# Abstract prime factorization proof

1. Sep 16, 2010

### kathrynag

1. The problem statement, all variables and given/known data
A positive integer a is called a square if a=n^2 for some n in Z. Show that the integer a>1 is a square iff every exponent in its prime factorization is even.

2. Relevant equations

3. The attempt at a solution
Well, I know a=p1^a1p2^a2....pn^a^n is the definition of prime factorization.
We let p=2n because any even number squared is an even numbers.
Not sure how to continue.

2. Sep 16, 2010

### kathrynag

I keep trying to work on this one on I'm getting nowhere. Am I on the right track or does anyone have a suggestion?

3. Sep 16, 2010

### hgfalling

It's not an even number; the exponents are even.

So in the prime factorization of a, ${a_1,a_2,...,a_n}$ are even. Now what does that mean about the square root of a?

4. Sep 16, 2010

### Petek

@kathrynag,

When you have an "iff" problem, you really have to solve two problems. Here's a restatement of the problem to make this more clear:
Let a > 1 be an integer.

1. Assume that a is a square. Prove that every exponent in its prime factorization is even.

2. Assume that every exponent in a's prime factorization is even. Prove that a is a square.

Does this help?

5. Sep 16, 2010

### kathrynag

Yeah, that does help. I fell better about the 2nd part of the proof.
The first part still worries me.
Assume n is a square. Then we have a=n^2
That's about as far as I get with that one.
I know what I need to prove, but it's getting there.
n^2=p1^a1p2^a2....pn^an

6. Sep 16, 2010

### kathrynag

It is also even.

7. Sep 17, 2010

### hgfalling

No. Consider 9. Its prime factorization is $3^2$. Therefore it satisfies the rules of your problem. But it's not even, nor is it the square of an even number.

8. Sep 17, 2010

### Petek

Hint: Write out the prime factorization for n and then use that to come up with another prime factorization for n^2.