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### Are All Perfect Numbers Even?

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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
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
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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/
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Date: 01/30/97 at 05:37:39
From: Anonymous
Subject: Re: 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).
```

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

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

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

Kind regards
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Associated Topics:
College Number Theory
High School Number Theory

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