An integer $a$ is called square-free if it is not divisible by the square of any integer greater than 1. Show that:

1. $a$ is square-free if and only if $a = \pm p_1 \dotsc p_r$, where the $p_i$’s are distinct primes;
2. every positive integer $n$ can be expressed uniquely as $n = ab^2$, where $a$ and $b$ are positive integers, and $a$ is square-free.

(1) seems to follow directly from the definition. Indeed, if the $p_i$'s are not distinct primes, then by definition there exists $p^e$ in $p_i$'s with $e \ge 2$. In other words, $a$ is not square-free.

Suppose $n$ is square-free. We can let $b := 1$ and $a := n$, so $n = n \cdot 1^2 = ab^2$.

Suppose $n$ is not square-free. Then $\displaystyle n = \prod_{i=1}^{r} p_i$, where there exists $p^e$ in $p_i$'s with $e > 1$. Start with $a := n$ and $c := 1$. For each of the $p^e$ in $p_i$'s with $e > 1$, there are only two possibilities. If $e$ is even, we can set $\displaystyle a := \frac{a}{p^e}$ and $c := {p^e}c$. If $e$ is odd (which must then be $> 2$), we can set $\displaystyle a := \frac{a}{p^{e-1}}$ and $c := {p^{e-1}}c$. Observe $a$ would end up as square-free and is unique, and c will become a product of distinct primes, each with an even exponent. Let $b = \sqrt{c}$, $n = ac = ab^2$.

$\Box$

# posted by rot13(Unafba Pune) @ 4:59 PM