Prove 2^n cannot be a perfect number
 

Given n is an integer, prove that 2^n cannot be a perfect number.

Ok, let's give it a try:

2^n is an even number.
Sum of its divisors (1 and many even numbers) is an odd number.
Hence, 2^n cannot be equal sum of its divisors. [TEX]\qed[/TEX]

Nice job Batalov. The proof looks valid to me.

Regards,
Matt

What about when n=0?

What about the more general k^n?

[size=1]A much more interesting proof would be to show that no odd number can be a perfect number.[/size]

[QUOTE=retina;404638]What about when n=0?

What about the more general k^n?

[size=1]A much more interesting proof would be to show that no odd number can be a perfect number.[/size][/QUOTE]

For n=0 k^n is odd but a list of proper divisors doesn't exist, the second statement can be partially worked out since odd +odd =even only odd numbers with an odd number of proper divisors can be perfect. Which also means that for k=odd only n=even need be considered, and yes I know this was likely all rhetorical

Since the sum of the smaller factors of 2^n equals 1 + 2 + 4 + ... + 2^(n-1) = 2^n - 1
it is never equal to 2^n, hence 2^n is never perfect.

But I like Batalov's parity explanation better.

Prove b^n (n > 1), and b is prime:

Proof:

1 is a divisor of b^n for all natural numbers.the sum of all the divisors of b^n not counting 1 is a multiple of b. Adding one gives us a non multiple of b, which in order for b^n to be a perfect number, the divisors must add up to b^n (which should give us a multiple of b of course) and does not.

[QUOTE=mathgrad;404623]Given n is an integer, prove that 2^n cannot be a perfect number.[/QUOTE]

Also see [URL="https://en.wikipedia.org/wiki/Almost_perfect_number"]Wikipedia: almost perfect number[/URL]. /JeppeSN