An integer is said to be square-free if it is not divisible by the square of
any integer greater than 1. Prove the following:
(a) An integer \(n>1\) is square-free if and only if \(n\) can be factored into a
product of distinct primes.
(b) Every integer \(n>1\) is the product of a square-free integer and a perfect
square. [Hint: If \(n=p_{1}^{k_{1}} p_{2}^{k_{2}} \cdots p_{s}^{k_{1}}\) is the
canonical factorization of \(n\), then write \(k_{i}=2 q_{i}+r_{i}\) where
\(r_{i}=0\) or 1 according as \(k_{i}\) is even or odd.]