Is this Goldbach-type problem easy to solve?

Question

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.

To help with people using Google: this is the same as finding two unique Prime Partitions of a number, each of only two elements. Related (not duplicate): http://math.stackexchange.com/questions/89240/prime-partition — apnorton, Jul 15 '14 at 18:09
@Bananarama : He is asking whether sum of the pair odd primes can be written as sum of another pair of odd primes. For example, $13+3=11+5$ — Swapnil Tripathi, Jul 15 '14 at 18:09
Yes. As I think was pointed out in the deleted answer, you can do this with any two sets of twin primes. So, in fact, there are probably infinite solutions, although this is unsolved. — hexaflexagonal, Jul 15 '14 at 18:15
@SwapnilTri: I'm asking something a bit weaker. Given an odd prime number $p$, is there an odd prime $q$ such that the sum $p+q$ can be written as a sum of another pair of odd primes. — Sebastien Palcoux, Jul 15 '14 at 18:17
@Divergent Queries: The argument given by bananarama is applicable only when $p$ itself is a twin prime. — Swapnil Tripathi, Jul 15 '14 at 18:17
@Sebastein : I understood that. Couldn't express it correctly. ;) — Swapnil Tripathi, Jul 15 '14 at 18:19
Sorry, you're right. I was being careless. I guess I misread the question as "Do such primes $p,q,p',q'$ exist?" but of course that's pretty obvious. — hexaflexagonal, Jul 15 '14 at 18:20
Do you mean "Are there odd prime numbers $q,p',q'$ such that..."? — , Jul 15 '14 at 18:29
@Bananarama: it's a corollary of the Polignac's conjecture, but the Polignac's conjecture is much stronger. I ask if my weaker problem can be solved (more) easily. — Sebastien Palcoux, Jul 15 '14 at 18:31

Asinomás · Answer 1 · 2014-07-15T18:25:18.243

2

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.

edited Jul 15 '14 at 18:25

answered Jul 15 '14 at 18:07

Asinomás

107,565

Is this an example of what you want? – Asinomás Jul 15 '14 at 18:11
1

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. – Squirtle Jul 15 '14 at 18:12
Oh, ok I understand – Asinomás Jul 15 '14 at 18:13
There is the answer! +1 for bouncing back! :D – Swapnil Tripathi Jul 15 '14 at 18:29
Same comment: it's a corollary of the Polignac's conjecture, but the Polignac's conjecture is much stronger and still open. I ask if my weaker problem can be solved (more) easily. – Sebastien Palcoux Jul 15 '14 at 18:36
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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. – Asinomás Jul 15 '14 at 18:37
It's true, you taught me the existence of this conjecture. Nevertheless, what you've posted is not an answer but a comment. – Sebastien Palcoux Jul 15 '14 at 18:42
I'm terribly sorry, it's just it's really hard for me to assess when something can be considered an answer or not when the OP doesn't give any information about what he has tried and doesn't even post the answer clearly. – Asinomás Jul 15 '14 at 18:45
It's ok. An answer could be an easy proof, or some evidence that it's as hard as the Polignac's or the Goldbach's conjectures. – Sebastien Palcoux Jul 15 '14 at 18:52

score 1 · Accepted Answer · answered Jul 15 '14 at 23:02

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.

What a coincidence! I had the same idea, I posted the question Is this weak asymptotic Goldbach's conjecture open? only 4 minutes before your answer. Thank you! — Sebastien Palcoux, Jul 15 '14 at 23:19

Is this Goldbach-type problem easy to solve?

2 Answers2