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Problem: Given an odd prime number $p$, are there odd prime numbers $q$, $p'$, $q'$ such that $\{p,q\} \neq \{ p',q'\}$ and $p+q = p'+q'$ ?

This comment informs that it's an obvious corollary of the Polignac's conjecture.
This conjecture is still open, and my problem seems much weaker, so that I ask for a proof.

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  • $\begingroup$ what is $\{p,q\}$? $\endgroup$ – Jorge Fernández Hidalgo Jul 15 '14 at 18:06
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    $\begingroup$ @Bananarama: it's a set. $\endgroup$ – Sebastien Palcoux Jul 15 '14 at 18:07
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    $\begingroup$ @Bananarama : He is asking whether sum of the pair odd primes can be written as sum of another pair of odd primes. For example, $13+3=11+5$ $\endgroup$ – Swapnil Tripathi Jul 15 '14 at 18:09
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    $\begingroup$ @Divergent Queries: The argument given by bananarama is applicable only when $p$ itself is a twin prime. $\endgroup$ – Swapnil Tripathi Jul 15 '14 at 18:17
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    $\begingroup$ @Sebastein : I understood that. Couldn't express it correctly. ;) $\endgroup$ – Swapnil Tripathi Jul 15 '14 at 18:19
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It seems likely that your result can be proven using methods like those used to bound the number of exceptions to the Goldbach conjecture. Let $E(x)$ be the number of even integers $\le x$ that cannot be written as a sum of two primes. It is known that $E(x) \in O(x^{1-\delta})$ for some $\delta>0$ (for instance, see references here). (That is, the number of exceptions, if there are any, grows relatively slowly.) Therefore, given a set $A\subseteq \mathbb{N}_{\text{even}}$ that is sufficiently dense (e.g., such that the number of its elements $\le x$ grows much faster than $x^{1-\delta}$), we can guarantee that some member of $A$ is a Goldbach number. In your case, let $A=\{p+q : q {\text{ is an odd prime}}\}$. This is a sufficiently dense set of even numbers: by the prime number theorem, the number of primes grows faster than $x^{1-\delta}$ for any $\delta>0$. So we have this:

For any odd composite integer $p$, there exist primes $q,p',q'$ such that $p+q=p'+q'$.

But to prove your statement when $p$ is prime, we need some member of $A$ to have not one but two distinct Goldbach partitions. Let $E_2(x)$ be the number of even integers $\le x$ that cannot be written as a sum of two distinct pairs of primes. (The only known exceptions are $6$, $8$, and $12$.) A proof that $E_2(x)\in O(x^{1-\delta})$ for some $\delta>0$ would imply your statement. Since the required bound is so weak, and the analogous result for $E(x)$ is long-known, it is plausible that this bound could be proven as well.

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Let $p$ be an odd prime and $q$ the next prime. We want to search in the natural numbers for a prime $p'$ such that $q'=p'+(q-p)$ is also a prime. If we find this $p'$ then we get

$p+q'=p'+q$.

This is probably easier to solve, but a temporary observation is that this would follow immediately from Polignac's conjecture.

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  • $\begingroup$ Is this an example of what you want? $\endgroup$ – Jorge Fernández Hidalgo Jul 15 '14 at 18:11
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    $\begingroup$ Well.... I'll change my vote back... Just seems like (1): You answered before you understood what was going on, (2) You have a ton of rep, so you should probably chill. $\endgroup$ – Squirtle Jul 15 '14 at 18:12
  • $\begingroup$ Oh, ok I understand $\endgroup$ – Jorge Fernández Hidalgo Jul 15 '14 at 18:13
  • $\begingroup$ There is the answer! +1 for bouncing back! :D $\endgroup$ – Swapnil Tripathi Jul 15 '14 at 18:29
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    $\begingroup$ I read it the first time, I thought perhaps you did not know of Polignac's conjecture, since you did not mention it in the original post. $\endgroup$ – Jorge Fernández Hidalgo Jul 15 '14 at 18:37

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