Associated Topics || Dr. Math Home || Search Dr. Math

### Odd Perfect Numbers

```
Date: 01/23/2001 at 11:12:24
From: chris billingham
Subject: Perfect numbers

My teacher has set me a question about the odd perfect numbers.

I have thought about it, and if there is a odd number, say 999, and
you want to see if it is a perfect number, you would need half of the
number:

999 / 2 = 499.5

499.5        all the other factors

All the other factors must equal 499.5 for it to work. But this can't
happen, as a factor has to be a whole number. So does this prove there
are no perfect numbers?
```

```
Date: 01/23/2001 at 15:45:35
From: Doctor Rob
Subject: Re: Perfect numbers

Thanks for writing to Ask Dr. Math, Chris.

Sorry, but this isn't a proof. It isn't even an example, because
2 is not a factor of 999, and neither is 499.5.

For n to be a factor of 999, it must be that 999 = n*m, where n and
m are both natural numbers. The factors of 999 are 999, 333, 111, 37,
27, 9, 3, and 1. The factors other than 999 add up to 521, which is
less than 999, so 999 is deficient, and not perfect.

If there is an odd perfect number N, it must have more than 300
decimal digits, and satisfy a large number of complicated
conditions.  Here are a few:

1. N must leave a remainder of 1 when divided by 12 or a
remainder of 9 when divided by 36.

2. It must have at least six different prime divisors.

3. It must have the form p^e*m^2, where p is a prime and both p
and e leave a remainder of 1 when divided by 4.

4. If m = q1^a1 * q2^a2 * ... * qn^an, each q a prime number,
then not all of the a's can equal 1.  If a2 = ... = an = 1,
then a1 > 2.  If a3 = ... = an = 1, then not both a1 and a2
can equal 2.

5. a1 = a2 = ... = an = 2 is impossible.

6. The set {1+a1, 1+a2, ..., 1+an} cannot have as a common
divisor 9, 15, 21, or 33.

7. If e = 5, then ai > 2 for all i's.

8. If N is not divisible by 3, it must have at least 9
different prime divisors.  If N is not divisible by 21, it
must have at least 11 such divisors.  If not divisible by 15,
it must have at least 14 different prime divisors, and if not
divisible by 105, it must have at least 27 such divisors.

9. If N has exactly r different prime divisors, the smallest of
them will be smaller than r+1.

Reference:
Beiler, Albert H., _Recreations in the Theory of Numbers - The
Queen of Mathematics Entertains_ (New York, NY: Dover Publications,
Inc., 1964), pp. 12-13.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search