Find all positve integers $n$ such that for all odd integers $a$, if $a^2 < n$ then $a | n $ ? (Ref. Titu Andreescu, Number theory, page. 5-6).
OUTLINE OF THE AUTHOR'S SOLUTION:
Consider a fixed positive integer $n$. Let $a$ be the greatest odd integer such that $a^2 < n$ and hence $n \leq (a+2)^2 $. If $a \geq 7$ then $a-4, a-2 $ and $a$ are odd integers that divide $n$. Any two of these numbers are relatively prime so $(a-4)(a-2)a | n$. It follows that $ (a-4)(a-2)a \leq (a+2)^2 $ . Then $a^2 (a-7) + 4(a-1) \leq 0$ which is false.Thus $a$ is $1,3$ or $5$ . If $a=1$ then $1^2 \leq n \leq 3^2 $ hence $n =\{1,2...,8\} $. Similarly for $a=3$ and $a=5$ . Thus $n = \{1,2,3,4,5,6,7,8,9,12,15,18,21,24,30,45\}$.
I HAVE THE FOLLOWING QUESTIONS IN RELATION TO THE SOLUTION GIVEN BY THE AUTHOR
- Why does the author choose $a$ as the greatest odd integer ?
- Why does the author use $\geq $ in the relation $a \geq 7$ and not other inequalities like $<$ etc. ?
- How did the author arrive at the number $7$ in the relation $a \geq 7$ ?
- How was the author able to choose $a-4, a-2$ and $a$ as the odd divisors of $n$ because if $a > 7$ then they are not the only odd divisors of $n$?