# Square free problem

• January 30th 2008, 12:07 PM
Square free problem
Prove that an integer n>1 is square-free iff n can be factored into a product of distinct primes.

My proof.

Assume $n=p_{1}p_{2}...p_{n}$ for p being distinct primes. Since all the factors are no equal to one another, n cannot be divided by a square of an integer.

conversely, assume n to be square free, then n cannot be divided by a square of an integer...
• January 30th 2008, 03:49 PM
ThePerfectHacker
Quote:

Originally Posted by tttcomrader
conversely, assume n to be square free, then n cannot be divided by a square of an integer...

Correct. Because if $n=p^2 p_1p_2...$ then $p^2$ divides $n$ which makes it non-square free.