Prove that every integer n > 1 is the product of a square-free integer and a perfect square.
proof.
Should I start by saying n = a product of primes by the fundamental theorem? But I can't seem to get anywhere after that.
Let us start easy. Suppose $\displaystyle p,q,r$ are prime and $\displaystyle n=p^2q^3 r$. Then we can write $\displaystyle n = (p^2 q^2) \cdot (qr)$ now $\displaystyle p^2 q^2 = (pq)^2$ is a square and $\displaystyle qr$ is square free. Another example, $\displaystyle n=p^2 q^4 s^3 r^5 = (p^2 q^4 s^2 r^4)(sr) = (pq^2sr^2)^2 (sr)$. Get the idea?
Because I am trying to show you an example. I am saying suppose that $\displaystyle n$ has that specific type of factorization, then we can do what I did above. Those were just examples. There is a general rule that we use, that rule is what you have to find and apply it to the general case.