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