# Study of existence of complementary space pairs based on second order dashed lines

Prerequisites:

The aim of the study that we are going to present here is to prove that the Hypothesis H.1.T.A (Hypothesis of existence of pairs of complementary spaces based on dashes) is valid for second order dashed lines. The strategy that will be adopted is the following:

• Reformulate the initial statement with respect to second order dashed lines, in order to make explicit which effective constraints must be satisfied by a possible solution;
• Prove that there is at least one solution which satisfies the constraints.

## Statement of the initial hypothesis for the second order

As already indicated, the first step is to rewrite the starting Hypothesis to relate it to second order dashed lines. Rewriting the statement of the Hypothesis for $k = 2$, we’ll obtain:

Hypothesis H.1.T.A for second order dashed lines

Let $n \gt 1$ be an integer, and $T = (n_1, n_2)$ a second order linear dashed line, the components of which are not necessarily consecutive prime numbers. Then there exist two positive integers $x$, $y$ such that:

1. the $x$th dash is in the first row of $T$;
2. the $y$th dash is also in the first row of $T$;
3. the $x$th dash precedes a space;
4. the $y$th dash follows a space;
5. $\mathrm{t\_value(x)} + \mathrm{t\_value(y)} = n_1 n$.

Once all the constraints have been made explicit using the known properties of second order dashed lines, following the approach of the characterization of spaces, the previous Hypothesis can be reformulated this way:

By the Corollary of Proposition L.C.4 (Spaces of a second order dashed line following a dash on the first row, second form), conditions 1. and 3. are equivalent to

$$n_1 x \mathrm{\ mod\ } (n_1 + n_2) \in S_T(1)$$
$$S_T(1) = \{1, \ldots, n_2 - 2, n_2\}$$

where the letter S stands for “subsequent”, but it is meant from the point of view of spaces, i.e. there is a space following the $x$-th dash.
Similarly, by the Corollary of Proposition L.C.3 (Spaces of a second order dashed line preceding a dash of the first row, second form), conditions 2. and 4. are equivalent to

$$n_1 y \mathrm{\ mod\ } (n_1 + n_2) \in P_T(1)$$
$$P_{T}(1) = \{2, \ldots, n_2\}$$
$$\mathrm{t\_value(x)} = n_1 \biggl \lceil \cfrac{n_2 \cdot x + 1}{ n_1 + n_2} \biggr \rceil \tag{1}$$
$$\mathrm{t\_value(y)} = n_1 \biggl \lceil \cfrac{n_2 \cdot y + 1}{ n_1 + n_2} \biggr \rceil \tag{1'}$$

so in condition 5. it is possible to simplify $n_1$ as follows:

\begin{aligned}\mathrm{t\_value(x)} + \mathrm{t\_value(y)} = n_1 n & \Leftrightarrow \\ n_1 \biggl \lceil \cfrac{n_2 \cdot x + 1}{n_1 + n_2} \biggr \rceil + n_1 \biggl \lceil \cfrac{n_2 \cdot y + 1}{n_1 + n_2} \biggr \rceil = n_1 n & \Leftrightarrow \\ \biggl \lceil \cfrac{n_2 \cdot x + 1}{n_1 + n_2} \biggr \rceil + \biggl \lceil \cfrac{n_2 \cdot y + 1}{n_1 + n_2} \biggr \rceil = n \end{aligned}

Explicit Hypothesis H.1.T.A for second order dashed lines

Let $n \gt 1$ be an integer, and $T = (n_1, n_2)$ a second order linear dashed line, the components of which are not necessarily consecutive prime numbers. Then there exist two positive integers $x$, $y$ such that:

$$n_1 x \mathrm{\ mod\ } (n_1 + n_2) \in S_T(1)$$
$$n_1 y \mathrm{\ mod\ } (n_1 + n_2) \in P_T(1)$$
$$S_T(1) = \{1, \ldots, n_2 - 2, n_2\}$$
$$P_{T}(1) = \{2, \ldots, n_2\}$$
$$\biggl \lceil \cfrac{n_2 \cdot x + 1}{n_1 + n_2} \biggl \rceil + \biggl \lceil \cfrac{n_2 \cdot y + 1}{n_1 + n_2} \biggl \rceil = n$$

## Search for solutions

We have thus rewritten the Hypothesis H.1.T.A (Hypothesis of existence of pairs of complementary spaces based on dashes) in order to refer to second order dashes; the next step is to prove that it is true, i.e. that there is at least one pair $(x, y)$ that satisfies the required criteria. The first obstacle in tackling the problem is given by the fact that, in the last equation, there is the operation $\lceil \cdot \rceil$, which is not easily treatable. However, the problem can be overcome through the properties of the operation $\mathrm{\ mod\ }$ and its variant $\mathrm{\ mod^{\star}\ }$, which allow us to move on to another reformulation. Indeed, the last equation can be rewritten as follows:

We’ll start by applying Definition 1.1 in Dashed line theory (Italian), on page 11:

$$\bigg \lceil \cfrac {a}{b} \bigg \rceil = \cfrac{a - a \mathrm{\ mod^{\star}\ } b}{b} + 1 \tag{3}$$

If, using a generic $w$ instead of $x$ and $y$, we set $a := n_2 w + 1$ and $b := n_1 + n_2$, the last equation of Hypothesis H.1.T.A.2 becomes

$$\cfrac {n_2 x + 1 - (n_2 x + 1) \mathrm{\ mod^{\star}\ } ( n_1 + n_2)}{n_1 + n_2} + 1 + \cfrac {n_2 y + 1 - (n_2 y + 1) \mathrm{\ mod^{\star}\ } (n_1 + n_2)}{n_1 + n_2} + 1 = n$$

that is, moving the two 1s to the right:

$$\cfrac {n_2 x + 1 - (n_2 x + 1) \mathrm{\ mod^{\star}\ } ( n_1 + n_2)}{n_1 + n_2} + \cfrac {n_2 y + 1 - (n_2 y + 1) \mathrm{\ mod^{\star}\ } (n_1 + n_2)}{n_1 + n_2} = n - 2 \tag{a}$$

In the following, we will indicate with the symbols $v$ and $w$ the two left-hand addends:

$$v := \cfrac {n_2 x + 1 - (n_2 x + 1) \mathrm{\ mod^{\star} \ } (n_1 + n_2)}{n_1 + n_2}$$
$$w := \cfrac {n_2 y + 1 - (n_2 y + 1) \mathrm{\ mod^{\star} \ } (n_1 + n_2)}{n_1 + n_2}$$

Therefore, taking into account how Hypothesis H.1.T.A.2 was obtained, we’ll have that:

$$\mathrm{t\_value}(x) = n_1 (v + 1) \tag{2}$$
$$\mathrm{t\_value}(y) = n_1 (w + 1) \tag{2'}$$
Equations (2) and (2′) derive respectively from formulas (1) and (1′) present in the first detail, taking into account that the integer part which multiplies $n_1$ is respectively equal to $v + 1$ and $w + 1$, by formula (3) of the previous detail.

From these equations it is clear that the two spaces of interest, $\mathrm{t\_value}(x) + 1$ and $\mathrm{t\_value}(y) - 1$, actually depend on $v$ and $w$, more than on $x$ and $y$ (in fact, like we will see later, different values of $x$ and of $y$ can determine the same value of $v$ and $w$, and then the same spaces). So even if, as the Hypothesis is formulated, the unknowns are $x$ and $y$, we can also consider $v$ and $w$ as the unknowns. We will assume that $v \geq 0$ and $w \geq 0$; in fact these conditions, by (2) and (2′), correspond to putting $\mathrm{t\_value}(x) \geq n_1$ and $\mathrm{t\_value}(y) \geq n_1$, so we’ll consider the dashes starting from the first one. Consequently, by formula (a) and by how we defined $v$ and $w$:

$$0 \leq v \leq n - 2 \tag{b}$$
$$0 \leq w \leq n - 2 \tag{b'}$$

## Proof of the existence of solutions

Using the formulation H.1.T.A.2, we can proceed with the final step, namely to prove that there is at least one $( x, y)$ which matches all constraints. The proof will start from an example, i.e. the Hypothesis developed for $n = 6$ with the dotted line $T = (3, 5)$, for which $n_1 = 3$, $n_2 = 5$, $n_1 + n_2 = 8$, and $n_1 n = 3 \cdot 6 = 18$:

Let $n = 6$ and $T = (3, 5)$. Then there exist two positive integers $x$, $y$ such that:

$$3 x \mathrm{\ mod\ } 8 \in S_T(1) \tag{4}$$
$$3 y \mathrm{\ mod\ } 8 \in P_T(1) \tag{5}$$
$$S_T(1) = \{1, 2, 3, 5\}$$
$$P_T(1) = \{2, 3, 4, 5\}$$
$$\cfrac {5 x + 1 - (5 x + 1) \mathrm{\ mod^{\star}\ } 8 }{8} + \cfrac {5 y + 1 - (5 y + 1) \mathrm{\ mod^{\star}\ } 8}{8} = n - 2 \tag{6}$$

Although the proof starts from this example, the conclusions that derive from it, as we will see, will go beyond it, being universally valid for all second order linear dashed lines that have prime numbers as components.

As a starting point for the proof, let’s build a table like this:

• In the first line are the values of $x$, starting from 1;
• In the second line there are the corresponding values of the quantity $n_1 x \mathrm{\ mod\ } (n_1 + n_2)$ which occurs in the equation and in the constraints, and which in this case is $3x \mathrm{\ mod\ } 8$;
• In the third line we’ll insert the value of $v := \cfrac {5 x + 1 - (5 x + 1) \mathrm{\ mod^{\star}\ } 8}{8}$, i.e. equal to the leftmost addend appearing in the first member of equation (6);
• By formula (b), the table continues up to a value of $v$ equal to $n - 2$, i.e. up to $6 - 2 = 4$. This corresponds to a maximum value of $x$ equal to 7, because for $x = 8$ you would get $v = 5$.

The resulting table looks as follows:

 $x$ 1 2 3 4 5 6 7 $3x \mathrm{\ mod\ } 8$ 3 6 1 4 7 2 5 $v$ 0 1 1 2 3 3 4

Next, let’s group the columns which have the same value of $v$:

 $x$ 1 2, 3 4 5, 6 7 $3x \mathrm{\ mod\ } 8$ 3 6, 1 4 7, 2 5 $v$ 0 1 2 3 4

Now we’ll do the same for $y$, but creating a table made in a different way:

• The first two lines are similar to the first two of the previous table, but they refer to $y$;
• In the third line, instead of $v$ we’ll insert the value of $w := \cfrac {5 y + 1 - (5 y + 1) \mathrm{\ mod^{\star }\ } 8}{8}$, i.e. equal to the other addend appearing in the first member of equation (6);
• The columns of the table are reversed with respect to the previous one, i.e. they start from the greatest value of $y$ instead of the smallest one.

The table looks as follows:

 $y$ 7 6, 5 4 3, 2 1 $3y \mathrm{\ mod\ } 8$ 5 2, 7 4 1, 6 3 $v$ 4 3 2 1 0

Starting from these two tables, we’ll create another one, like this:

• At the top there are the first two rows of the first table;
• Then there are the first two rows of the second table, swapped places;
• Then there is a row containing the sum between $v$ and $w$, taken respectively from the two tables.

The resulting table is the following:

 $x$ 1 2, 3 4 5, 6 7 $3x \mathrm{\ mod\ } 8$ 3 6, 1 4 7, 2 5 $3y \mathrm{\ mod\ } 8$ 5 2, 7 4 1, 6 3 $y$ 7 6, 5 4 3, 2 1 $v + w$ 0 + 4 1 + 3 2 + 2 3 + 1 4 + 0

It can be seen that $v + w$ is always 4, as predicted by equation (6) from which we started to construct $v$ and $w$.

Now, we’ll highlight with a black border all the values of the second row that respect the constraint (4), i.e. that belong to $S_T(1) = \{1, 2, 3, 5\}$, and those of the third row that respect the constraint (5), i.e. that belong to $P_T(1) = \{2, 3, 4, 5\}$:

 $x$ 1 2, 3 4 5, 6 7 $3x \mathrm{\ mod\ } 8$ 3 6, 1 4 7, 2 5 $3y \mathrm{\ mod\ } 8$ 5 2, 7 4 1, 6 3 $y$ 7 6, 5 4 3, 2 1 $v + w$ 0 + 4 1 + 3 2 + 2 3 + 1 4 + 0

Then, we’ll indicate in yellow the columns which have at least one number highlighted both on the second and on the third row, i.e. such that at least one $x$ and at least one $y$ of the column satisfy the respective constraints (4) and (5):

 $x$ 1 2, 3 4 5, 6 7 $3x \mathrm{\ mod\ } 8$ 3 6, 1 4 7, 2 5 $3y \mathrm{\ mod\ } 8$ 5 2, 7 4 1, 6 3 $y$ 7 6, 5 4 3, 2 1 $v + w$ 0 + 4 1 + 3 2 + 2 3 + 1 4 + 0

All the pairs $(x, y)$ corresponding to the numbers highlighted in the columns in yellow, i.e. (1, 7), (3, 6) and (7, 1), are therefore, by construction, all solutions of equation (6) satisfying constraints (4) and (5). Equation (6) is represented in the last row as the sums $v + w = n - 2$.

The pairs (2, 6), (2, 5) and (3, 5) give rise to the same values of $v$ and $w$ as the pair (3, 6), however they are not feasible solutions because if it were $x = 2$ it would be $3x \mathrm{\ mod\ } 8 = 6 \notin S_T(1)$ (so the pairs (2, 6) and (2, 5) are to be discarded), while if it were $y = 5$ it would be $3y \mathrm{\ mod\ } 8 = 7 \notin P_T(1 )$ (therefore also the pair (3, 5) is to be discarded).

Once the solutions are represented this way, it still remains to understand under what conditions they actually exist. We’ll begin by observing that, extending the second row of the previous table:

 $3x \mathrm{\ mod\ } 8$ 3 6, 1 4 7, 2 5 0, 3 6, 1 4 7, 2 5

we can notice how the unhighlighted cell which contains no bordered numbers (i.e. the cell containing 4) appears once every 5 cells, and there are, in turn, 5 possible cells in total, i.e. (0, 3), (6, 1), (4), (7, 2) and (5), given that, if we also consider $x = 0$, the first cell (3) would become in turn (0, 3). We’ll represent this situation using dashed line notation; that is, we’ll transform the previous table into a new one, in which we place a dash in the cells that correspond to cell (4):

 $3x \mathrm{\ mod\ } 8$ — —

This table has a dash every 5 cells, as in the representation of the dashed line $T = (5)$:

 1 2 3 4 5 6 7 8 9 10 $\ldots$ 5 — — $\ldots$

The row of $3x \mathrm{\ mod\ } 8$, therefore, is a “shifted” dashed line $T = (5)$, in the sense that its dashes alternate with empty cells in the same way, but their sequence starts in a different cell.

Now, let’s apply the same principle to the row of $3y \mathrm{\ mod\ } 8$ of the previous table:

 $3y \mathrm{\ mod\ } 8$ — —

The result is similar to the previous one: also in this case we have obtained a shifted dashed line, the dashes of which are at the same distance as before, even if the sequence starts in a still different cell.

Now let’s go back to the table we started from, but with dashes instead of numbers, and indicating in yellow the columns that were previously highlighted in the same way:

 $x$ 1 2, 3 4 5, 6 7 $3x \mathrm{\ mod\ } 8$ — $3y \mathrm{\ mod\ } 8$ — $y$ 7 6, 5 4 3, 2 1 $v + w$ 0 + 4 1 + 3 2 + 2 3 + 1 4 + 0

The second and third lines of the table constitute a dashed line, which is of the second order, but is shifted with respect to the linear dashed lines we know. The interesting aspect is that the pairs of valid solutions, i.e. (1, 7), (3, 6) and (7, 1), are all found in correspondence with the yellow columns, which correspond to the spaces of this dashed line. Therefore, in order to prove that there are solutions, it is sufficient to show that there are spaces in the following shifted dashed line (the representation of which has been extended with respect to the previous one, so as to highlight the repetition of the dashes):

 $3x \mathrm{\ mod\ } 8$ — — $3y \mathrm{\ mod\ } 8$ — —
It can be seen that the dashed line displayed in the table above is very similar to what we called double dashed line in the proof strategy based on spaces. In fact, a first order double dashed line would be obtained from the table if the rows were overlapped and if one of the two groups of equidistant dashes started from the first column. The fact that double dashed lines somehow occur in different proof strategies makes us think that they are one of the key elements behind the statement of Goldbach’s conjecture.

Finding the spaces of a dashed line made this way, and/or proof that they exist, is very simple, because the dashes are repeated both on the first and on the second row with the same periodicity, and the distance between them, in each row, is equal to $5 = n_2$.

It still remains to be proved that in this table the distance between consecutive dashes of the same row, starting from a generic second order dashed line $(n_1, n_2)$, is equal to $n_2$.

In general, whatever $n_1$, $n_2$ and $n$ be, we can state that the first space cannot be beyond the third column. In fact the worst case, i.e. the one in which the first space is as far away as possible from the start of the dashed line, is the one in which:

• The first dash of the first row is in the first cell of the row;
• The first dash of the second row is in the second cell of the row.

The dashed line corresponding to this case would be the following:

 $x$ 1 2 3 $\ldots$ $n_1 x \mathrm{\ mod\ } (n_1 + n_2)$ — $\ldots$ $n_1 y \mathrm{\ mod\ } (n_1 + n_2)$ — $\ldots$ $v + w$ $0 + (n - 2)$ $1 + (n - 3)$ $2 + (n - 4)$ $\ldots$

In this case, or in the equivalent case where the two rows are swapped, the first space is in the third column. In general, also in any other arrangement of the dashes, provided that they are spaced apart, on the same row, by $n_2$, and taking into account that $n_2 \geq 3$ (being $n_2 \gt n_1 \geq 2$), the first space would never be beyond the third column. This is proved by contradiction: if the space furthest from the beginning of the dash were on the fourth column, for example, it would be preceded by three columns that are not spaces, which would mean that, in these columns, there would be at least three dashes. But, whatever their mutual position (all three on the first row, or all three on the second, or alternating), on at least one row the distance between two consecutive dashes should be at most 2, which would be less than $n_2 \geq 3$, which is not possible. The reasoning is similar if we assume that there are more than three dashes, or that the first space is in the fifth or sixth column, and so on.

Now, always considering the worst case, we’ll insert again the row with $v$:

 $x$ 1 2 3 $\ldots$ $n_1 x \mathrm{\ mod\ } (n_1 + n_2)$ — $\ldots$ $n_1 y \mathrm{\ mod\ } (n_1 + n_2)$ — $\ldots$ $v + w$ $0 + (n - 2)$ $1 + (n - 3)$ $2 + (n - 4)$ $\ldots$ $v$ $0$ $1$ $2$ $\ldots$

The value of $v$ corresponding to the first space of the shifted double dashed line, i.e. to the first feasible solution of equation (6), is at most 2, for the previous discussion. So we can say that there is at least one solution with $v \leq 2$, provided that the table has at least three columns, because the value $v = 2$ is on the third column. By construction, the number of columns in the table is $n - 1$ (from $v = 0$ to $v = n - 2$); therefore, a sufficient condition for the existence of a value of $v$ that satisfies our hypothesis is that $n - 1 \geq 3$, i.e. $n \geq 4$.
Now that we have found a condition of existence of the solution, to find the two spaces of interest it is sufficient to remember that one follows the $x$-th dash and the other precedes the $y$-th, so the solutions corresponding to a feasible value of $v$, and to the corresponding value of $w = n - 2 - v$, are:

\begin{aligned} p & := \\ \mathrm{t\_value}_T(x) + 1 & = \text{[for the (2)]} \\ n_1 (v + 1) + 1\end{aligned} \tag{7}
\begin{aligned} q & := \\ \mathrm{t\_value}_T(y) - 1 & = \text{[for the (2')]} \\ n_1 (w + 1) - 1\end{aligned}

We can also observe that $q$, once $p$ has been calculated, can be calculated simply as $n_1 n - p$, by the last equation of the Hypothesis H.1.T.A.1.
Having established that there exists at least one solution with $v \leq 2$, by imposing this constraint in formula (7), we can deduce that there exists a space $p$, corresponding to a solution of Hypothesis H.1.T.A.1, such that

$$p \leq 3 n_1 - 1$$

So not only we have found the solution, but we’ve also found an upper bound for one of the two corresponding spaces.
Summarizing, we can say that Hypothesis H.1.T.A.1 always admits a solution for $n \geq 4$, and in this case there exists a solution such that one of the two corresponding spaces is less than or equal to $3 n_1 - 1$.

We’ll verify if the statement from which we started, i.e. the Hypothesis H.1.T.A (Hypothesis of existence of pairs of complementary spaces based on dashes), is actually valid for the pairs we found. That is, we have to determine if, taking $n = 6$, the dashed line $T = (3, 5)$ and the pairs $(x, y)$ equal to (1, 7), (3, 6) and (7, 1), each of them satisfies the equation

$$\mathrm{t\_value}(x) + \mathrm{t\_value}(y) = n_1 n$$

We also need to verify that $\mathrm{t\_value}(x)$ precedes a space and that $\mathrm{t\_value}(y)$ follows a space.
Applying the Corollary of Theorem T.8 (Formula for calculation of linear second order function $\mathrm{t\_value}$ for the first row) to replace $\mathrm{t\_value}$ with its second order expression, and replacing $n$ with 6, $n_1$ with 3 and $n_2$ with 5, we’ll obtain the following equation equivalent to the previous one:

$$3 \biggl \lceil \cfrac{5 \cdot x + 1}{8} \biggl \rceil + 3 \biggl \lceil \cfrac{5 \cdot y + 1}{8} \biggl \rceil = 18$$

Let’s start with the first pair (1, 7). Replacing $x$ with 1 and $y$ with 7 we’ll get

$$3 \biggl \lceil \cfrac{5 \cdot 1 + 1}{8} \biggl \rceil + 3 \biggl \lceil \cfrac{5 \cdot 7 + 1}{8} \biggl \rceil = 3 \biggl \lceil \cfrac{6}{8} \biggl \rceil + 3 \biggl \lceil \cfrac{36}{8} \biggl \rceil = 3 \cdot 1 + 3 \cdot 5 = 3 + 15 = 18$$

as expected.
Also $\mathrm{t\_value}(x) + 1 = 3 + 1 = 4$ and $\mathrm{t\_value}(y) - 1 = 15 - 1 = 14$ are spaces ($\mathrm{t\_value}(x)$ and $\mathrm{t\_value}(y)$ are respectively the expressions $3 \biggl \lceil \cfrac{5 \cdot 1 + 1}{8} \biggl \rceil$ and $3 \biggl \lceil \cfrac{5 \cdot 7 + 1}{8} \biggl \rceil$, then precisely 3 and 15; the corresponding dashes can be viewed using the dashed line viewer by selecting the options $n = 15$, “These components” = “3,5”, “Show dash number” selected).
Now we’ll do the same thing with (3, 6):

$$3 \biggl \lceil \cfrac{5 \cdot 3 + 1}{8} \biggl \rceil + 3 \biggl \lceil \cfrac{5 \cdot 6 + 1}{8} \biggl \rceil = 3 \biggl \lceil \cfrac{16}{8} \biggl \rceil + 3 \biggl \lceil \cfrac{31}{8} \biggl \rceil = 3 \cdot 2 + 3 \cdot 4 = 6 + 12 = 18$$

as expected. Again $\mathrm{t\_value}(x) + 1 = 6 + 1 = 7$ and $\mathrm{t\_value}(y) - 1 = 12 - 1 = 11$ are spaces.
The verification for the third pair (7, 1) mirrors that for (1, 7), given that the addends at the first member of the expression, if you exchange $x$ with $y$, are identical.