# Thread: Square free problem 2

1. ## Square free problem 2

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.

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 $p,q,r$ are prime and $n=p^2q^3 r$. Then we can write $n = (p^2 q^2) \cdot (qr)$ now $p^2 q^2 = (pq)^2$ is a square and $qr$ is square free. Another example, $n=p^2 q^4 s^3 r^5 = (p^2 q^4 s^2 r^4)(sr) = (pq^2sr^2)^2 (sr)$. Get the idea?

3. I don't understand that you can set $
n=p^2q^3 r
$
, what allows an integer to be a product of two or more of the same prime number?

I don't understand that you can set $
n=p^2q^3 r
$
, what allows an integer to be a product of two or more of the same prime number?
Because I am trying to show you an example. I am saying suppose that $n$ 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.