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I've stumbled across this playing around and summing primes at random during a boring lecture. Is this a known conjecture? Can it be proven?

My conjecture: There exists at least one non trivial solution such that $2p_n = p_a + p_b$ (the trivial being obviously $a=b=n$) for $n > 2$.

Tested by starting at the trivial $2p_n = p_n + p_n$ then incremented the right as I decremented the left until both were prime again or I've ran below $2$ with the left number. The second condition never occurred though, and I've tested for the first $1000$ primes by writing a simple program.

It fascinated me for the fact that this would mean that $p_b = 2p_n - p_a$ where $b > n > a$ so by knowing primes up to the $n$th you would have enough information to evaluate the $(n + 1)$-th?

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    $\begingroup$ This would follow from Goldbach's conjecture, though of course that itself is currently unproven. $\endgroup$ – Zev Chonoles Jun 17 '16 at 23:16
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    $\begingroup$ Are there no conditions on $a$ and $b$? So you're just conjecturing that every prime is the average of two other primes? This is quite close to the Goldbach conjecture -- presumably true, but apparently intractable to prove. $\endgroup$ – mjqxxxx Jun 17 '16 at 23:20
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    $\begingroup$ You don't actually know in advance which $p_a$ is going to partner with some $p_b > p_n$ to make this identity, even if (as seems strongly likely) Goldbach's conjecture is true. So no - you can't evaluate the $(n+1)$th prime from this. (example $59 \times 2 = 47+71$ which doesn't tell us about $61$ or $67$, or show that $59 \times 2 -53 = 65$ is not prime) $\endgroup$ – Joffan Jun 17 '16 at 23:22
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    $\begingroup$ It does not seem to follow directly from the Goldbach Conjecture. $\endgroup$ – André Nicolas Jun 17 '16 at 23:27
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    $\begingroup$ @AndréNicolas yes, strictly you need the extended Goldbach Conjecture - which is also regarded as true-but-unproven. $\endgroup$ – Joffan Jun 17 '16 at 23:34
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This is equivalent to saying that there are infinitely many triples of primes $(p_a,p_n,p_b)$ which form an arithmetic progression. While this is a consequence of the Green-Tao theorem, it was shown in 1939 itself by van der Corput. In the paper "Linear equations in primes" (Mathematika, 39 (1992), pages 367-378), Balog showed that there exists for every $n$, $n$ primes such that the average of every two of them is prime.

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    $\begingroup$ This does not answer the original question however, which asks whether every prime is the middle element of such an arithmetic progression. $\endgroup$ – Aravind Jun 19 '16 at 12:57
  • $\begingroup$ I've overlooked this when I accepted the answer. Now I'm not so sure anymore. The "better formated" question would be: There are infinitely many triples of primes ($p_a, p_n, p_b)$ which form an arithmetic progression such that every prime is the middle element of such an arithmetic progression. then? (looking for terminology help here to maybe update the question title) @Aravind $\endgroup$ – Ilhan Jun 19 '16 at 13:45
  • $\begingroup$ It was sufficiently close of an answer for me to accept it as one, thanks for your pinpoint though. I decided on keeping it. Though it would be lovely if you could help me with terminology to sum this idea up into one simple sentence @Aravind $\endgroup$ – Ilhan Jun 19 '16 at 13:56

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