# Show that no perfect number is a square.

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### 1 Answer

Let n be a perfect number with factorization

`2^(a_0)p_1^(a_1)...p_k^(a_k)`

The number of odd divisors (including n) is

`(a_1+1)(a_2+1)...(a_k+1)`

Because the sum of an sequence of numbers is even only if there are an even number of odd numbers in the sequence and odd only if there are an odd number of odd numbers in the sequence we can make the following observations.

If n is an even perfect number the number of odd divisors must be even.

If n is an odd perfect number then the number of odd proper divisors must be odd.

Now if n is a square then each of the exponents in it's factorization must be even. The number of odd divisors must be odd because a product of odd numbers is odd. With n even we must have an odd number of divisors, but when n is odd, and n is not considered a proper divisor, there are an even number of odd proper divisors.

To sumerize if n is a square and even, it has an odd number of odd proper divisors. If n is a square and odd it has an even number of odd proper divisors.

If n is an even perfect number it cannot be a square because the number of odd divisors must be even.

If n is an odd perfect number it cannot be a square because the number of odd proper divisors must be odd.

So a perfect number cannot be a square.