### Mersenne Primes and Perfect Numbers (09 Sep 2002)

- A good argument against Dave Winer's position on software copyright
- KCachegrind, a KDE frontend for the superb Valgrind
- Suggested answering machine greetings
- HP fires Bruce Perens for being nasty to Microsoft
- Indroduction to Geometric Algebra

Mersenne (named after a french monk) primes are of the form 2^{n}-1 where n is an integer, greater then 0. There is a distributed.net like effort to find them called GIMPS (search Google).

A perfect number is a number where the sum of its divisors (excluding itself, but including 1) equals that number. For example 6 is perfect because (1 + 2 + 3) = 6. Thus the sum of all the divisors is twice the number.

Now, I read a while back in a book that (2^{n}-1)(2^{n-1}) was proven to be a perfect number, but the book didn't have the proof. Thankfully, I ran across the proof today. That proof leaves out a number of steps thou, so here's a better one:

*p*= 2^{n}- 1, and is prime*m*= (2^{n-1})p- sigma(
*x*) is the sum of all the positive divisors of*x* - sigma(
*a***b*) = sigma(*a*)*sigma(*b*) where gcd (*a*,*b*) = 1 (think about it) - sigma(
*m*) = sigma (2^{n-1}*p*) - = sigma(2
^{n-1}) * sigma (*p*) - Now,
*p*was defined to be prime, thus its only divisors are 1 and itself, thus the sum of those divisors must be*p*+ 1 - Also, by thinking of the prime factorisation of 2
^{a}, the divisors of 2^{a}must include all lesser powers of 2 (where the power is greater then, or equal to, 0). By still considering the prime factorisation there can be no other divisors. Thus sigma (2^{n-1}) must be 2^{n}-1. It might help to think of the numbers in binary form to see this. - thus sigma(
*m*) = (2^{n}-1)(*p*+1) - expanding this gives 2
^{n}(2^{n}-1) which is equal to 2*m*. - Thus the sum of all the divisors of
*m*is 2*m*. Thus*m*is perfect.

One from The Book