From: Dave Rusin <rusin@math.niu.edu>
Date: Wed, 16 Dec 1998 14:35:17 -0600 (CST)
To: Luxemburger@snafu.de
Subject: Re:  how to find prefect numbers ?

You wrote:

>Please can anyone help me to find a good method to find "vollkommene
>Zahlen"(or Perfect Numbers)
>
>an example is 6 -> is divided by 1 +2+3 
>the sum of these divisors is 6 again !

No odd perfect numbers are known. It has not been proven that none exist,
but results have been proven which would require such a number to have
many prime factors to high exponents, so it is generally assumed that none 
exist. (I believe that the question of whether odd perfect numbers exist
is the oldest open question in mathematics, now about 2 and a half millenia 
old!)

It is known that an even number is perfect iff it is of the form
	2^(p-1) * ((2^p) - 1)
where the second factor is prime. Primes of the form  2^p-1  are known as
Mersenne primes. It is known that this expression can only be prime if  p  
itself is prime, and there are fairly efficient ways to determine whether a
number  2^p - 1  is prime, so very large primen numbers, and thus very
large perfect numbers, have been found in this way. Indeed, what is usually
true and is true today is that the largest number known to be prime is a
Mersenne prime. About 3 dozen are known, with up to a million digits
(giving a two-million digit perfect number).

For more information, consult any elementary number theory book. You
might be interested in Ribenboim's Book of Prime Number Records, or Guy's
Unsolved Problems in Number Theory, for example.

More information about testing Mersenne numbers for primality is at
	index/11Y05.html
and its parent page.

[deletia -- djr]
==============================================================================

Subject: Perfect numbers (from Mathematical Reviews on the Web)
From: rusin
Date: Dec 18 1998 08:52 (datestamp)
To: get current bounds on odd-perfect-number searches :-)

   Selected Matches for: Anywhere=(odd perfect number)

   98k:11002 11A25 (11Y70)
   Hagis, Peter, Jr.(1-TMPL); Cohen, Graeme L.(5-UTSY)
   Every odd perfect number has a prime factor which exceeds $10\sp 6$.
   (English. English summary)
   Math. Comp. 67 (1998), no. 223, 1323--1330. [ORIGINAL ARTICLE] 
   
   The authors prove that if $N$ is an odd perfect number then $N$ has a
   prime factor which exceeds $10\sp 6$. This result is an improvement of
   all known results in this subject, such as, among others, the result
   of M. S. Brandstein announced in [Abstracts Amer. Math. Soc. 3 (1982),
   no. 1, 257, Abstract 82T-10-240] that at least one of the primes such
   that $p\sb i\mid N$ is greater than $5·10\sp 5$. In the proof of this
   result the authors use some arithmetical properties of the cyclotomic
   polynomial and several interesting lemmas and propositions whose
   proofs are given by theoretical considerations and computer
   calculations.
   
                      Reviewed by Aleksander Grytczuk
     _________________________________________________________________
   
   96f:11009 11A25 (11Y70)
   Cook, Roger(4-SHEF-SM)
   Factors of odd perfect numbers. (English. English summary) Number
   theory (Halifax, NS, 1994), 123--131,
   CMS Conf. Proc., 15,
   Amer. Math. Soc., Providence, RI, 1995.
   
   If $N$ is odd, let $p\sb 1<p\sb 2<\cdots <p\sb r$ denote the prime
   factors of $N$. Let $S(N)=\sum (1/p\sb i)$. Chein and Hagis have shown
   that if $N$ is perfect, then $r\geq 8$. It is known that if $r=8$,
   then $p\sb 1=3$ and $p\sb 2=5$ or 7. In each of these cases, the
   author finds an upper bound for $S(N)$, thus extending prior results
   by other researchers.
   
                        Reviewed by Neville Robbins
     _________________________________________________________________

   96b:11130 11N64 (11A25)
   Heath-Brown, D. R.(4-OXM)
   Odd perfect numbers. 
   Math. Proc. Cambridge Philos. Soc. 115 (1994), no. 2, 191--196.
   
   It is not known whether or not odd perfect numbers can exist. C.
   Pomerance [Math. Ann. 226 (1977), no. 3, 195--206; MR 55 \#12616]
   showed that if $N$ is an odd perfect number with at most $k$ prime
   factors, then $N \le (4k)\sp {(4k)\sp {2\sp {k\sp 2}}}$. The author
   sharpens the above estimate. In fact, he derives an estimate for odd
   "multiply perfect" numbers which in the ordinary case reduces to
   $N<4\sp {4\sp k}$. The estimate shows that odd perfect numbers cannot
   have a smaller than average number of prime factors.
   
                          Reviewed by P. Haukkanen
     _________________________________________________________________

   94m:11003 11-02
   Schinzel, Andrzej(PL-PAN)
   Progress in number theory during the years 1989--1992. 
   Discuss. Math. 13 (1993), 75--80.
   
   The author gives a brief survey of progress made in the past few years
   on twelve selected problems in number theory. The problems include
   such well-known problems as Fermat's conjecture, the Riemann
   hypothesis, and the question whether there exist odd perfect numbers,
   as well as some lesser known problems. Among the more spectacular
   advances that are described here are the proof by Alford, Granville,
   and Pomerance that there exist infinitely many Carmichael numbers, and
   Wooley's work on Waring's problem. In connection with the Fermat
   conjecture, the author only mentions some recent computational work;
   Wiles' breakthrough on this problem apparently came too late to be
   included in this survey.
   
                        Reviewed by Adolf Hildebrand
     _________________________________________________________________
   
   94c:11003 11A25
   Starni, Paolo
   Odd perfect numbers: a divisor related to the Euler's factor.
   (English. English summary)
   J. Number Theory 44 (1993), no. 1, 58--59.
   
   Let $\sigma(n)$ be the sum of the positive divisors of the natural
   number $n$; $n$ is said to be perfect if $\sigma(n)=2n$. In this
   paper, the author proves the following result: If $n=\pi\sp \alpha
   M\sp 2$ is an odd perfect number ($\pi$ is prime, $(\pi,M)=1$ and
   $\pi\equiv\alpha\equiv 1\bmod 4$), $\alpha+2$ is prime and
   $(\alpha+2,\pi-1)=1$ then $M\sp 2\equiv 0\bmod {\alpha+2}$.
   
                         Reviewed by Armel Mercier
     _________________________________________________________________
   
   92c:11004 11A25 (11Y70)
   Brent, R. P.(5-ANUIP); Cohen, G. L.(5-NSWIT); te Riele, H. J.
   J.(NL-MATH)
   Improved techniques for lower bounds for odd perfect numbers. 
   Math. Comp. 57 (1991), no. 196, 857--868.
   
   A positive integer $N$ is called perfect if $\sigma(N)=2N$, where
   $\sigma$ is the sum of divisors function. It is not known whether any
   odd perfect numbers exist, though it has long been known that if there
   is any number of this type, then it must have many prime factors.
   Brent and Cohen [same journal 53 (1989), no. 187, 431--437, S7--S24;
   MR 89m:11008] have developed an algorithm to check whether an odd
   number $N$ is perfect using explicit factorization of $\sigma(q\sp
   k)$, where $q\sp k\mid\mid N$. It is elementary that such a factor
   satisfies $q\sp {2k}<N$; here the authors show---subject to a
   numerical check---how to establish such an inequality with a higher
   exponent than $2k$. This estimate leads in turn to a quicker test for
   perfection, and with it the authors show that there is no odd perfect
   number less than $10\sp {300}$.
   
                         Reviewed by H. G. Diamond
     _________________________________________________________________
   
   89m:11008 11A25 (11Y05 11Y70)
   Brent, Richard P.(5-ANU-PS); Cohen, Graeme L.(5-NSWIT)
   A new lower bound for odd perfect numbers. 
   Math. Comp. 53 (1989), no. 187, 431--437, S7--S24.

   Some recent "proofs" of lower bounds for odd perfect numbers have been
   questioned [see, e.g., B. Stubblefield, in Black mathematicians and
   their works, 211--222, Dorrance, Ardmore, PA, 1980; MR 81m:10011].
   Wary of the fate of these "proofs", the authors give a
   straightforward, algorithmic proof that there is no odd perfect number
   less than $10\sp {160}$. The 3133-line proof was largely
   computer-generated and appears in a supplement to the paper. The proof
   uses the customary factor-chain method for problems involving odd
   perfect numbers. The elliptic curve integer factoring method was used
   to furnish a large number of factorizations of numbers of the form
   $\sigma(p\sp a)$.
   
                    Reviewed by Samuel S. Wagstaff, Jr.
     _________________________________________________________________

   87m:11005 11A25
   Cohen, Graeme L.(5-NSWIT)
   On the largest component of an odd perfect number. 
   J. Austral. Math. Soc. Ser. A 42 (1987), no. 2, 280--286.
   
   The author proves that every odd perfect number $n$ has a prime factor
   $p\sp \alpha\Vert n$ for which $p\sp \alpha>10\sp {30}$. The proof
   requires a great deal of computation, which is not carried out in the
   paper but can be obtained from the author. He also obtains other
   inequalities for odd perfect numbers.
   
                            Reviewed by P. Erdos
     _________________________________________________________________
   
   86f:11010 11A25
   Wagon, Stan(1-SMTH)
   The evidence: perfect numbers. 
   Math. Intelligencer 7 (1985), no. 2, 66--68.
   
   The author briefly surveys the history of the problem of the
   nonexistence of odd perfect numbers.
     _________________________________________________________________
   
   86f:11009 11A25
   Cohen, G. L.(5-NSWIT); Williams, R. J.(5-NSWIT)
   Extensions of some results concerning odd perfect numbers. 
   Fibonacci Quart. 23 (1985), no. 1, 70--76.
   
   As is well known, an odd perfect number $N$, if one exists, must be of
   the form $N=p\sp \alpha q\sp {2\beta\sb 1}\sb 1 q\sp {2\beta\sb 2}\sb
   2\cdots q\sp {2\beta\sb \tau}\sb t$ for distinct odd primes $p,q\sb
   1,q\sb 2,\cdots,q\sb t$, with $p\equiv\alpha\equiv1\bmod4$. The
   authors add the following to previously known results: If $\beta\sb
   1=\beta\sb 2=\cdots=\beta\sb t=\beta$, then $\beta\ne6,8,11,14$ or 18;
   and if $\beta\sb 2=\cdots=\beta\sb t=1$, then $\beta\sb 1\ne5$ or 6.
   They also show that if $x$ is the number of prime powers $q\sp
   {2\beta\sb i}\sb i$ for which both $q\sb i\equiv1\bmod 4$ and
   $\beta\sb i\equiv1\bmod2$, then $p-\alpha\equiv4x\bmod8$.
   
                          Reviewed by V. C. Harris
   
   © Copyright American Mathematical Society 1998