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I was thinking about twin primes and I came to ask this question:

If we have two distinct primes $P_1$ and $P_2$ which are both greater than $3$, then does there always exist a prime $P_3$ such that

$$\begin{cases}P_1+6 P_3 \in \mathbb P \\ P_2+6 P_3 \in \mathbb P \\ \end{cases}$$

I tested this for all prime pairs up to $3581$ (500th prime) and the only one for which the $P_3$ was greater than $541$ (100th prime) was pair $\{1451,3163\}$ (230th and 447th primes). But $P_3 = 631$ works.

I might add that this is, to my mind, equivalent to twin prime conjecture. Why? Well every twin prime pair is of the form $(6k-1,6k+1).$ Now, for some given prime $P$, this can be written as $$(6(n P+v)-1,6(n P+v)+1)$$ and this, by expansion, is $$(6 n P+6 v-1,6 n P+6 v +1).$$ But there is a potential twin prime here too, namely $(6v-1,6v+1).$ So by setting $n=1,$ and $$\begin{cases}6 v-1 \in \mathbb P \\ 6v+1 \in \mathbb P \\ \end{cases}$$ which is clearly possible, then, if the above is true, we can select $P$ so that both of the originals are prime and we have gotten ourselves a new and bigger twin prime pair. So then there would be infinitely many twin primes.

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    $\begingroup$ Is it even known whether for every prime $P_1>3$ there exists a prime $P_3$ such that $P_1+6P_3 \in \Bbb P$? $\endgroup$ – Jaycob Coleman Oct 24 '13 at 20:33
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    $\begingroup$ If you require this only for $P_1=5$ and $P_2=7$ while at the same time generously allowing natural numbers for $P_3$, you have the twin prime conjecture. $\endgroup$ – Hagen von Eitzen Oct 24 '13 at 20:56
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    $\begingroup$ @HagenvonEitzen I don't see how this is the case when he only conjectures the existence of at least one such $P_3$ for each pair. $\endgroup$ – Jaycob Coleman Oct 24 '13 at 21:09
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    $\begingroup$ Taking $P_1,P_2$ to be a twin prime pair, this conjecture implies there is a larger twin prime pair. $\endgroup$ – Wojowu Oct 13 '16 at 15:55
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    $\begingroup$ "We can select" is a very problematic phrase. $\endgroup$ – JustKevin Dec 8 '16 at 19:08
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A few words, and a few search heuristics:

Third paragraph, about all twin prime pairs having that form, not strictly true (3,5) . Secondly, the conjecture, is closer to the prime gap conjecture about any even gap infinitely often. Third, for 6v-1,6v+1 to both be prime, v can't be one of 3 forms, set out by a similar argument to the Sieve of Sundaram ( exception: 0 as either input to those forms).

Now on to Search heuristics :

  • Mod 7,$p_3$ is not the same as either of $p_1$ or $p_2$ if so one of the expressions goes to 0 mod 7, so would have to sum to 7 or not be prime.(Also shows no set of 3 pairs with all distinct non-zero remainders, will have the same $p_3$ without it being 7)
  • Mod 5, you can show that no 2 prime pairs with all distinct non-zero remainders can have the same $p_3$ except if it equals 5
  • Mod 3, shows us that any pair without distinct non-zero remainders, will never find a twin prime pair.
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  • $\begingroup$ I don't understand what you're trying to say about search heuristics. Could you clarify? $\endgroup$ – Robert Shore Feb 20 at 1:45
  • $\begingroup$ Mod 7 the expressions become p1-p3 and p2-p3, Mod 5 the expression become p1+p3 and p2+p3, Mod 3 the expressions become p1 and p2. the first ones solve to 0 remainder if they are the same. The second ones solve to 0 if they are additive inverses. The last ask if primes exist with same remainder mod 3. $\endgroup$ – Roddy MacPhee Feb 20 at 1:52
  • $\begingroup$ since there are 7 remainders Mod 7 any 3 all distinct pairs of remainders use up all non-zero remainders forcing p3 to have 0 remainder. $\endgroup$ – Roddy MacPhee Feb 20 at 1:55
  • $\begingroup$ I'm suggesting that you edit your answer to include a lot more exposition. How are you suggesting one should use these heuristics? $\endgroup$ – Robert Shore Feb 20 at 1:55
  • $\begingroup$ Mod 5 works similar, except now they can't have additive inverse remainders. There are only 4 non-zero remainders to have 2 pairs of distinct remainders can use up all non-zero remainders, forcing p3 to have 0 remainder. $\endgroup$ – Roddy MacPhee Feb 20 at 1:59

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