1
$\begingroup$

Is it true that between any non-prime square-free number and it's double is another non-prime square-free number?

$\endgroup$
1
$\begingroup$

Let the non-prime square-free number be denoted as $\prod_{n=1}^{k} p_n$, where $k > 1$, and $p_{n} > p_{n-1}$ for all $n$. By Bertrand's postulate, we are guaranteed a prime $p$ such that $p_k < p < 2p_k$. Then we must have

$$\prod_{n=1}^{k} p_n < p\prod_{n=1}^{k-1} p_n < 2\prod_{n=1}^{k} p_n,$$

and since $p > p_k$, $p > p_n$ for all $n$, meaning $p$ is distinct from all $p_n$; therefore, there exists a composite square-free number between any composite square-free number and its double.

$\endgroup$
4
  • $\begingroup$ Is this a partly non-trivial correct observation? $\endgroup$ – user128932 Aug 15 '14 at 2:34
  • $\begingroup$ @user128932 What do you mean by that? $\endgroup$ – Fargle Aug 15 '14 at 3:35
  • $\begingroup$ Is this a useful observation that isn't easily dismissed as trivial? $\endgroup$ – user128932 Aug 15 '14 at 6:01
  • $\begingroup$ I'm not sure how useful it is, but it's definitely not trivial. It follows rather immediately from Bertrand's postulate, but I don't think that makes it trivial. $\endgroup$ – Fargle Aug 15 '14 at 18:02
0
$\begingroup$

Aside from the trivial pair 1, 2, yes: Bertrand's Postulate tells you there is not merely a squarefree number, but a prime number, which is trivially squarefree.

$\endgroup$
1
  • $\begingroup$ The recent edit to the OP clarifies that both the original and sought square-free numbers are composite. $\endgroup$ – Fargle Aug 14 '14 at 7:49

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.