The Search for an Odd Perfect Number Is Over

– there is none.

By Yale Schwartz

The simple definition for a perfect number is a number N whose proper divisors add up to the number itself. As an example we see that 6 is the smallest perfect number because the sum of its divisors (1, 2, 3) add up to 6.

6 = 1 + 2 + 3

Note that with this simple definition, the number 6 itself is not included as one of the divisors. This simple definition unfortunately leaves us with an ambiguous position when we consider the number 1. Since all numbers are divisible by 1, 1 is always listed as a divisor. However, according to this simple definition, the number N itself is not to be included. So it is unclear whether 1 should be included as a proper divisor of 1. Hence, the definition for a perfect number was modified as follows.

A perfect number is a number whose divisors add up to twice the number.

N is a perfect number
if…

2*N = ∑ all divisors of N

Thus, for N = 6 and all its divisors (1, 6) and (2, 3) here is the summation:

(1 + 6) + (2 + 3) = 12 = 2*6

Note that divisors usually come in pairs. After all, if N is divisible by divisor d, then N must also be divisible by N/d. For example, since N = 6 is divisible by d = 2, it follows that N = 6 will also be divisible by N/d = 3. The only exception to this is when N is a perfect square.

Since divisors are only listed once, the divisors for N = 9 are (1 + 9) and (3).

With this in mind, we’d like to write the general form for the perfect Odd Number with all its divisors as follows.

**2*N = (1 + N) + (d _{1}
+ N/d_{1}) + (d_{2} + N/d_{2}) + . . .
+ (d_{x} + N/d_{x})**

However, before we can do this we must deal with the exceptions.

If N is an odd prime number, can it ever be a perfect number?

If so, then. . .

2*N = ∑ all divisors of N

Where N is prime the only divisors are only (1, N), so we have

2*N = 1 + N

Solving for N, we see this can only be true where N=1.

But N=1 is not a perfect number because divisors can only be listed once.

So, no odd prime number is a perfect number.

If N is an odd perfect square, the general form would be written as follows:

**2*N = (1 + N) + (d _{1}
+ N/d_{1}) + (d_{2} + N/d_{2}) + . . .
+ (d_{x})**

**where d _{x}
is the sqare root of N.**

If N is an odd perfect square, can it ever be a perfect number?

No. The reason here is simple, consider the sums of the divisor pairs.

**2*N = (1 + N) + (d _{1}
+ N/d_{1}) + (d_{2} + N/d_{2}) + . . .
+ (d_{x})**

2*N yields an even number.

(1 + N) yields an even number because N is odd.

(d_{i} + N/d_{i}) yields an even number
because both divisors must be odd numbers.

(d_{x}) is an odd number because it is a divisor of
an odd number.

** 2*N
= (1+N) + (d _{1}+N/d_{1})
+ (d_{2}+N/d_{2}) +
... + (d_{x})**

** even = (even) + ( _{ }even_{ } ) + (_{ }even_{ } ) +
... + (odd)**

Hence we have a contraction.

So, if N is an odd perfect square, it cannot be a perfect number.

Having eliminated all prime numbers and all perfect squares, we can write the general form of an odd perfect number as follows.

**2*N = (1 + N) + (d _{1}
+ N/d_{1}) + (d_{2} + N/d_{2}) + . . .
+ (d_{x} + N/d_{x})**

Now let’s consider the value of N mod_{4}.

Since N is an odd number, N ≡ +1 mod_{4} or N ≡
-1 mod_{4}.

Let’s first consider the case where N ≡ -1 mod_{4}

When N ≡ -1 mod_{4}, consider the sums of the
divisor pairs.

**2*N = (1 + N) + (d _{1}
+ N/d_{1}) + (d_{2} + N/d_{2}) + . . .
+ (d_{x} + N/d_{x})**

2*N is equivalent to –2 mod_{4}.

(1 + N) is equivalent to 0 mod_{4}.

Since N ≡ -1 mod_{4} its divisors must be
paired as (+1, -1) or (-1, +1).

Therefore, the sum of all remaining pairs of divisors is
equivalent to 0 mod_{4}.

(d_{x} + N/d_{x}) is equivalent to 0 mod_{4}.

Expressing this relationship in equivalent terms of mod_{4}
we have:

** 2*N
= (1+N) + (d _{1}+N/d_{1}) + (d_{2}+N/d_{2})
+ ...
+ (d_{x}+N/d_{x})**

** ≡-2 = (≡ 0) + (+1 _{ } –1_{ }) + (-1_{ } +1_{ }) + ...
+ (≡ 0)**

** ≡-2 = (≡ 0)**

Again, we have a contraction.

So, if N ≡ -1 mod_{4} it cannot be a perfect
number.

Consider the case where N ≡ +1 mod_{4} and x
is an odd number.

**2*N = (1 + N) + (d _{1}
+ N/d_{1}) + (d_{2} + N/d_{2}) + . . .
+ (d_{x} + N/d_{x})**

2*N is equivalent to +2 mod_{4}.

(1 + N) is equivalent to +2 mod_{4}.

In each divisor pair d_{n} is equivalent to -1 or
+1.

However, since N ≡ +1, and (d_{n}*N/d_{n})
= N, then

if d_{n}
≡ -1 then N/d_{n} ≡
-1

if d_{n}
≡ +1 then N/d_{n} ≡
+1

Therefore, each divisor pair (d_{n} + N/d_{n})
is equivalent to either +2 or –2.

And each pair of divisor pairs (d_{n} + N/d_{n})
+ (d_{n+1} + N/d_{n+1}) is equivalent to 0 mod_{4.}

So for the case where x is an odd number, the sum of all
pairs of divisor pairs is equivalent to 0 mod_{4.}

(d_{1} + N/d_{1}) + (d_{2} + N/d_{2})
+ . . .
+ (d_{x-2} + N/d_{x-2}) + (d_{x-1} + N/d_{x-1})

Leaving only the last pair of divisors

(d_{x} + N/d_{x}) which is equivalent to +2
of –2 mod_{4.}

In either case, the formula is equivalent to this form.

** 2*N = (1+N) + 0 + 0 + 0 + . . . + (dx+N/dx)**

** ≡+2 = (≡+2) + 0 + 0 + 0 +
. . . + (≡+2 or ≡-2)**

** ≡+2 = (≡ 0)**

Again, we have a contraction.

So, if N ≡ +1 mod_{4} with an odd number of
pairs of proper divisors, it cannot be a perfect number.

Consider the case where N ≡ +1 mod_{4} and x
is an even number.

**2*N = (1 + N) + (d _{1}
+ N/d_{1}) + (d_{2} + N/d_{2}) + . . .
+ (d_{x} + N/d_{x})**

This is the last case we need to consider.

I use induction to prove this last case.

Assumption:

For N equals an odd number with an even number of pairs of proper divisors, there is no value for x such that the following equation is true.

**2*N = (1 + N) + (d _{1}
+ N/d_{1}) + (d_{2} + N/d_{2}) + . . .
+ (d_{x} + N/d_{x})**

We begin with the base case where N = 45 having an even pair of proper divisors (3, 15) and (5, 9). The summation of its divisors is not equal to 2*N.

**2*45 < ? >
(1 + 45) + (3 + 15) + (5 + 9)**

**90 <> 78**

Now consider the sum of the divisors for N+4. In the resulting equation, the value of N+4
is the next odd number ≡ +1 mod_{4}. However, it may be a prime number, a perfect
square, or neither. The divisors of N+4
are shown as **e _{y}** because they bear no relationship to the
divisors of N. And the number of pairs
of divisors is shown as y because it has no relationship to x.

**2*(**N+4**) =
(1 + (**N+4**)) + (e _{1} + (**N+4

We can conclude the following.

If N+4 is a prime, it cannot be a perfect number (Case #1.).

If N+4 is a perfect square, it cannot be a perfect number (Case #2).

If N+4 ≡ -1, it cannot be a perfect number (Case #4).

If N+4 ≡ +1 with an odd number of pairs of proper divisors, it cannot be a perfect number (Case #5).

Thus the only case we need to consider is where N+4 ≡ +1 with an even number of pairs of proper divisors, for which we have the form.

**2*(**N+4**) =
(1 + (**N+4**)) + (e _{1} + (**N+4

But this form matches the assumption, namely that N+4 is an odd number with an even number of pairs of proper divisors. Hence, it too is never true.

Q.E.D.

There is no odd perfect number.

~Yale Schwartz

Version 20051002.18:11

Copyright © Yale Schwartz, 2005

=== Review comments by Lucan

From: <me@lucasoman.com>

To: "Yale Schwartz" <yalesafe.com>

Sent: Thursday, September 29, 2005 9:21 AM

Subject: Re: odd perfect number (all unicorns are blue and all odd perfectnumbers are round)

Mr.
Schwartz:

I enjoyed reading your proof. Your approach was an interesting one. I think

you've definitely narrowed the solution space for this problem, and I'll have

to research to see if anyone else has narrowed it that far. However, I don't

believe you've proven that there is no odd perfect number. In particular, I

don't believe that Case 6 holds up.

First, there is no "base case" for your inductive proof. Although it
is

trivial, you need to find an odd number that is +1=mod4 with an even number of
pairs of

proper divisors and prove that it is not perfect, even if only as a formality.

Second, and more importantly, you have not proven the very key to this proof:

that if N is not perfect, then N+4, also, cannot be perfect. You've simply

shown that Cases 1-5 apply to N+4. The assumption that N is not perfect can be

made from the start, if you have a base case. However, you cannot keep making

that assumption for each N+4x (where x is an integer).

Again, I've enjoyed your proof, and you may yet prove Case 6. I think it's rude

that you've received no replies; professors can be so snobbish. I'm guessing,

though, that they did not reply because they saw that there was a fatal error

(Case 6) in the proof.

I hope you don't take my criticism personally. I just figured you'd actually
want some honest feedback on your proof (which it seems you have not gotten to
date).

Best of luck to you in your (ambitious) mathematical endeavors.

Lucas Oman

=== My Response to Lucan

Sent: Sunday, October 02, 2005 7:37 PM

Hi, Lucan.

Thanks for your review comments.

Regarding
your first point, I've corrected my proof by adding the base case where N = 45
(+1mod4) and x = 2 (the lowest even number of pairs of proper divisors). as
follows.

We begin with the base case where N = 45 having an even pair of proper divisors (3, 15) and (5, 9). The summation of its divisors is not equal to 2*N.

**2*45 <?> (1 + 45) + (3 + 15) +
(5 + 9)**

**90
< > 78**

Regarding
your second point (and I admit this is the part of the proof that's
questionable)...

I'm using
induction on the value of x, the number of pairs of divisors, with any value of
N+4 wher N = +1mod4. The only failure of this method that I can see is
that I may have to demonstrate the base case for "every" value of N =
+1mod4 where x=1 pair of proper divisors is not a perfect number. Because
for all other values of x, the resulting summation must take on one of the
previous forms which have been proven to be not perfect.

Alas,
that's as far as I'm taking it for now.

I'm pleased
that my method discloses many findings that I've seen which use more elaborate
methods. For example, if N is a multiple of 3 it cannot be perfect.

And here's an insight that may help someone else to finish the work.

When N is an even number, e.g. 6, it is of course divisible by 2. As such we have 1/2 of the summation completed. Now all we need for the number to be perfect is to have the remaining bits to add up to the remaining 1/2. In this case 1/3 + 1/6 manage to do this.

When I
think of this as an area problem, it seems more reasonable that the remaining
bits for an even number have a better chance of filling the remaining space
because the “eveness” of the values are more compatible with other divisions.

Whereas,
when N is an odd number, the area puzzle “feels” like there can never be a
combination of odd numbered fractional parts that will fill the remaining space
exactly. I know this line of thinking
has no mathematical merit, bit it’s what led me to think of the problem in mod
4 terms, i.e. extra eveness because it’s divisible by 2 twice.

Thanks
again for your review comments.