20150623, 01:26  #1 
Jun 2015
1_{16} Posts 
Prove 2^n cannot be a perfect number
Given n is an integer, prove that 2^n cannot be a perfect number.

20150623, 02:02  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10010010101011_{2} 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. Last fiddled with by Batalov on 20150623 at 03:28 Reason: its != it's 
20150623, 10:09  #3 
"Matthew Anderson"
Dec 2010
Oregon, USA
712_{10} Posts 
Nice job Batalov. The proof looks valid to me.
Regards, Matt 
20150623, 10:20  #4 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
17E1_{16} 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. 
20150623, 11:44  #5 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
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
Last fiddled with by science_man_88 on 20150623 at 11:47 
20150706, 19:45  #6 
May 2004
New York City
2^{3}·23^{2} Posts 
Since the sum of the smaller factors of 2^n equals 1 + 2 + 4 + ... + 2^(n1) = 2^n  1
it is never equal to 2^n, hence 2^n is never perfect. But I like Batalov's parity explanation better. 
20160318, 16:35  #7 
"NOT A TROLL"
Mar 2016
California
197 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. 
20160319, 01:30  #8  
"Jeppe"
Jan 2016
Denmark
2×83 Posts 
Quote:


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