
#1
Nov303, 11:03 PM

P: 6

I have an assignment free night and thought I would make a post that hopefully hasn't been made before.
A discussion of perfect numbers: A positive integer n is said to be perfect if n is equal to the sum of all its positive divisors, not including n itself Alternatively, if σ(n) is the sum of all divisors of n, then a perfect number is defined as σ(n) = 2n As an example, the number 6 is perfect, since: σ(6) = 1+2+3+6 = 12 = 2(6) Indeed, such numbers were known in the time of Euclid and before. The perfect numbers known were: P1 = 6 P2 = 28 P3 = 496 P4 = 8128 The search for perfect numbers eventually gave rise to the following theorem, which is attributed to Euclid. If 2^(k)  1 is prime, k>1, then n = 2^(k1)(2^(k)1) is a perfect number (Euler later asserted that all perfect numbers are of this form) Traditionally, numbers defined as Mn = 2^(n)  1 are called Mersenne numbers (or primes) after Marin Mersenne, who showed that the determination of such numbers, and therefore the quest to discover more perfect numbers, can be narrowed to the case where n is prime. There are infinitely many primes, but are there infinitely many Mersenne primes? From here, many mathematicians searched for larger perfect numbers, and even larger prime numbers. Finding perfect numbers is no doubt one of the oldest problems in number theory. There is no known odd perfect number, probably due to the enormous size of perfect numbers (for example, the perfect number P9 is 22 digits long and P37 is almost 2 million digits long). Clearly we cannot find an odd perfect number in an exhaustive manner. However, the part that I find interesting is that it has not been proven that an odd perfect number does not or can not exist; although, it has been proven that any even perfect number ends in the digit 6 or the digit 8. But why is it so difficult to show that an odd perfect prime number cannot exist, or why can it not be proven that an odd perfect number may exist? It has not even been shown that there exists infinitely many prime numbers p for which Mp is prime (which in turn *could* show there exists infinitely many perfect numbers). Intuitively one could infer that there must be infinitely many perfect numbers, but it has not been rigorously shown. One of those unsolved problems... but wherein lies the difficulty? Of course this kind of problem is one of curiousity, not usefulness. pardon my notsoclear notation. I am not sure how to apply super and subscripts. perhaps if no one can contribute to this discussion, they can at least help me with this [:)] 



#2
Nov403, 06:20 AM

P: 3,175

an odd perfect number might not exist but a lot of work has been done to determine it's propeties. (which is weird because if it's not exist then the propeties are meaningless.)
here's where you can read a little about the work that has been done on the subject: http://mathworld.wolfram.com/OddPerfectNumber.html btw, can someone show me the proof of Stuyvaert that an odd perfect number must be a sum of squares? 



#3
Nov503, 11:32 AM

P: 3,175

σ(n)=2n+1 and σ(n)=a+b+...+n=2n+1 and according to Stuyvaert (which i couldnt find anything about him) n=c^2+d^2+... now if some of the squares are his divisors then a and b (the divisors) will be put like this n=a^2+b^2+c^2+d^2 where c and d arent particularly his divisors. now something that i think will be remarkable is if such an odd perfect number does exists and it is the sum of squares and those squares are his divisors not all of them ofcourse (because then the sum would be bigger then the number). 



#4
Nov503, 04:32 PM

P: 6

Perfect Numbers
thx for the link! it's interesting that so much can be known about something that may or may not exist [6)]
so far I have been unable to find anything on Stuyvaert proof, or any other documentation, other than what you stated . Maybe I'll make a trip to the library if I'm feeling motivated. 



#5
Nov503, 05:45 PM

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PF Gold
P: 16,101

If σ(n) = 2n + 1, that would mean the sum of all the positive divisors of n (not including n) would be n + 1. 



#6
Nov703, 12:23 PM

P: 3,175

and prime numbers are semi perfect numbers the sum of there divisors is always n+1.
for example 2: 2+1=3 wich is n+1. 



#7
Nov1303, 05:09 AM

P: 3,175

has there been a search of the number of the divisors of the perfect numbers besides 1 and the perfect number itself?
because those of 6 are 2 and 3, 2 numbers those of 28 are 2,14,4 and 7, four numbers. is the number of the divisors of an even perfect number (not include 1 and itself) also even? it's easy to proove it check the divisors of the next even perfect number (im too lazy to do so ). [:))] 



#8
Nov1303, 06:13 AM

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A number n has an odd number of divisors iff it is a perfect square.
I don't know whether perfect numbers can be perfect squares or not, though. My gut says no, but no proof leaps to mind immediately. 



#9
Nov1303, 07:09 AM

P: 3,175

anyway it will still be odd because substracting an even number from an odd number equals odd number. 6 isnt a perfect square number so your guts are true, i dont know about other perfect numbers. btw if your gutsare true it still doesnt mean the number of divisors arent odd. 



#10
Nov1303, 11:41 PM

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If my guts are true that no perfect number is a perfect square, then the theorem I quoted says that no perfect number has an odd number of divisors... Anyways, I have proven the following statement: For any perfect number n: n is odd if and only if n is a perfect square. (I'm tired, it's too late to write the proof, I'll post it tomorrow if you want me to) 



#11
Nov1403, 10:18 AM

P: 3,175

so you prooved that an odd perfect number if existed has an odd number of divisors?
now let's say there arent odd perfect numbers how will you proove that an even perfect number has an even number of divisors? 



#12
Nov1403, 05:11 PM

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Whoops, I made a mistake, I have proof that all perfect numbers cannot be perfect squares.
For a given perfect number n, the key is to count how many odd divisors n has. Let the prime factorization of n be n = 2^e_{0} p_{1}^e_{1} p_{2}^e_{2} ... p_{k}^e_{k} The number of odd divisors of n is (e_{1} + 1) (e_{2} + 1) ... (e_{k} + 1) (this includes n itself) If n is a perfect square, then each of the exponents in its prime factorization must be even. Thus, the number of odd divisors of n must be odd. If n is even, then this implies the sum of all of the proper divisors of n must be an odd number, which contradicts the hypothesis that n is perfect. If n is odd, then this implies n has an even number of odd proper divisors, thus the sum of all of the proper divisors of n is an even number, which contradicts the hypothesis that n is perfect. Thus, if n is a perfect number, n cannot be a perfect square. 



#13
Nov1503, 04:36 AM

P: 3,175





#14
Nov1503, 11:00 AM

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P: 16,101

Because a number, n, has an odd number of divisors if and only if n is a perfect square, this proves that no perfect number can have an odd number of divisors. 



#15
Dec2603, 11:10 AM

P: 10

.
Perhaps I misunderstand, but isn't it enough to show that an odd perfect number cannot be square, to note that in most numbers the proper factors can be paired as a, total/a; b, total/b; c, total/c etc, and that in a square there is one pair [sqroot, total/sqroot] where the two are not distinct factors but the same number? Thus giving an odd number of proper factors, but an even total sum of factors counting 1, so ruling out a sum to an odd total? Did I get mixed up somewhere? . 



#16
Dec2603, 11:19 AM

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The problem with learning high powered math is that you tend to forget the easy ways to prove things. [:)]




#17
Dec2603, 01:00 PM

P: 10

.
Ah, thanks, Hurkyl!  so I didn't get that wrong then? . 



#18
Dec2603, 01:13 PM

P: 10

.
Loop quantum & kebz33, the condition that sounds most interesting to me is that one about any odd perfect having to be, should there be any, a product of a square number and a prime number raised to an odd power. If I remembered that right. . 


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