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As one can notice every integer greater than $1$ is a sum of two squarefree numbers.(numbers that are not divided by some prime square power). Can we prove that?

Edit: Can we have bounds for the length of these numbers? (meaning the number of the primes that divide it)

Chen'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.

And one 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

do i really need the prime plus 2 or the 1 plus the double of a prime in oredr to have all the odd numbers?I think i do not need them but can we prove that??

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

EDIT Searching wikipedia i realised that this is a well-known conjecture, for more details see http://en.wikipedia.org/wiki/Lemoine%27s_conjecture

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3 Answers 3

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A majority of numbers are squarefree. More precisely, you can show that there is a constant $c<\frac{1}{2}$ such that for each positive integer $n$, less than $cn$ of the numbers $1$ through $n$ are nonsquarefree.

One way to get a rough bound for $c$ is to observe that at most $n/4$ of the numbers from $1$ to $n$ are divisible by $4$, at most $n/9$ divisible by $9$, at most $n/25$ divisible by $25$, etc. If we add up all these inequalities, there are less than $(1/4+1/9+1/25+\cdots)n\lt .46n$ nonsquarefree numbers from $1$ to $n$. The sum $\displaystyle{\sum_{p\text{ prime}}\frac{1}{p^2}=.4522474200410654985065...}$ is the prime zeta function evaluated at $2$.

So you can take $c=.46$. For $n\geq 13$, $\lfloor n/2\rfloor \gt .46n$, so not all of the $\lfloor n/2\rfloor$ pairwise disjoint sets $\{1,n-1\},\{2,n-2\},\ldots,\{\lfloor n/2\rfloor,n-\lfloor n/2\rfloor\}$ can contain a nonsquarefree number.

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    $\begingroup$ +1. Sorry, I didn't see your answer. But I will leave mine for the links :-) $\endgroup$
    – Aryabhata
    Feb 5, 2011 at 21:35
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    $\begingroup$ No need to apologize. I'm glad you answered. $\endgroup$ Feb 5, 2011 at 21:41
  • $\begingroup$ This is definitely a nice answer. Let me just say that I would not say that "most number are squarefree", since that makes me think you mean that the density of the set of squarefree numbers is one. I think "More than half of all numbers are squarefree" is less misleading. $\endgroup$ Feb 5, 2011 at 22:25
  • $\begingroup$ @Pete: Thanks for the suggestion. I will edit to make it less misleading. $\endgroup$ Feb 5, 2011 at 22:27
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    $\begingroup$ @Moron: Ah, good point, that was careless of me. No indication is given there of how large $n$ must be to apply the bounds. I'll update with a different approach in a bit. $\endgroup$ Feb 6, 2011 at 3:52
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It is known that the Schnirelmann Density of square free numbers is $\displaystyle \frac{53}{88}$ for instance, as mentioned here: http://www.jstor.org/pss/2040089. Note this is different from the natural density which is known to be $\displaystyle \frac{6}{\pi^2}$.

This implies that for any $n$, the number of square free numbers $\displaystyle \le n$ is at least $\displaystyle \frac{53n}{88}$. Since $\displaystyle \frac{53}{88} \gt \frac{1}{2}$, this implies that $n+1$ can be written as the sum of exactly two squarefree numbers.

btw, your title does not match the question.

Additive Basis means that you also include $\displaystyle 0$ in the set which you seem to exclude by saying two instead of at most two.

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  • $\begingroup$ thanx, of course the first part was just an exercise, what about the second part of the question? $\endgroup$
    – minasteris
    Feb 6, 2011 at 15:41
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    $\begingroup$ @minasteris: Please don't change the question like that. You can always open a new question, linking that to this one. Also, if the first question was just an "exercise" what was the point in asking it? If you knew the answer already, you have just wasted the time of a few people on this site, when they didn't have to. $\endgroup$
    – Aryabhata
    Feb 7, 2011 at 8:16
  • $\begingroup$ do you think that it is better to reedit it in the first form and make a new question for the second part? $\endgroup$
    – minasteris
    Feb 7, 2011 at 9:56
  • $\begingroup$ It would have been better if you had done that before editing this question. Now I expect many people have seen this newer version, and creating a new question will probably only cause more confusion. I would suggest leave it like it is. $\endgroup$
    – Aryabhata
    Feb 7, 2011 at 9:59
  • $\begingroup$ Sorry ,i just needed some points in order to offer them as a bounty to math.stackexchange.com/questions/15831/… . After i thought the second part and i thought that it would be an interesting question.Thank you $\endgroup$
    – minasteris
    Feb 7, 2011 at 10:37
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According to this paper, the number of ways to represent a positive integer $N$ as a sum of a prime and a $k$-free integer is asymptotic to

$$ {N\over\log N}\prod_{p\nmid N}\left(1-{1\over p^{k-1}(p-1)}\right) $$

as $N\to+\infty$.

By looking into the details of Chen's argument (see last chapter of Halberstam & Richert's Sieve Methods published in 1974), one can observe what Chen actually proved is that for a given large even integer $N$, the number of primes $p\le N$ such that

$$ N-p=\begin{cases} p_1p_2 & N^{1/10}<p_1<p_2<N & p_1,p_2\text{ prime} \\ p_1 & N^{1/10}<p_1<N & p_1\text{ prime} \end{cases} $$

is greater than

$$ {0.67N\over\log^2N}\prod_{\substack{p|N\\p>2}}{p-1\over p-2}\prod_{p>2}\left(1-{1\over(p-1)^2}\right). $$

I hope these two results can answer your question.

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