Thread: 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.

Originally Posted by tttcomrader

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 are prime and . Then we can write now is a square and is square free. Another example, . Get the idea?

I don't understand that you can set , what allows an integer to be a product of two or more of the same prime number?

Originally Posted by tttcomrader

I don't understand that you can set , 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 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.