Confusing abstract algebra question?

Question

A natural number n>1 is said to be square free if d is in the set of natural numbers with d^2|n implies d = 1. Show that n is square free if and only if n = p1....pk for distinct primes p1....pk.

I'm not really sure how to go about this... any help would really be appreciated!

Part of my problem is that I don't fully understand what it means for a number to be square free.... I've never seen or heard this before...

Answer 1

Basically, if n is square free, then the only perfect square number that divides it is d^2, where d has to be 1.

To say that d^2 divides n is the same as saying there exists a constant k where k(d^2) = n.

This is a two way proof because of the "if and only if" phrase.
Forward -->
First, you have to prove that if n = p1...pk for distinct primes p1 ...pk, then n is square free.
In other words, under that condition that n = p1...pk for distinct primes p1 ...pk, you must show that the only perfect square number that divides n is 1.
k(d^2) = n
d^2 = n/k

Since we are under the condition that n is prime, k must be equal to 1 or n.
If k = n, then d^2 = 1
If k = 1, then d^2 = n This case implies that n is factorable by sqrt(n), which contradicts our condition that n is prime because the prime numbers by definition does not have any factors other than 1 and itself.
So d = 1 is the only possible case here.

Backwards <---
Now, you have to prove that if n is square free, then n = p1...pk for distinct primes p1 ...pk.

This is where I question the problem, 10 is square free and yet it's not a prime number. That is a counterexample against the problem. I could be really wrong, but hope my input helps

Answer 2

I was following you until you got to

iff n=p1....pk for distinct primes p1....pk

In the first expression, are you saying that n is the product of k distinct prime numbers, each with exponent 1?

Must be. Recall the Fundamental Therorm of Arithmetic:

n can be factored into primes with exponents. If the exponent of pk, call it mk, is greater than 1 then d=pk and so d^2|n

That's all that you're doing: fooling around with unique factorization using the Fund. Theo. of Arith.