Are All Perfect Numbers Even?

Date: 01/16/97 at 10:48:35
From: Anonymous
Subject: Perfect Numbers

Are all perfect numbers even?  Has it been proved that perfect numbers 
MUST be even?  I believe they must be, which I'll discuss below, but I 
saw your "Ask Dr. Math" service, and this question immediately came to 
mind.

About ten years ago, I discovered on my own a formula for generating 
perfect numbers, which are natural numbers whose proper factors add up 
to the number itself.  (Proper factors of a number are factors less 
than the number itself.)  If I recall properly, the formula is: 
P = [2^(n-1)] x [(2^n) - 1], where n is a prime number.  

I was a bit disappointed to read within a few months that Gauss or 
Euler had already discovered the same formula several hundred years 
earlier, but I was still pleased with my own discovery.  It seems to 
me that a structural examination of why the proper factors of perfect 
numbers add up to the number itself reveals that it has to do with the 
perfect number apparently needing to have as one of its factors a 
number that is exactly half of itself, which means that the perfect 
number must of necessity be an even number.  

Anything you can provide on this would be great.  I teach junior high 
math in St. Paul, MN.  

Thanks,
Bob Hazen

Date: 01/27/97 at 10:40:54
From: Doctor Lorenzo
Subject: Re: Perfect Numbers

No one knows whether all perfect numbers are even, which I will talk 
more about below.  In your formula P = 2^(n-1) x (2^n - 1), you 
actually want (2^n)-1 to be prime, not n to be prime.  A prime number 
of the form M_n = 2^n -1 is called a Mersenne Prime.  M_2, M_3, M_5 
and M_7 are prime, which led Mersenne to conjecture that all such 
numbers were prime, but that's not the case.  For example, 
M_11 = 2047 = (23)(89).  As of the late 60's (when my reference was 
written), there were only 23 known Mersenne primes, the largest being 
M_11213.  (I think they've found a couple more since then, but the 
list is still short). Let me quote a paragraph from Oystein Ore's 
_Invitation to Number Theory_:

"This result shows that each Mersenne prime gives rise to a perfect
number.  In Section 2.2 we mentioned that so far 23 Mersenne primes 
are known, so we also know 23 perfect numbers.  Are there any other 
types of perfect numbers?  All the perfect numbers of the form 3.4.1 
[your formula] are even, and it is possible to prove that if a perfect 
number is even it is of the form (3.4.1).  This leaves us with the 
question: ARE THERE ANY ODD PERFECT NUMBERS?  Presently we know of 
none and it is one of the outstanding puzzles of number theory to 
determine whether an odd perfect number can exist.  It would be quite 
an achievement to come up with one and you may be tempted to try out 
various odd numbers.  We should advise against it; according to a 
recent announcement by Bryant Tuckerman at IBM (1968), an odd perfect 
number must have at least 36 digits."

A result that's NOT hard to prove is that an odd perfect number must 
be of the form (odd prime) x (perfect square), but that doesn't narrow 
things down much.

-Doctor Lorenzo,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 01/30/97 at 05:37:39
From: Anonymous
Subject: Re: Perfect Numbers

Thank you for your reply about perfect numbers.

Here are some of my ideas about a proof regarding the existence of odd 
perfect numbers. A look at the known perfect numbers suggests to me a 
global approach to why perfect numbers must be even. 

Consider the proper factors of 28 and 496:

28: {1,2,4,7, 14}  		496: {1,2,4,8,16,31,62,124,248}  
			
The largest proper factor of 496 is 248, which gets us sum-wise 
halfway to our desired sum of 496.  

Observation 1: Any perfect number P that is not even has a smallest 
factor of n not equal to 2, so n = 3 at least.  This means that the 
largest factor of P is the number P/n, which (if n = 3) is only 1/3 of 
P, which sumwise doesn't "get far" toward the desired sum of P.  At 
least with even P, the largest proper factor gets us halfway to P.  
Another way of saying this is to think of the sequence 
P = P/2 + P/4 + P/8 + ... + P/(2^m), plus exactly another of the last 
term "P/(2^m)" to get us that last missing part of the desired sum.   
 
496's next proper factor of 124, when added to 248, gets us sumwise 
3/4 of the way, leaving 1/4 to add.  Then 62 gets us 7/8 of the 
way, and the 31 (which is 1/16 of 496) gets us 15/16 of the way.  Thus 
we are "missing" 31 = 1/16 of the desired sum of 496.  But quite 
nicely, we can sum all the consecutive powers of 2 that are less than 
31 to actually get a sum of the missing 31: 1 + 2 + 4 + 8 + 16 = 31.

A similar process happens with the perfect number 28.  The factors of 
14 + 7 = 21, so we are "missing" a 7, for which we can use all the 
consecutive powers of 2 that are less than 7, to obtain the sum of 7: 
1 + 2 + 4 = 7.

Observation 2: These desired powers of 2, which always add up to 1 
less than the next power of 2 (1 + 2 + 4 = 7, which is 
8 - 1 = (2^3) - 1, and 1 + 2 + 4 + 8 + 16 = 31, which is 
32-1 = (2^5) - 1, will be unavailable to sum up that last missing 
chunk of the partial sum equal to P UNLESS that P is an even number in 
the first place, or unless P has lots of factors that are consecutive 
powers of 2, as made available by the Mersenne prime formula for 
perfect numbers.

The part I'll keep brief is that the first factor higher than the 
highest power of 2 must be a prime number, in order to avoid the 
proper factor sum from being so large that P becomes an abundant 
number (see definition below).
 
This is not a rigorous appraoch, and I'm sure it's somewhat sketchy, 
but at a global level, it convinces me that perfect numbers must be 
even, and moreover, must be of that Mersenne form, in order to avoid 
being either deficient (having a proper factor sum less than P) or 
abundant (having a proper factor sum greater than P).  

Date: 01/30/97 at 08:50:15
From: Doctor Steven
Subject: Re: Perfect Numbers

Your global approach is nice but it doesn't really contain enough.  
It seems you're trying to show that odd numbers "can't" (read not as 
can't but as improbable) be perfect numbers, because the factors will 
sum up to less than the number.  Say 3, 5, 7, 9, 11, 13, and 15 are 
all factors of our number, Q.  So:

      3R = Q
      5T = Q
      7S = Q
      9U = Q
     11V = Q
     13W = Q
     15Z = Q.

So R, S, T, U, V, W, and Z are also factors. Solve these equations for 
the unknown factors:

     R = Q/3
     T = Q/5
     S = Q/7
     U = Q/9
     V = Q/11
     W = Q/13
     Z = Q/15

This means that:

R + T + S + U + V + W + Z 
   = (Q/3) + (Q/5) + (Q/7) + (Q/9) + (Q/11) + (Q/13) + (Q/15)
   = (15015Q/45045) + (9009Q/45045) + (6435Q/45045) + (5005Q/45045) +
     (4095Q/45045) + (3465Q/45045) +(3003Q/45045)       

Add these fractions up to get:

     R + T + S + U + V + W + Z = 46027Q/45045 > Q

So we should have no problem finding an odd number Q whose factors 
will add up to at least Q.  Likewise we should have no problem finding 
odd numbers whose factors won't add up to Q.  

-Doctor Steven,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 01/30/97 at 21:50:24
From: Anonymous
Subject: Re: Perfect Numbers

Thanks for your reply.  Your remark that "trying to show that odd 
numbers "can't" be perfect numbers, because the factors will sum up to 
less than the number" isn't quite accurate.  I am aware that some odd 
numbers are deficient while others are abundant, so I'm not trying to 
show that the proper factors of odd numbers will sum up to less than 
the number.  

I'm also aware that my global approach indeed doesn't do it at all 
because it definitely is not a proof as such.  Perhaps what my remarks 
and insights were getting at is that a detailed analysis of the nature 
of the factors of these known perfect numbers - about "half" of which 
are powers of two, and the other "half" begin with a prime number that 
is 1 less than the next higher power of 2, along with doubles (and 
quadruples and octuples...) of that prime number.  This analysis 
helped me see why the sum of the various factors of the known perfect 
numbers actually sum to be precisely, no more and no less, the perfect 
number itself!  It can't be a strict power of 2, because the sum of
any consecutive powers of 2 is always going to be 1 less than the next 
power of 2.  However, the powers of 2 get us close to summing to the 
original number.

By now, I may be rambling.  I checked a few high odd numbers 
(3 x 5 x 7 x 11 x 13) and the sum of the proper factors gets close to 
the number.  I think my insight into the marvelous structure of 
perfect numbers, and how their proper factors so neatly sum to the 
number, became overgeneralized in my mind.  I'll fiddle sometime with 
some odd numbers and examine their structure.  Fear not - I recall 
your comment about how long an odd perfect number must be!

In the meantime, I have papers to correct, lesson plans to write, and 
kids to feed dinner to.  

Kind regards