mersenneforum.org Prove 2^n cannot be a perfect number
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 2015-06-23, 01:26 #1 mathgrad   Jun 2015 18 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 3×3,041 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
 2015-06-23, 10:09 #3 MattcAnderson     "Matthew Anderson" Dec 2010 Oregon, USA 10010110002 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 169C16 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.
2015-06-23, 11:44   #5
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts

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

Last fiddled with by science_man_88 on 2015-06-23 at 11:47

 2015-07-06, 19:45 #6 davar55     May 2004 New York City 101468 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.
2016-03-19, 01:30   #8
JeppeSN

"Jeppe"
Jan 2016
Denmark

25·5 Posts

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

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