2$ is $2n = 3^2 + 1 = 10$ and the greatest is $2n = 5^2 - 1 = 24$. Then, for searching Goldbach’s pairs fir all even numbers $10 \leq 2n \leq 24$, we will always use the same dashed line, which, in this case, is $T_2 = (2, 3)$.

Using these observations, we can determine for example which are the even numbers related to $k = 1, \ldots, 5$ and which are the corresponding dashed lines $T_k$:

 $2n$ interval Order $k$ Dashed line $T_k$ From 4 to 8 1 (2) From 10 to 24 2 (2, 3) From 26 to 46 3 (2, 3, 5) From 48 ato118 4 (2, 3, 5, 7) From 122 to 168 5 (2, 3, 5, 7, 11)

### Looking for Goldbach’s pairs

Going back to Hypothesis H.1 (Hypothesis of existence of Goldbach’s pairs based on dashed lines), next step is attempting to prove it, i.e. try to understand within what conditions, betweeen the spaces within the validity interval, which are also prime numbers, we can find at least two ones having the starting number as their sum. About this, we are trying to explore two strategies: one of them uses properties of spaces, another instead uses the relations between spaces and the dashes being in columns adjacent to them. These two strategies are described in two dedicated pages:

Neither of these strategies has still been revealed as resolutive, because they are investigations still in progress, so they are still incomplete; but, they contain some partial results, which could bring to a final solution.

## Strategy based on prime factors decomposition

Another possibility for using dashed lines for proving Goldbach’s conjecture is using factorization dashed lines, i.e. dashed lines which have, as components, the prime factors of a given number. Starting with this type of dashed lines, we can formulate a new specific hypothesis:

Hypothesis of existence od Goldbach’s pairs based on factorization dashed lines

Given an even number $2n \gt 6$, its factorization dashed line contains at least two spaces $p$ and $q$, such as $(p, q)$ is a Goldbach’s pair for $2n$.

Also in Hypothesis H.2, we did not start from $2n = 4$, how original Goldbach’s conjecture statement would want, but we had to skip 4 and 6, starting directly from 8. This is necessary because the spaces of factorization dashed lines od $2n = 4$ and $2n = 6$, respectively $(2)$ and $(2, 3)$, do not allow to find Goldbach’s pairs; but the hypothesis, until now, seems to be verified for all even numbers starting from 8 which have been tried.

The purpose of proof strategy based on factorization is just proving Hypothesis H.2: if it were true, Goldbach’s conjecture would be a direct consequence of it, then we would automatically succeed into proving it. Advancements in this investigation, which is still in progress, are described in a dedicated page, whose content will be modified and expanded as analysis proceeds.