# Regarding 'non-square- free ' numbers.

Call an integer 'n' that is not a square or a prime power or a square-free a 'square-in'.Let n be square-in. Then between n and (2 n) is there another square-in? This is a kind of 'variation' on Bertrand's Postulate.

• Let there exist Sr and S(r+1) ; two consecutive square-ins where S(r+1)/ S(r) > 2. So S(r) < 2S(r) < S(r+1) yet (2 S(r)) would be square-in. Contradiction. Therefore for any square-ins S(r) and S(r+1) ; S(r+1)/ S(r) <= 2 . If there is an integer n such that between n and 2n there is no square-in then there exists square-ins S(t) and S(t+1) where S(t) < n < 2n < S(t+1) yet S(t+1)/S(t) <= 2 , contradiction. Does this prove the above? – user128932 Sep 21 '14 at 4:38

For $n > 8$ the answer is certainly, as we just need to look at multiples of four, and at least one of them is not a square.
$n=1$ false, $n = 2$ false, $n = 3$ false, $n = 4$ true if we allow the upper bound of the range. $5 \leq n \leq 8$ all true due to either $8,12$. So depending on the definition, take $n \geq 4$ or $n > 4$.
EDIT: It has been noted that you require $n$ to be square-in, the first of which is $8$. So the claim is true.
• Due to the restriction that $n$ is square-in, the hypothesis is true for all positive $n$, since the first positive square-in $n$ is $8$. – Fengyang Wang Aug 15 '14 at 3:09
• $8$ is a prime power, so isn't the first square-in actually $12$? – Barry Cipra Sep 8 '14 at 8:16