5
$\begingroup$

This is related to the conjecture that all odd integers greater than $17$ can be written as the sum of 3 distinct primes.

Schinzel showed that the Goldbach conjecture implied this in 1959 and as the Goldbach conjecture has been verified up to $4\times10^{18}$ by Oliveria e Silva, Herzog and Pardi, this conjecture holds up to there as well.

Vinogradov's proof that all sufficiently large odd integers are the sum of three primes implies that as the number of representations of a sufficiently large odd integer as the sum of three primes is large enough, it must be the sum of 3 distinct primes.

Harald Helfgott's proof of the weak Goldbach conjecture also implies that all odd integers greater than $10^{27}$ can be represented as the sum of 3 distinct primes (assuming I've understood it correctly).

So the interval between $4\times 10^{18}$ and $10^{27}$ is the remaining interval on which to verify whether all odd integers greater than $17$ can be written as the sum of 3 distinct primes.

In Helfgott's proof, he uses the fact that the Goldbach conjecture is verified up to $4\times 10^{18}$ along with a prime ladder to show that the odd integers in this interval are the sum of 3 primes. But that prime ladder is a list of primes from 3 to beyond $10^{27}$ such that consecutive primes have difference at least 6 and at most $4\times 10^{18}$. Thus for odd $n$ between $4\times 10^{18}$ and $10^{27}$, there is always a $p$ in the ladder such that $n-p$ is equal to an even number less than $4\times 10^{18}$ and is therefore the sum of 2 primes, $q$ and $r$, so $n = p + q + r$.

As the verification of the Goldbach conjecture up to $4\times 10^{18}$ showed that each even integer was the sum of 2 distinct primes, it would be necessary to create a similar prime ladder such that consecutive primes had difference at least $8$ (as $n-p=6$ could only be written as 3 primes as $n = p + 3 + 3$ which is not a representation with distinct primes) and ensuring that odd integers between $4\times 10^{18}$ and $8\times 10^{18}$ are the sum of 3 distinct primes (as $p$ could potentially equal $q$ or $r$ for $n$ in this interval).

How difficult is it to make such a ladder and verify that primes between $4\times 10^{18}$ and $8\times 10^{18}$ are the sum of three distinct primes? For the second part I imagine finding a prime, $s$, just below $4\times 10^{18} - 8$ and a prime, $t$, just above $4\times 10^{18}$ would solve it as by adding any even number less than $s$ to $s$ you could reach any odd number up to just below $8\times 10^{18}$ and by adding even numbers less than $t$ to $t$ you could reach the other odd numbers less than $8\times 10^{18}$.

$\endgroup$
6
  • 5
    $\begingroup$ For the ladder, surely the first step is to take Helfgott's latter, note instances of gap $6$, and adjust them by searchin for a slightly smaller or larger prime? For the second part, can't one use the fact that there exists a prime between $n/2$ and $n$ (or more precise bounds known), and then represent $n-p$ as a sum of two distinct primes? $\endgroup$
    – Will Sawin
    Apr 13 at 1:06
  • $\begingroup$ Harald Helfgott is on this site. Maybe he can weigh in? $\endgroup$
    – sharpend
    Apr 13 at 18:49
  • $\begingroup$ @Sarosh: At the bottom left of this page there is a link named "Contact": click it, and in the form on the page that you are taken to select "I need to merge user profiles". List the user profiles that you have used and want merged, and the StackExchange team will readily do it for you. $\endgroup$
    – Alex M.
    Apr 13 at 19:11
  • $\begingroup$ I am not at all against the use of computers for number theoretic problems, but in this specific case, I would say it is a complete waste of worldwide computing resources, given the fact that even if someone does find one or two or even a whole bunch of counterexamples in that range, it would not have the least effect on any other mathematical question... sorry! Resources can be used better than that. $\endgroup$
    – Wolfgang
    Apr 14 at 8:48
  • $\begingroup$ In Helfgott's proof he says a similar prime ladder can be constructed taking only about 25 hours on a single processor core on a modern computer. Additionally I believe the prime ladder he created is suitable to answer this question and have answered so below $\endgroup$
    – Sarosh
    Apr 15 at 8:50
1
$\begingroup$

The slight issue with Will Sawin's answer above for the second part where he suggests Bertrand's postulate is the case where the prime in between n/2 and n equals n-6, n-4 or n-2. Therefore it seems better to use the fact that the largest prime gap less than 15x10^18 is 1510 so for odd 4x10^18 < n < 15x10^18 there is definitely a prime, p, such that n - 3024 < p < n - 1510 and as 6 < n - p < 4x10^18, n - p is the sum of 2 distinct primes, neither of which could equal q as they would have to be much smaller than q as n is greater than 4x10^18. This implies that all odd numbers less than 15x10^18 are the sum of 3 distinct primes.

For the first part to do with prime ladders, I misstated the problem. The difference between consecutive primes on the ladder could be between 4 and 4*10^18 and the issue was that if the gaps between consecutive primes in the ladder, p and q, p > q, were 4x10^18 apart then if n equals p+2, p+4 or p+6, n-q > 4x10^18 and n-p = 2, 4 or 6 none of which are the sum of 2 distinct primes. This is all from page 28 of Helfgott's arxiv pdf released in 2015. He shows that 4x10^18 + 2 is the sum of two distinct primes on pg 28 and on pg 305 states that there is a prime ladder from 3 to 8.8x10^30 with consecutive primes on the ladder having difference less than or equal to 4x10^18 - 6. So if p and q are any consecutive primes on the ladder, as they have difference less than or equal to 4x10^18 - 6 then for any odd n between q + 8 and p + 6 inclusive, n - q is less than 4x10^18 and greater than 6. So n - q is the sum of 2 distinct primes so n is the sum of three distinct primes (as q cannot be equal to the 2 distinct primes that sum to n - q as n - q < 4x10^18 so if n > 8x10^18 then q > n - q and if n < 8x10^18 then the argument at the beginning of this shows that n is the sum of 3 distinct primes). I believe that this shows that all odd numbers in the interval 4x10^18 and 10^27 are the sum of 3 distinct odd numbers (the result pretty much follows from the work done within Helfgott's proof).

$\endgroup$
1
  • $\begingroup$ Why doesn't this answer the question? $\endgroup$
    – Sarosh
    Apr 15 at 8:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.