Solved

# A question about the number of divisors for perfect squares

Posted on 2010-11-09

Hey,

I'm trying to prove the number of distinct positive divisors of a positive integer n is odd if and only if n is a perfect square.

I have an idea on how to go about this... but its not quite as robust as I'd like it to be, and there are a lot of holes in it, this is what I have so far:

Since n is a square, you would be able to find at least 2 factors for it, and then you would be able to take at least 2 factors of the original factors of n, and repeat this process till you run out of integers to factor. Since you are factoring each divisor into 2 parts, you would always have an even number of divisors, and then n^2 divides n^2, so you have even number of divisors + 1 which would equal an odd divisor.

The hole I'm running into is... when I have 36, I take 6 & 6, and I take the factors of 6, I get 2 & 3, and I get 2, 3, 6 & 36... which is not odd... so... that's where my theory kinda fails.

Do I need to set some restrictions? Or am I going about this the wrong way?

Appreciate any help on this.