ID  Problem  Prize  State 

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 xth positive integer not divisible by neither n_1, nor n_2, nor n_3. 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 solutions received 
4 
Let n_1, n_2 and n_3 be prime numbers such that n_1 \lt n_2 \lt n_3; 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:
This statement is a more general form of another one which has been proved by one of our readers (see Study about the existence of complementary space pairs based on second order dashed lines). However, the proof of the latter is rather complex because it distinguishes between many cases and subcases; we hope that instead some reader will be able to prove this more general statement by reasoning in a different way. This problem is part of one of our proof strategies, the one based on dashes. The proof of this statement would be an important step for generalizing it further, up to prove Hypothesis H.1.T, of which Goldbach’s Conjecture is a direct consequence. 
100€  No solutions 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:
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 solutions 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 AttributionShareAlike 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 AttributionShareAlike 3.0 Unported License.. At the same time, we will remove the problem from this page.