# Regarding square-free numbers and their doubles.

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

## 2 Answers

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.

• Is this a partly non-trivial correct observation? – user128932 Aug 15 '14 at 2:34
• @user128932 What do you mean by that? – Fargle Aug 15 '14 at 3:35
• Is this a useful observation that isn't easily dismissed as trivial? – user128932 Aug 15 '14 at 6:01
• 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. – Fargle Aug 15 '14 at 18:02

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.

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