Prizes

Following the example of the mathematician Paul Erdős, we have decided to give away prizes for the solution of our unsolved problems. These are some “open points” of our proof strategies, the ones that we consider most significant to get closer to the final goal, that is the proof of Goldbach’s Conjecture:

ID	Problem	Prize	Status
1	Let $n_1$ and $n_2$ be prime numbers such that $n_1 \lt n_2$ ; let $n$ be an integer greater than 1. We want to prove that there exist two multiples of $n_1$ , which we will denote by $n_1 h$ and $n_1 k$ with $h$ and $k$ positive, such that: $n_1 h - 1$ and $n_1 k + 1$ cannot be divided by $n_2$ $h + k = n$ This statement is part of a larger problem, which has been tackled using the dashed line theory (see Study of existence of complementary space pairs based on second order dashed lines) and it has been proved except for one point left open, highlighted within the cited page: the aim is to prove precisely this open point, thus completing the proof. This problem is part of one of our proof strategies, the one based on dashes. Completing the proof would be the first step to prove analogous properties, where the integers involved are more than two.	40€	No solution received
2	Let $n_1$ , $n_2$ and $n_3$ be integers such that $1 \lt n_1 \lt n_2 \lt n_3$ . It is required to find a function $f: \mathbb{N}^{\star} \Rightarrow \mathbb{N}$ such that, for each $x \gt 0$ : $f(x) - \frac{(n_3)^2}{2} \leq \mathrm{t\_space}_{(n_1, n_2, n_3)}(x) \leq f(x) + \frac{(n_3)^2}{2} \tag{1}$ where $\mathrm{t\_space}_{(n_1, n_2, n_3)}(x)$ indicates the $x$ -th integer not divisible by neither $n_1$ , nor $n_2$ , nor $n_3$ . The function $f$ must be expressed by a formula that includes only operators of sum, difference, product, division and distinction between a finite number of cases. It is required not only to find $f$ , but also to prove that (1) holds whatever are $n_1$ , $n_2$ , $n_3$ and $x$ . As a starting point, the article Calculation of $\mathrm{t\_space}$ for dashed lines of arbitrary order can be considered. The solution to this problem would solve the last open point of the cited article, with $\delta = \frac{(n_3)^2}{2}$ . This choice of $\delta$ is motivated by the fact that, if $(n_1, n_2, n_3) = (2, 3, 5)$ , then $\frac{(n_3)^2}{2} = \frac{25}{2}$ is smaller than half the width of validity interval, which in this case ranges from $5 + 1 = 6$ to $7^2 - 1 = 48$ included (so half the width is $(48 - 6)/2 = 21$ ). This ensures that if $f(x)$ belongs to the validity interval, $\mathrm{t\_space}_{(n_1, n_2, n_3)}(x)$ must belong to it, because the difference in absolute value $\|f(x) - \mathrm{t\_space}_{(n_1, n_2, n_3)}(x)\|$ is less than or equal to $\frac{(n_3)^2}{2}$ , which itself is less than half the width of the interval (in the worst case, $f(x)$ can be in the middle of the interval and $\mathrm{t\_space}_{(n_1, n_2, n_3)}(x)$ may be near one of the bounds, but will always be within the interval). In this way it could therefore be proved that there are spaces in the validity interval of the dashed line $T_3$ (same definition as before) based solely on the $f(x)$ function. Clearly, if we limit ourselves to the $T_3$ dashed line alone, the verification of the existence of spaces in the validity interval can be performed directly, but we hope that the solutions we will receive are potentially extendable to all dashed lines $T_k$ .	100€	No solution received
3	Let $n_1, n_2, \ldots, n_k$ be integers such that $1 \lt n_1 \lt n_2 \lt \ldots \lt n_k$ . Consider the set $S := \left\{x > 0 \mid \begin{aligned}&x \textrm{ it is not divisible by any of the integers } n_1, n_2, ..., n_k;\\& x + 1 \textrm{ is divisible by } n_1 \end{aligned}\right\}$ It is required to prove that this set can be expressed as $S = \{f(x)\ \|\ g(x) \in \{n_2 a_2 + n_3 a_3 + ... + n_k a_k\ \|\ a_2 \in I_2, a_3 \in I_3, ..., a_k \in I_k\}\}$ where: $f(x)$ and $g(x)$ are functions from $\mathbb{N}^{\star}$ in $\mathbb{N}$ defined only using operators of sum, difference, product, division, modulo, integer part rounded up or down, distinction between a finite number of cases. $I_2, I_3, \ldots, I_k$ are finite subsets of $\mathbb{N}^{\star}$ It is advisable to start from the proof of the Proposition LC.5 (Spaces of a dashed line of the third order that precede a dash of the first row). Proving a theorem of this type would be essential to make progress in one of our proof strategies, the one based on dashes.	200€	No solution received

Problem

Prize

Status

Let $n_1$ and $n_2$ be prime numbers such that $n_1 \lt n_2$ ; let $n$ be an integer greater than 1. We want to prove that there exist two multiples of $n_1$ , which we will denote by $n_1 h$ and $n_1 k$ with $h$ and $k$ positive, such that:

$n_1 h - 1$ and $n_1 k + 1$ cannot be divided by $n_2$
$h + k = n$

This statement is part of a larger problem, which has been tackled using the dashed line theory (see Study of existence of complementary space pairs based on second order dashed lines) and it has been proved except for one point left open, highlighted within the cited page: the aim is to prove precisely this open point, thus completing the proof.

This problem is part of one of our proof strategies, the one based on dashes. Completing the proof would be the first step to prove analogous properties, where the integers involved are more than two.

40€

No solution received

Let

n_1

n_2

and

n_3

be integers such that

1 \lt n_1 \lt n_2 \lt n_3

. It is required to find a function

f: \mathbb{N}^{\star} \Rightarrow \mathbb{N}

such that, for each

x \gt 0

f(x) - \frac{(n_3)^2}{2} \leq \mathrm{t\_space}_{(n_1, n_2, n_3)}(x) \leq f(x) + \frac{(n_3)^2}{2} \tag{1}

where $\mathrm{t\_space}_{(n_1, n_2, n_3)}(x)$ indicates the $x$ -th integer not divisible by neither $n_1$ , nor $n_2$ , nor $n_3$ .
The function $f$ must be expressed by a formula that includes only operators of sum, difference, product, division and distinction between a finite number of cases. It is required not only to find $f$ , but also to prove that (1) holds whatever are $n_1$ , $n_2$ , $n_3$ and $x$ .

As a starting point, the article Calculation of $\mathrm{t\_space}$ for dashed lines of arbitrary order can be considered. The solution to this problem would solve the last open point of the cited article, with $\delta = \frac{(n_3)^2}{2}$ .

This choice of

\delta

is motivated by the fact that, if

(n_1, n_2, n_3) = (2, 3, 5)

, then

\frac{(n_3)^2}{2} = \frac{25}{2}

is smaller than half the width of validity interval, which in this case ranges from

5 + 1 = 6

7^2 - 1 = 48

included (so half the width is

(48 - 6)/2 = 21

). This ensures that if

f(x)

belongs to the validity interval,

\mathrm{t\_space}_{(n_1, n_2, n_3)}(x)

must belong to it, because the difference in absolute value

|f(x) - \mathrm{t\_space}_{(n_1, n_2, n_3)}(x)|

is less than or equal to

\frac{(n_3)^2}{2}

, which itself is less than half the width of the interval (in the worst case,

f(x)

can be in the middle of the interval and

\mathrm{t\_space}_{(n_1, n_2, n_3)}(x)

may be near one of the bounds, but will always be within the interval). In this way it could therefore be proved that there are spaces in the validity interval of the dashed line

T_3

(same definition as before) based solely on the

f(x)

function. Clearly, if we limit ourselves to the

T_3

dashed line alone, the verification of the existence of spaces in the validity interval can be performed directly, but we hope that the solutions we will receive are potentially extendable to all dashed lines

T_k

100€

No solution received

Let

n_1, n_2, \ldots, n_k

be integers such that

1 \lt n_1 \lt n_2 \lt \ldots \lt n_k

. Consider the set

S := \left\{x > 0 \mid \begin{aligned}&x \textrm{ it is not divisible by any of the integers } n_1, n_2, ..., n_k;\\& x + 1 \textrm{ is divisible by } n_1 \end{aligned}\right\}

It is required to prove that this set can be expressed as

S = \{f(x)\ |\ g(x) \in \{n_2 a_2 + n_3 a_3 + ... + n_k a_k\ |\ a_2 \in I_2, a_3 \in I_3, ..., a_k \in I_k\}\}

where:

$f(x)$ and $g(x)$ are functions from $\mathbb{N}^{\star}$ in $\mathbb{N}$ defined only using operators of sum, difference, product, division, modulo, integer part rounded up or down, distinction between a finite number of cases.
$I_2, I_3, \ldots, I_k$ are finite subsets of $\mathbb{N}^{\star}$

It is advisable to start from the proof of the Proposition LC.5 (Spaces of a dashed line of the third order that precede a dash of the first row).

Proving a theorem of this type would be essential to make progress in one of our proof strategies, the one based on dashes.

200€

No solution received

If you think you have found the solution for one of these problems:

Write it down in all the details, in Italian or English, and send it to . You can use the format you prefer, as long as it is easily readable, for example Word with formulas written in MathType, PDF generated by LaTeX, or HTML pages that make use of the KaTeX or MathJax libraries; we prefer the HTML format, because it would simplify the publication on our site if the solution is accepted.
We will reply in a short time confirming the receipt of your email. At the same time, we will update the status of the problem in this page, to make public the fact that we have received a solution.
We will review your solution in order to check if it is valid; this phase, which is the most delicate, may take some time. If we have received multiple solutions for the same problem, we will examine them all in order of receipt. Only the first solution that we consider valid will be entitled to the prize, but all valid solutions can be published on our website.
If we consider your solution invalid, for example because it is incorrect or incomplete, we will notify you, possibly proposing improvements, so that you can send us a new version. If you send it modified to us, we will examine it again, restarting from step 3.
If we consider your solution valid, we will ask you for your consent to publish it on our site under the Creative Commons Attribution-ShareAlike 3.0 Unported License, quoting your name (or the pseudonym you prefer). This consent is mandatory for the purpose of receiving the prize.
Once received the requested consent, we will update the status of the problem with the wording “Solved, to be published”
We will ask you for your bank account details to proceed with the payment of the prize. You will receive a bank transfer from an Italian account, with reason for payment: “solution for problem X www.dimostriamogoldbach.it”, where X is the ID of the problem. At the moment there are no other ways to pay the prize.
We will publish your solution on our website, respecting the Creative Commons Attribution-ShareAlike 3.0 Unported License.. At the same time, we will remove the problem from this page.