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Problem
let $\mathrm P$ be the set of prime numbers ; $\forall(p,p+2)\in\mathrm P^2: p\times(p+2)=A_1 ,\quad A\in\mathrm N$
where $A_1$is written on the figure : $A_1=a_1^1a_2^1a_3^1....a_{n_0}^1$ where $(a_i^1)_{1\leq i\leq n_0}\subset\mathrm N $
So we put : $$g(n_0)=\displaystyle\sum_{i=1}^{i=n_0}a_i^1=a_1^1..a_{n_0}^1=A_2\quad ,A_2\in\mathrm N$$ Where $A_2$ is written on the figure :
$A_2=a_1^2a_2^2a_3^2...a_{n_1}^2,\quad (a_i^2)_{1\leq i\leq n_1} \subset \mathrm N \quad \quad \text{and}\quad n_1\leq n_0$
So we find : $$g(n_1)=\displaystyle\sum_{i=1}^{i=n_1}a_i^2$$ we reput the step until we finally get :
$$g(1)=A_k=a_1^k=8 $$ for any $(p,p+2)\in \mathrm P^2$
Exemples: $$5\times7=35,\quad g(1)=3+5=8$$ $$11\times13=143,\quad g(1)=1+4+3=8$$ $$29\times31=899,\quad g(2)=8+9+9=26\implies g(1)=2+6=8$$ $$2027\times2029=4204088,\quad g(2)=6+8+5+5+8+3=35\implies g(1)=3+5=8$$ $$7877\times7879=62062883,\quad g(2)=6+2+6+2+8+8+3=35\implies g(1)=3+5=8$$

I applied these steps to almost all pairs of less than $3000$ and other pairs greater than $5000$ and i did not find any pair of primes refuting this property
I found this problem whill studying one of this articles to prove thet prime dont'end using the topologies in this article

I wanted to ask if this problem existes before or not because i searched and found nothing about it

So this problem did not exist, how then can I prove or disprove this problem ,even though I did not find any in these results I await your suggestions and information about this ,thank you

Bernard
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Bachamohamed
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  • Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Aug 10 '20 at 22:32
  • Ok thank's @Shaun – Bachamohamed Aug 10 '20 at 22:34
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    $g(1)$ is simply remainder modulo $9$. So the entire question is "Why the remainder modulo $9$ of $pq$ is always $8$ where $p,,q$ are primes and $q=p+2$?". – Alexey Burdin Aug 10 '20 at 22:34

2 Answers2

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Your notation is baffling. You seem to say $3 \times 5 = A_1$ and $5 \times 7 = A_1$, which both cannot be simultaneously true. Also, it seems that $A_2$ is the product of a bunch of squares, as you have written it.

Your examples, however make it clear that you intend to take any pair of twin primes, $p, p+2$, then find the digital root of $p(p+2)$.

Your question is: is that digital root always $8$. This question is answered affirmatively here: Digital root of twin prime semiprimes .

Eric Towers
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As noted by Alexey Burdin in the comments, the $g(1)=8$ is the same as noting that $A\equiv8\mod9$. Also, from the definition, $A$ is the product of a pair twin primes.

Let's consider the following possibilities:

  1. $A=3\times5$
  2. All other pairs of twin primes

In the first case, we get that $g(1)=6$. This is an exception to your pattern.

For other cases, we can write the pair of twin primes as $6k-1$ and $6k+1$ for some integer $k$. Their product is therefore $36k^2-1$. As $k$ is an integer, $36k^2$ is always divisible by 9. Therefore, $36k^2-1\equiv-1\equiv8\mod9$, and the pattern will hold for all cases except for the pair $3$ and $5$.

Moko19
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