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Curious about the Goldbach conjecture, and reading about twin primes, I was wondering if it is possible that every prime number as $P$, can be expressed as a prime factor of at least one $E$ such that $E$ is the sum of a pair of twin primes?

The probability that a number $n$ is prime is about $1/\log(n)$, therefore, if the probability that $n+2$ is also prime was independent of the probability for $n$, we should have the approximation $p_2(n) \sim \frac{n}{ \log^2(n)}$ (In a simplified way).

So far checking the first $5000$ twin primes seems encouraging, but as far as large numbers, and with twin primes being more rare, I am wondering if it is even possible?

Examples ($P$ is any prime. $E$ is the sum of a pair of twins $T_1$ and $T_2$):

$P$ $E$ (one example). $E = T_1 +T_2$
$2$ $24 = 2 ⋅ 12$ $11+13$
$3$ $36 = 3 ⋅ 12$ $17+19$
$5$ $60 = 5 ⋅ 12$ $29+31$
$7$ $84 = 7 ⋅ 12$ $41+43$
$11$ $396 = 11 ⋅ 36$ $197+199$
$13$ $624 = 13 ⋅ 48$ $311+313$
$17$ $2040 = 17 ⋅ 120$ $1019+1021$
$19$ $1140 = 19 ⋅ 60$ $569+571$
$23$ $276 = 23 ⋅ 12$ $137+139$
$\,\vdots$ $\,\,\,\,\vdots$ $\,\,\,\,\vdots$
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    $\begingroup$ Two obvious comments: Heuristically, the answer is yes: there are (heuristically) infinitely many pairs of twin primes, and for a fixed $P > 2$ the probability of the sum of such a pair being a multiple of $P$ is $1/P$. On the other hand, your conjecture implies the twin prime conjecture, so you cannot expect an unconditional positive answer. $\endgroup$ Jun 21 at 9:26
  • $\begingroup$ @Mees de Vries. Thanks, yes I was aware that proofs are not involved, but something didn’t click with me with regards to twin primes being more rare, but in the other hand when summing them we have more possibilities for more prime factors (especially for larger numbers). I wasn’t sure if it was even possible. $\endgroup$ Jun 21 at 9:35
  • $\begingroup$ "I was wondering if it is possible that any prime number as $P,$ can be expressed as..." $${}$$ This could be read as "I was wondering if it is possible that there is any prime number $P$ that can be expressed as..." $${}$$ Or it could be read as "I was wondering if it is possible that any prime number $P$ at all, no matter which one, can be expressed as..." $${}$$ If the latter was intended, then just changing "any" to "every" would make it unambiguous. $\endgroup$ Jun 21 at 10:55
  • $\begingroup$ If only finitely many twin primes exist, then this statement is false. And nobody has yet proved that infinitely many twin primes exist. $\endgroup$ Jun 21 at 11:00
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    $\begingroup$ @Michael Hardy I have modified “every” instead of “any” per your comment $\endgroup$ Jun 21 at 15:28
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To add to what's been said in the comments, here are some further thoughts.

Case 1: The twin prime conjecture is false. This is unlikely to be true, but in the event that somehow there are only finitely many twin primes, then as Michael Hardy said in their comment, the statement is false. To be rigorous, let $A = \left\{p+1: p\ \text{and}\ p+2\ \text{are prime} \right\}$. The sum of a pair of twin primes $p$ and $p+2$ is $2p+2=2(p+1)$. If $B$ denotes the collection of all prime divisors of elements of $A$, then the falsehood of the twin prime conjecture implies that $A$ is finite, and so $B$ must also be finite. Thus the claim would be false, since there are infinitely many primes and $B$ is a finite subset of primes.

Case 2: The twin prime conjecture is true. This is much more likely the case. With the same definitions of the sets $A$ and $B$, one needs to show that $B$ is the set of primes. That is, every prime $q$ (other than 2, obviously) is a divisor of at least one number of the form $p+1$, where $p$ and $p+2$ are both primes. This is an interesting sieve-theoretic problem that might be phrased as showing that, for a given odd prime $q$, the sum $$ \sum_{\substack{p\\ p,p+2\ \text{are prime}\\ q\mid p+1}} 1 = \sum_{\substack{p \equiv -1\ (\text{mod}\ q)\\ p,p+2\ \text{are prime}}} 1 $$ is nonzero. The sum on the right is a question about twin primes in arithmetic progressions. As Mees de Vries said in their comment, this result implies the twin prime conjecture, so certainly no proof of this claim currently exists (nor is likely to exist for quite a long time).

However, there are averaged versions of this that do exist completely unconditionally. Let $\Lambda(n)$ denote the von-Mangoldt function (which is essentially a weighted version of the prime counting function). For coprime integers $a$ and $q$, define $$ \psi(x;q,a,2k) = \sum_{\substack{m,n\leq x\\ m-n=2k\\n\equiv a\ (\text{mod}\ q)}} \Lambda(m)\Lambda(n). $$ That is, $\psi(x;q,a,2k)$ counts the number of primes $p$ such that $p+2k$ is also prime and $p\equiv a\ (\text{mod}\ q)$. For your question, we take $q$ a prime, $a=q-1$, and $k=1$. If we denote by $H(x;q,a,2k)$ the expected main term of this sum, then there are several results which prove that the error term $$ \left|\psi(x;q,a,2k) - H(x;q,a,2k) \right| $$ is small on average over $k$. For results of this kind, as well as precise definitions of the heuristic main term $H$ and the phrase "small on average", see this paper by Mikawa and the references mentioned in the introduction.

Also, to see that your statement implies the twin prime conjecture, suppose there are only finitely many twin primes and let $L$ be the largest of them. Choose a prime $q > L$. Then by your claim, there exist twin primes $p$ and $p+2$ such that $q$ divides their sum. But clearly $p > q > L$, contradicting the fact that $L$ was the largest twin prime. Thus there are infinitely many twin primes.

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  • $\begingroup$ An alternative version of the last paragraph is to take the contrapositive of the "case 1" analysis, which concludes by noting that if TP is false, then C is false (where C is the OP's claim is false. By the contrapositve, if C is true, then TP must be true. $\endgroup$ Jun 21 at 22:27

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