-1
$\begingroup$

I recently noticed that many even numbers formed as 2p (p prime number) have at least two pairs of prime numbers that they dercribe that even number via Goldbach's Conjecture.

Examples: 10=2*5=5+5=7+3 34=2*17=17+17=23+11=29+5=31+3

I wanted to know if there's any research on how many pairs exist that they describe an even number. I think that, equilevantly, my question is : Is there at least one k >=1 for every p so that p-2k and p+2k are prime at the same time?

$\endgroup$
0

1 Answer 1

5
$\begingroup$

All approaches to Goldbach's problem give not only the existence of solutions, but lower bounds for the number of solutions.

Using the circle method one can show that the asymptotic formula $$ \#\{(p,q):p+q=n, p, q\mbox{ prime}\} \sim \mathfrak{S}(n)\frac{n}{\log^2 n} $$ holds for almost all integers $n$. The first result of this type is probably due to Cudakov (Dokl. Akad. Nauk. SSSR 17, 1937). Replacing the asymptotic formula by a lower bound Montgomery and Vaughan (Acta Arith. 27, 1975) showed with the exception of $\mathcal{O}(X^{1-\delta})$ even integers $n$ the number of representations is not too much smaller than the expected size. As far as I know, the best results in this direction are due to Pintz, but not completely published.

Sieve methods, which can prove the existence of solutions of the equation $n=p+q$, where $q$ has at most two prime factors, also prove the existence of many such solutions. However, here the proven lower bounds differ a lot from the expected magnitude, because sieves prior to GPY did not deal with "at most 2 prime factors" directly, but with "the smallest prime factor is at least $n^{1/3}$". Although all integers with no small prime factor has at most two prime factors, most integers with at most two prime factors have one very small and one very large prime factor.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .