Prove 2^n cannot be a perfect number

mathgrad · 2015-06-23, 01:26

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

Batalov · 2015-06-23, 02:02

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. $\qed$

MattcAnderson · 2015-06-23, 10:09

Nice job Batalov. The proof looks valid to me.

Regards,
Matt

retina · 2015-06-23, 10:20

What about when n=0?

What about the more general k^n?

A much more interesting proof would be to show that no odd number can be a perfect number.

science_man_88 · 2015-06-23, 11:44

Quote:

Originally Posted by retina

What about when n=0?

What about the more general k^n?

A much more interesting proof would be to show that no odd number can be a perfect number.

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

davar55 · 2015-07-06, 19:45

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.

PawnProver44 · 2016-03-18, 16:35

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.

JeppeSN · 2016-03-19, 01:30

Quote:

Originally Posted by mathgrad

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

Also see Wikipedia: almost perfect number. /JeppeSN

2015-06-23, 01:26	#1
mathgrad Jun 2015 1 Posts	Prove 2^n cannot be a perfect number Given n is an integer, prove that 2^n cannot be a perfect number.

2015-06-23, 02:02	#2
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 19×491 Posts	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. $\qed$ Last fiddled with by Batalov on 2015-06-23 at 03:28 Reason: its != it's

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Is this a Perfect Number ?	Godzilla	Miscellaneous Math	8	2016-09-05 05:56
Down to a perfect number	fivemack	Aliquot Sequences	0	2014-12-23 09:47
Odd Perfect Number is 36k+9 ?	isaac	Miscellaneous Math	5	2014-07-22 22:18
How do I prove a 4000 digit number is prime??	VJS	Lounge	4	2005-05-09 20:56
How do you prove a number is prime?	Alien	Math	12	2004-01-07 11:36

2015-06-23, 10:09	#3
MattcAnderson "Matthew Anderson" Dec 2010 Oregon, USA 5·7·19 Posts	Nice job Batalov. The proof looks valid to me. Regards, Matt

2015-06-23, 10:20	#4
retina Undefined "The unspeakable one" Jun 2006 My evil lair 3×2,027 Posts	What about when n=0? What about the more general k^n? A much more interesting proof would be to show that no odd number can be a perfect number.

2015-07-06, 19:45	#6
davar55 May 2004 New York City 2³×23² Posts	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.

2016-03-18, 16:35	#7
PawnProver44 "NOT A TROLL" Mar 2016 California C5₁₆ Posts	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.