In this article, we prove that only perfect squares have an odd number of divisors
This behavior can be readily observed in the first few natural numbers
Let’s start with the proposition that
Now, let’s take a slight segue and establish a method to count the number of divisors there are for any given natural . Suppose we uniquely factor into the product of powers of its primes , then it is clear that
We can show this by rewriting as
Because , each of , hence must still be a natural number.
Now let’s establish a counting argument on the divisors of based on the powers of each of the primes in its factorization. Informally, suppose there are prime factors of , then in order for another number to be a divisor of , must be a product of the powers of the prime factors such that . Suppose , there are 3 ways to fill the exponent for :
This suggests that in general, for each prime, there are ways of filling its exponent without considering 0 and ways considering 0. If there are such exponents to be filled, then the number of total divisors is described by the expression
Going back to the proof, the rest is trivial given the above expression.
We first consider the forward direction
If is odd, then it must be the case that all . If all exponents of the primes are even, then naturally is a perfect square.
Next, we consider the reverse direction
If and , then . Therefore, the number of divisors of n is
Because will always be odd, the product of a finite number of odd natural numbers will still be odd, hence is odd whenever is a perfect square.