Is this equivalent to Goldbach's conjecture?

2011-02-08T09:18:20Z

As one can easily prove http://math.stackexchange.com/questions/20564/sums-of-square-free-numbers-is-this-conjecture-equivalent-to-goldbachs-conjec every integer greater than $1$ is a sum of two squarefree numbers.

Can we have bounds for the length of these numbers? I write $(n,m)$ to denote the sum of a squarefree number of n prime factors to one of m prime factors, when $n$ or $m$ $=0$ then i mean that the summand is $1$.

Chenn's theorem asserts that for large enough even numbers the length $(2,1)$ is enough Goldbach's conjecture says that $(1,1)$ would be enough too.

CONJECTURE: Every odd number can be written as a sum of two squarefree numbers of length at most $(2,1)$ (meaning as a sum of a prime and a double of a prime or a sum of a prime plus 2 or as a sum of 1 plus a double of a prime)

Questions

1 Is there any easy counterexample?

2 do i really need the prime plus 2 or the 1 plus the double of a prime in order to have all the odd numbers? It seems too difficult to me to prove that i do not need them.

3 What is the relation of this conjecture to Goldbach's conjecture? does the one implies the other?

I apologise for the elementary style of my question , i think that this conjecture is well known but i haven't met it. If it is well known maybe it is known the relation to the Goldbach Conjecture. Maybe i miss something obvious...

NOTE: From the second question we have one new conjecture

CONJECTURE:Every prime $p$ is $p=p1+2(p2-1)$ and $p=p3+1/2(p4-1)$ for some primes $p1,p2,p3,p4$ . Of course someone could ask many questions about these form for the consecutive primes ,etc.

EDIT:after asking this question i found this related article en.wikipedia.org/wiki/Lemoine%27s_conjecture at wikipedia.

ADDED i think that one can easily see that if every even number is the sum of a prime and a Sophie Germain prime or his pair (meaning a prime of the form $2p+1$ ) this would be too strong to implie both do we have a counterexample to this??

Answer by quid for Is this equivalent to Goldbach's conjecture?

2011-02-10T14:18:48Z

The information given in the linked page on Wikipedia (to a certain extent) answers all three questions of the OP.

In more detail; It is a conjecture of Lemoine (1894) that every odd number (ignoring immediate and negligible size constraints) can be written in the form $p + 2q$ with (odd) primes $p,q$.

Thus, conjecturally the additional numbers of the form $p+2$ and $2p+1$ are not needed (cf. question 2).

It is also said there that this conjecture has been verified up to $10^9$; so there are no easy counter examples to Lemoine's conjecture (and thus also not to the weaker question asked, cf. question 1).

Finally, it is discussed there that Lemoine's conjecture is similar to but stronger than Golbach's weak conjecture, also called ternary Goldbach's conjecture; i.e., the assertion that every odd number (again, except very small exception) is the sum of three primes.

Goldbach's weak conjecture is in fact (almost) proved; Vinogradov showed that every sufficiently large odd number is the sum of three primes; however, the constant is very large so that a (computational) verification of the finitely many remaining values is open, yet under the generalized Riemmann Hypothesis Deshouillers et al. (1997) were able to fully prove Goldbach's weak conjecture.

There seems to be not direct link to the Goldbach conjecture (in the sense of an equivalence or implication); for example, if there were one it seems feasible it would also be discussed on that page (cf. question 3), and in the absence of an 'obvious' equivalence/implication it seems hard (for me) to envision a non-obvious one (except possibly at the level of actual strategies of proof).

From the general level of difficulty Lemoine's conjecture seems closer to the Goldbach conjecture, than to the weak Goldbach's conjecture; as in both one has only two primes to choose, opposed to three in the weak Goldbach's conjecture. (Note the parallelity to the twin prime conjecture and Chen's theorem.)

Whether or not the assymetrie in the equation $p+2q$ opposed to $p+q$ is rather helpful or an obstacle, is something I am not really competent to judge. My guess is that it is rather helpful, making Lemoine's conjecture possibly slightly more accessible than Goldbach's conjecture. Finally, I believe (though again I cannot tell for sure) that to additionally allow $p+2$ and $2p+1$ should not change much the general level of difficulty.

Technical remark: This is an expansion of my comment. I performed this expansion after the question reappeared following the suggestions to that extent recently expressed on meta (not specific to this question, but as a general policy). I am not an expert on this type of questions; mainly, I tried to summarize the information that I could find easily on the web, and to supplement them with some speculations based on cursory knowledge of the subject.

Is this equivalent to Goldbach's conjecture? - MathOverflow

Is this equivalent to Goldbach's conjecture?

Answer by quid for Is this equivalent to Goldbach's conjecture?