Is this Goldbach-type problem easy to solve? 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.
 A: 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.
A: 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.
