I have thought of a conjecture similar to Goldbach Conjecture. I have shown the result to be true with a program in C++ up until $n=30000$.

$\forall n>2$ with $n$ even,there exists two primes $p,q$ with $n<p,q<2n$ $\;$ and $\;$ $p+q=3n$

We know Betrand's postulate:

For any integer $n > 3$, there always exists at least one prime number $p$ with $n < p < 2n - 2$.

But I don't know if this postulate can help me.

I wanted to place it somewhere public so people can think about it. Are there any results similar to this?

  • 1
    $\begingroup$ Side-note: Get rid of that "$|$", it normally stands for a "divides" notation. $\endgroup$ – barak manos Feb 11 '15 at 12:49

There are various known modifications of the Golbach conjecture were restrictions are imposed on the size of the summands.

This idea is not really original.

Bertrand's postulate is a very weak result; it is extremely unlikley to be of direct use in the proof of a conjecture that should be rather more difficult than the Goldbach conjecture.


Sure, of course this conjecture is similar to the Goldbach Conjecture.

GC: If $n>2$ is even then there are primes $p,q$ with $p+q=n$.

Yours: If $n>2$ is divisible by 6 then there are primes $p,q$with $p+q=n$ and $\frac{n}{3} < p,q < \frac{2n}{3}$.

I'm certain that the following strengthening of your conjecture is also true:

Yours++: if $n>2$ is even then there are primes $p,q$ with $p+q=n$ and $\frac{n}{3} < p,q < \frac{2n}{3}$.

The only difference is that I'm not requiring that $n$ is divisible by 6, only that it's even. Now, this stronger version is actually the same as GC except that it makes a further claim about the size of the primes.

We believe that not only GC is true, it is in fact true "in abundance": there are likely many, many pairs of primes that add up to any given even number. In particular, there's no reason to believe that such a pair cannot be found in the interval between one third and two thirds of n. Nevertheless, this is a stronger conjecture, and it is conceivable that we will prove GC without proving Yours++, and it's even possible (though extremely unlikely) that GC is true while Yours++ is false.

Now, your original conjecture (Yours) does not imply GC because it doesn't make any claim about even numbers that don't also happen to be divisible by 3. I find it very unlikely that this restriction makes any difference; in other words I would regard Yours and Yours++ as essentially equivalent.

The main question I would propose is - what do you expect would be the value of exploring this variation of GC? It's fine to suggest refinements or versions of open problems, but this is usually done with an eye towards making things clearer or more specific or easier to prove.

For example, if you had reason to believe that for large, even numbers, the only possible primes that add up to those numbers are in the middle-third interval you suggest, that would be quite sensational and an important thing to focus on. But I'm fairly certain this isn't true, so that for all even numbers there are sum-of-primes representations that lie all over the place. Therefore I'm afraid it seems unlikely that these variations would help in the quest of resolving GC itself.


the only way to get p + q = 3n is to add one prime of the form 6k+1 to one of the form 6k-1. so we must have:

(6x+1) + (6y-1) = 6(x+y) = 3n which gives n=2(x+y). So if we take p=13 and q=11, we have n=2(2+2)=8. So n < p becomes 8 < 13 and q < 2n becomes q < 16.

So we can see that there are many pairs (x,y) that can work.


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