Newsgroups: bit.listserv.christia
Date:         Thu, 20 Aug 1992 19:01:24 EDT
From:         JEK@NIHCU.BITNET
Subject:      Perfect numbers

     Explaining perfect numbers? My turn to try.

     In what follows, "number" always means a whole number (integer)
greater than zero.
     By M<n> (that should be M subscript n) we denote the number
(2^n)-1. (That's two raised to the power n, and then minus one.)
Thus, M<1> = 1, M<2> = 3; M<3> = 7, M<4> = 15, and so on.  These
numbers are called Mersenne numbers. Now M<rs> is divisible by M<r>
and by M
, so that if n is not prime, then M<n> will not be prime
either. The converse atatement, that if n is prime then M<n> will
always be prime, does not (alas) hold. For example, M<11> = 2047 =
23 * 89 .
     Given a number m, we say that n is a divisor of m if there is a
number k such that n*k = m. Clearly, m is always a divisor of
itself. We call the divisors of m that are less then m "proper
divisors." The proper divisors of 12, for example, are 1, 2, 3, 4,
and 6.
     A number is said to be "perfect" if the sum of its proper
divisors equals the original number.  It is not known whether there
are any odd perfect numbers. (I once had a proof that there are
none, discovered at 3 am. I found the flaw in the proof at 4 am.) It
is known that there are none that are smaller than 10^36.
     If p is a prime number, and 2^p - 1 is a prime number, then
2^(p-1) * (2^p - 1) is a perfect number. Moreover, all even perfect
numbers are of this form.

     In order to search for even perfect numbers, one therefore
proceeds as follows:
     [1] Generate the sequence of prime numbers, one at a time,
beginning 2, 3, 5, 7, 11, etc.
     [2] Take each prime p in turn, and calculate the Mersenne
number M<p> = 2^p - 1. Test M<p> to see whether it is a prime
number. Once you get past the first few numbers, M<p> will be very
large, so that testing it for primeness by dividing in turn by all
the prime numbers less than its square root is not practical, even
with a computer. Fortunately, there are short cuts, fancy tests like
the Lucas criterion (don't ask).
     [3] If M<p> is not prime, discard it and go to the next prime.
If M<p> is prime, then 2^(p-1) * M<p> is a perfect number.

This process will enable you to find all the even perfect numbers
between 5 and the point at which you get tired or run out of
computer time.

The first few primes p for which M<p> is prime are:

p = 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279,
2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701,
23209, 44497

As of 1982, M<44497> was the largest known prime number. I am fairly
sure that I have heard of progress since then, but I have no handy
reference. It is not known whether the number of Mersenne primes
(and hence the number of even perfect numbers) is finite or
infinite.

For a good, pleasantly written introduction for the non-specialist
(one term of calculus would help, but if you liked high-school
algebra or geometry you are okay), I recommend Daniel Shanks, SOLVED
AND UNSOLVED PROBLEMS IN NUMBER THEORY, 3rd edition, 1985, ISBN
0-8284-1287-9,297; $20 from the Chelsea Press, 15 E 26th Street, NY,
NY 10010, Telephone (212) 889-8095. (If you don't have $20, talk to
your librarian about a purchase or an inter-library loan.)

 Yours,
 James Kiefer