Talk:Perfect number

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Contents 1 References 2 Odd perfect numbers 3 About Ore's proof 3.1 new argument 4 About Stuyvaert's proof 5 Is 1 a perfect Number? 6 Number of Distinct Prime Factors 7 Pomerance Heuristic 8 Undecidability 9 Ohno's result trivia vs. trivial 10 Rare 11 Special term? 12 The first two lines in disagreement. 13 Hare's result; WAREL's edits 14 Dead link 15 "not distinct prime factors" vs not 'necessarily distinct prime factors for odd perf #s 16 McDaniel claim 17 Perisastri and Grun 18 Nielsen's result 19 I'm confused 20 June 28 21 Odd perfect numbers 22 Yamada's statement 23 Curtiss result 24 What do you mean unsupported? 25 Application? 26 These are all proven 27 Minor results ? 28 OPN proofs 29 Interesting Product 30 Why Even Perfect Numbers Work (and Odd Perfect Numbers may not!) 31 Nitpick!

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Only a finite number (39) of Mersenne primes (hence perfect numbers) are presently known

Is it true that every even perfect number comes from a Mersenne prime? If so, then this should be stated on the page. If not, then the above statement is not cogent.

Also, if there is indeed such a one-to-one connection between Mersenne primes and perfect numbers, then the Great Internet Mersenne Prime search should probably be mentioned.--AxelBoldt

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It's true that every known perfect number comes from a Mersenne prime. All even perfect numbers come from Mersenne primes, and no odd perfect numbers are known. I believe this is stated on the page, but I'll try to make it clearer.

The GIMP is referenced on the Mersenne prime page. I'm intending to eliminate as much of the redundancy between this page and that one as possible, and I'll include a more emphatic pointer from here to there. -- Hank Ramsey

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"The GIMP" is an image-processing program. "GIMPS" is the Mersenne Prime search.

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Yes, of course. Sorry about that. - Hank

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Ok, I see it now: Euler proved that the even perfect numbers come from Mersenne primes.

Eliminating redundancy is of course good; the paragraph about 10:00 am Los Angeles campus appears on both pages and I think it has a bit too much detail :-) --AxelBoldt

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In base 6 senary number system all even perfect numbers besides 6 itself end in 44. The first of these is senary 44 itself which is 28.

That may be correct if you had written numeral system instead of number system. Michael Hardy 00:45, 1 Dec 2003 (UTC)

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It hasn't been proven that all perfect even numbers always end in a 6 or 8, but so far, the first thirty do. Also, each of these perfect numbers that ends in 8, really ends in 28.

Contrary to what is said in this article, the even perfect numbers only end in 6 or 28. This is true since perfect numbers are also triangular (and thus end in either 0, 1, 3, 5, 6, or 8 - which can be easily verified). Also note that a perfect number is a product of a power of 2 and a mersenne prime. If a perfect number ended in a 0, the mersenne prime would end in a 5, and the only prime number that ends in 5 (5 itself) isn't a mersenne prime.

kelvSYC 05:18, 28 Feb 2004 (UTC)

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Content of the perfect numbers page, now redirected here. Charles Matthews 18:55, 11 Jul 2004 (UTC)

In mathematics, a perfect number is a positive integer that is equal to the sum of all its proper divisors.

For example, 6 is a perfect number, its proper divisors being 3, 2 and 1; with 3 + 2 + 1 = 6.

The first eight perfect numbers are:

• 6
• 28
• 496
• 8128
• 33550336
• 8589869056
• 2305843008139952128.

The first three perfect numbers were known to Euclid. Euler discovered the eigth perfect number in 1732, after showing that even perfect numbers can be constructed from Mersenne primes.

There are no known odd perfect numbers, although all numbers with 300 digits or less have been checked.

[edit] References

MathWorld, Perfect numbers http://mathworld.wolfram.com/PerfectNumber.html

[edit] Odd perfect numbers

Why can't we say :

                 N =   (p-1)   p       =    (p-1) p     (p-1)  =        (p-1) (p-1)      (p-2)
                      2      (2- 1)       (2   * 2 )  - 2         2((  2  *  2      ) -  2   )

                            (p-1) (p-1)      (p-2)
                N/2 = 2((  2  *  2      ) -  2   )      

And that proof that N is always even and consequently there are no odd percfect numbers. ?!?!?!?!?!?!?!?!?!?!?!?!

You can't say that because that form is only for even perfect numbers and has nothing to do with OPNs. CRGreathouse (t | c) 02:47, 8 January 2007 (UTC)

That formula only works for the first 4 perfect numbers....

This is a proof that there are no odd perfect numbers.

I want it to be checked before I put it on the website (and mabey cleaned up a bit).

First some Definitions: d(n)=sum of the factors of n P(n)= d(n)-n

Lemma: If n has any odd factors, P(n) is even.

p is an odd prime P(p)=1-p the lemma is true for primes.

p is an odd prime and c is an odd # d(p*c)=d(c)+p*d(c)+c

A, B, C,…. Factors of c +p*A, p*B, p*C,…..p*Factors of c +c=d(c)+p*d(c)+c

P(p*c)=d(c)+p*d(c)+c-p*c=(1+p)*d(c)+(1-p)*c P(p*c)=(1+p)*d(c)+(1-p)*c=(1+1)*d(c)+(1-1)*c=0*d(c)+0*c=0 (mod 2) /*because p is odd*/ The lemma is true for composites with odd factors

Now the proof:

Since p is odd p= 1 or 3 (mod 4) By the above lemma P(c)=d(c)-c=0 or 2 (mod 4) d(c)=c(+2) (mod 4) so d(c) is odd P(p*c)=(1+p)*d(c)+(1-p)*c=(1+3)*d(c)+(1-3)*c=0+2*c=2*c=2 (mod 4) /*If p=3 (mod 4)*/ /*since c is odd*/ P(p*c)=(1+p)*d(c)+(1-p)*c=(1+1)*d(c)+(1-1)*c=2*d(c)+0=2*d(c)=2 (mod 4) /*If p=1 (mod 4)*/ /*since d(c) is odd (by above)*/

If n is an odd composite, P(n)=2 (mod 4) and it can’t be equal to 0. If n is an odd prime, P(n)=1-p, which is definitely not 0 .

If n is odd, P(n) is not =0 and so can’t be perfect!

This completes the proof that there are no odd perfect numbers !--SurrealWarrior 23:42, 18 Apr 2005 (UTC)

This is full of errors, first of all the lemma is false as P(15) = 1+3+5 = 9 which is odd. Also where p is a prime surely P(p) = 1 which is definitely not even! 172.188.68.250 (talk) 13:43, 4 May 2008 (UTC)

You should write it out in more detail. What exactly is the condition in the lemma? It seems that n=9 contradicts it: d(9) = 1+3 = 4 and P(9) = -5 is odd.

By the way, does anybody know about Simon Davis' proof that there are no odd perfect numbers in arXiV e-print hep-th/0401052? It looks to me like serious mathematics, but I'm not into number theory, and it is strange to publish it in hep-th. -- Jitse Niesen 16:55, 19 Apr 2005 (UTC)

Simon Davis' proof looks like nonsense to me. The definitions fall apart in the first page. --C

The problem is with my formula for d(p*c) only works when p and c have no common factors. The lemma and the proof only works when the number is square-free.--SurrealWarrior 02:31, 23 Apr 2005 (UTC)

There is a second article of Simon Davis' at Mathew Watkins' page http://www.math.ex.ac.uk/~mwatkins/zeta/davis.pdf where Davis formulates the theorem in a similar way, but with a shorter proof than the former article. I would like to know what definitions are meant that 'fall apart' (as C mentions above). For instance, a 'repunit' is just a sum of repetitive powers of a prime, thus sum(i=0..k, p^i) for p prime.

The idea of Davis seems to be to look at divisors of repunits and he notes that d(n) is the product of repunits. So, if the repunits in d(n) have other primefactors than n, it means that n can not be perfect. He explores a rationality condition to find the only candidates for odd perfect numbers, but concludes that they are not perfect, hence there are no odd perfect numbers. - Bcurfs 17:40, 30 December 2005 (UTC)

I've got a question; it seems absurd to say that an odd perfect number can have a prime factorization including a prime to the power of 4n+1. Here's my reasoning: 1) An odd number's prime factorization includes only odd numbers. 2) if an even number of odd numbers is added together, the result is an even number. 3) Given the form , the divisors of N can be exhaustively listed as the product of some power of q ranging from 0 to 4n+1 inclusive, and a divisor of the number M = N / q^{4n + 1}. 4) However, by #3, there are 4n+2 divisors of N for each divisor of M. (0, 1, 2, 3, 4, 5 ... 4n, 4n+1) 5) By #4, then, N has an even number of divisors. 6) By #5, #2, #1 and the definition of a perfect number, N has to be even; and therefore isn't an odd perfect number. I'm aware this reasoning must have a problem with it - otherwise the question of odd perfect numbers would have been closed already. But what's the problem? Michael Ralston 11:44, 3 January 2006 (UTC)

You have to exclude the number N itself from the list of divisors. If N is a perfect number, then the sum of all its divisors is 2N; the sum of all divisors excluding the number itself is N. -- Jitse Niesen (talk) 18:05, 3 January 2006 (UTC)

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• If an odd perfect number exists, it will not be of the form as shown in the Article; i.e.,

N=P4p+1Q2qR2r.... , with 4p+1, 2q+1, 2r+1, ..., each being a prime.

If such an N is perfect, we ought to have

(a) 2N=(1+P+P2+P3+...+P4p+1)(1+Q+Q2+...+Q2q)(1+R+R2+...+R2r)×.... , or

(b) 2N=(1+P)[1+(P2)+(P2)2+...+(P2)2p](1+Q+Q2+...)(1+R+R2+...)×... 

While the expression in (a) is acceptable, that in (b) is not. The latter is tantamount to treating N as N=P(P²)^2pQ^2qR^2r.... P² is neither a prime nor relatively prime to P. A true statement cannot have an unacceptable alternative expression. 69.158.119.184 02:50, 17 September 2006 (UTC)

I don't follow you. The expression in (b) above makes no sense, whether or not the initial expansion for N does. — Arthur Rubin | (talk) 21:07, 18 September 2006 (UTC)

• The sum of the proper divisors of P^4p+1 is 1+P+P²+P³+...+P^4p+1. The sum can also be rewritten as (1+P)×[1+(P²)+(P²)²+....+(P²)^2p]. The latter is the same as saying P^4p+1=P×(P²)^2p, with P² being treated as a prime and also as relatively prime to P, which of course is unacceptable. 209.167.89.139 18:26, 19 September 2006 (UTC)

I agree that (b) is unacceptable. But it also doesn't appear in the article. — Arthur Rubin | (talk) 19:23, 19 September 2006 (UTC)

• What appeared in the Article is the claim that an odd perfect number has the form of:

N=P4p+1Q2qR2r.... , with 4p+1, 2q+1, 2r+1, ..., each being a prime.

I showed that such an N cannot be perfect, using the argument I presented. 209.167.89.139 21:41, 19 September 2006 (UTC)

Nope. Just because σ(P^4p+1) is (1+P+P²+P³+...+P^4p+1) and is equal to (1+P)[1+(P²)+(P²)²+...+(P²)^2p], which is σ(P)"σ"((P²)^2p) (where "σ" means that we are pretending that P² is prime) doesn't mean the first or second expression makes no sense. — Arthur Rubin | (talk) 21:43, 20 September 2006 (UTC)

• Of the expressions, in your notation: σ(P^4p+1)=σ(P)"σ"[(P²)^2p]=σ(P^2p)"σ"(P^2p+1)=σ(P)σ(P^2p)σ[(–P)^2p], only the first; i.e., σ(P^4p+1), is acceptable as the sum of the proper divisors of P^4p+1 and the rest are not. Can a statement still be accepted as true despite it has unacceptable alternative representations? I think not.209.167.89.139 21:00, 21 September 2006 (UTC)

• • Only the first equality is correct, as well you should know. The rest are just wrong, even if not necessarily meaningless. But the existance of unacceptable alternative representations has nothing to do with the correct formulas. — Arthur Rubin | (talk) 00:14, 22 September 2006 (UTC)
  • I don't get it. A "true" statement is not supposed to admit any unacceptable representation.69.158.127.134 20:02, 24 September 2006 (UTC)
    • Any statement can be misinterpreted. I'm not at all sure what you're trying to say, but it looks wrong. (I assume that 69.158.127.134 and 209.167.89.139 are the same person.) — Arthur Rubin | (talk) 20:50, 30 September 2006 (UTC)
• Yes, that's me. My point is, suppose statement S means E₁, also E₂, also ..., etc. If some E_k are inadmissible, then S cannot be true. Conversely, for S to be true, all E_n have to be admissible. 209.167.89.139 15:45, 2 October 2006 (UTC)
  • As I said, any statement can be misinterpreted. In this case, the correct' statement would be "S can be misinterpreted as E, but E is clearly false. Hence, S is false?". — Arthur Rubin | (talk) 16:40, 2 October 2006 (UTC)
  • You accept what you believe, and I reject what I don't. I'll leave it at that. 209.167.89.139 18:08, 2 October 2006 (UTC)

[edit] About Ore's proof

What Ore did is proving every perfect number is harmonic. Not only even numbers but every number that is perfect.WAREL 18:53, 27 March 2006 (UTC)

• Suppose odd integer N is perfect. Then σ(N)=2N. Denote σ(N)=M. Then σ(M)=σ(2)σ(N)=3M, 2 and N being mutually prime. Is there an M such that σ(M)=3M? 69.158.119.151 02:01, 24 September 2006 (UTC)

Well there are known such n that are 0 mod 4, but none that are known that are 1 mod 2 or 2 mod 4 (2 mod 4 is easily seen to be equivalent to the existence of an OPN). The smallest n such that σ(n)=3n is n=120. JoshuaZ 02:48, 24 September 2006 (UTC)

You are careful to say "there are known ....", but not "it has been shown ....". Yes, all such M are 0 mod 4. If it can be proved that all σ(M)=3M cannot be 2 mod 4, then non-existence of OPN is also proved. Isn't it? Donkey2ft 22:34, 30 November 2006 (UTC)

I think so. JoshuaZ 22:38, 30 November 2006 (UTC)

• If OPN have the form N=X^4x+1Y^2yZ^2z… , then OPN can also have the form N=AB^2bC^2c… Donkey2ft 22:50, 29 November 2006 (UTC)

Yes, but in the first form X, Y, Z, and the others can be assumed to be pairwise relatively prime, while that cannot be assumed with the latter form. CRGreathouse (t | c) 13:13, 30 November 2006 (UTC)

Thanks. Can you explain why we cannot say A, B, C, ...., are mutually relatively prime? Or conversely, why can't x in the exponent be zero? I am interested in knowing if there is a proof that the second expression cannot be an OPN.Donkey2ft 21:35, 30 November 2006 (UTC)

Well, we just don't get that much info. The proof comes from looking at how σ(p^k) behaves mod 4 and uses the multiplicativity of σ. If you work out the proof you'll see why you could have a highper power on A. JoshuaZ 21:42, 30 November 2006 (UTC)

It's not that x couldn't be zero, it's that we can't prove that it must be zero (when the X, Y, Z, ... are pairwise relatively prime). CRGreathouse (t | c) 00:07, 1 December 2006 (UTC)

Please correct me if I am wrong. Suppose X, Y, Z, … are distinct odd primes. So are A, B, C, …. Multiplicativity of σ allows us to consider just the behavior of σ(X^4x+1) mod 4, and σ(A) mod 4, in both expressions for OPN. σ(N)=2N requires that both X and A be primes of the form 1 mod 4. Since both σ(X^4x+1) and σ(A) are 2 mod 4, existence of OPN as X^4x+1Y^2yZ^2z… does not preclude existence of OPN as AB^2bC^2c… Donkey2ft 00:46, 1 December 2006 (UTC)

I don't know what you're trying to show. Certainly I admit that the exponent on the special prime could be zero, I just don't agree that it need be zero. Any OPN can be written in the form with S, A1, ..., Ak distinct primes, whereas there's no reason to assume that the same is true of . (It would be trivially true if there were no OPNs.) CRGreathouse (t | c) 05:03, 1 December 2006 (UTC)

Here's my train of thought. A composite odd number, in the most general form, can be expressed as N=ABC…P^pQ^qR^r…, where A, B, C, … , P, Q, R, … are all distinct odd primes. For N to be an OPN, there is no reason to preclude N from being N=AP^2pQ^2qR^2r… , where 2p+1, 2q+1, 2r+1, …, are each a prime, and A congruent 1 mod 4. To me, the issue of perfectioness or imperfectioness of N=P^4p+1Q^2qR^2r… , with P congruent 1 mod 4, does not imply the perfectioness or imperfectioness of N=AP^2pQ^2qR^2r… I do not understand why we should overlook the second expression. Donkey2ft 15:28, 1 December 2006 (UTC)

Well, for one we can't prove that an odd perfect number must be of that form. In that form the prime which occurs an odd number of times in the factorization (sometimes called the "Euler factor") must occur to exactly the first power. However, we can't prove that's the case. The best we can prove is that the power it is raised to is 1 mod 4. JoshuaZ 15:40, 1 December 2006 (UTC)

I try to follow what you said. Can I infer from your first sentence that it has been proved that an OPN must not be of the form in which one prime occurs to exactly the first power? In my view, until that is proved, then and only then will the popular form in which all primes are powered be the only candidate left. Donkey2ft 16:33, 1 December 2006 (UTC)

No, for the third time, you can't assume that. CRGreathouse (t | c) 16:40, 1 December 2006 (UTC)

[edit] new argument

Let A>b>0 be integers. We can view A=Σa_kb^k as representation of A in base b. The representation is unique, and all a_k are non-negative integers.

Suppose N=X^4x+1Y^2yZ^2z… is an OPN. It can be shown X+1=[2(X²)^2xY^2yZ^2z…]/D, and X=σ[(X²)^2xY^2yZ^2z…]/D, where D=2(X²)^2xY^2yZ^2z…–σ[(X²)^2xY^2yZ^2z…]. The first expression implies X=2Y^βZ^γ…–1. Not all β, γ, … are equal to zero. Say, β≠0. So X=2SY^β–1, where S is an integer. The relationship indicates X>Y. Expansion of X in base Y is thus guaranteed; i.e., X=Σa_kY^k exists. Recall all a_k are unique, non-negative integers. The supposition of existence of N as an OPN, with use made of multiplicavity of σ, also leads to expansion of X in powers of Y, given by the second expression as: X=Kσ(Y^2y).

Wrong again. (If you can remove the places where I said you wrong before, I can do the same for you.) K is NOT an integer, so is not equal to a_k. — Arthur Rubin | (talk) 02:26, 9 December 2006 (UTC)

Several observations can be made. If 2y≠max.(k), then the supposition is false.

??? — Arthur Rubin | (talk) 02:26, 9 December 2006 (UTC)

If one of a_k is distinct, the supposition is also false, because its coefficients are all equal to K. If all a_k are identical, then K=a_k is an integer. But then K=1 in view of X being a prime. Now we are led to a contradictory relationship of 2SY^β–2=Y+Y²+…+Y^2y–1. Y divides into the RHS but not the LHS. So such N cannot be an OPN. The same approach can be used to prove that N=AB^2bC^2c… cannot be an OPN. We have exhausted all possible forms of OPN.Donkey2ft 00:22, 9 December 2006 (UTC)

Guys, this is all interesting but is WP:OR and not very relevant to Wikipedia. Maybe discuss by email or take it to sci.math? JoshuaZ 03:12, 9 December 2006 (UTC)

[edit] About Stuyvaert's proof

There is a statement in Mathworld that Stuyvaert (1896) proved that an odd perfect number must be a sum of squares. Are any of you familiar with this topic? Do you think it's saying that an odd perfect number must be a sum of "two" squares? Please give me any comment.--User:DYLAN LENNON 02:31, 10 Jul 2005 (UTC)

It must be the sum of two squares because every number is the sum of any number of squares.--SurrealWarrior 22:29, 11 July 2005 (UTC)

Thanks for your reply,SurrealWarrior.Do you think,then,that it should be restated as so? Or is it already saying? This might be more of a question about English. Could you teach me? --User:DYLAN LENNON 02:31, 12 Jul 2005 (UTC)

I am sorry to say that I have deleted it Dylan's edit. I think the sentence at the Odd Perfect Number article at MathWorld is not enough evidence. Firstly, it is not clear exactly what Stuyvaert proved; secondly, the webpage does not include a reference to Stuyvaerts work; and thirdly, the only thing I could find is on the Net (though I didn't try hard) is this thread, where nobody could find a reference either. -- Jitse Niesen (talk) 14:03, 12 July 2005 (UTC)

I've just received a mail from MathWorld.com.It says that they meant to say "two" squares.It also says they'll fix to it soon. It seems most likely Stuyvaert did prove that kind of result.Is it OK for all of you to edit after I researched about it? --User:DYLAN LENNON 12 July 2005 (UTC)

I would prefer if you asked the folks at MathWorld to provide a reference, but you may put it in the article if you wish. -- Jitse Niesen (talk) 21:49, 13 July 2005 (UTC)

I am opposed including this. Any odd perfect number N must be the sum of two squares, for it must be of the form q times a square, where q is of the form 4n + 1. Then q must be the sum of two squares, as is well known. Multiplying through, so must N - but this is trivial. Septentrionalis 21:06, 28 March 2006 (UTC)

I'm not sure is makes sense to include it either, although if memory serves me Stuyvaert had a weird proof rather than the simple one above. I haven't been able to track down Stuyvaert's original paper though. If there is some interesting way of proving the result then it should probably stay in. Does anyone else have access to the original paper? JoshuaZ 21:28, 28 March 2006 (UTC)

Stuyvaert only mentioned what Pmanderson just said in the textbook he wrote.WAREL 03:44, 29 March 2006 (UTC)

In that case, it should probably be removed. Incidentally, the comment you added about Makowski is also pretty trivial and should probably be removed also. JoshuaZ 04:39, 29 March 2006 (UTC)

• It is claimed that an OPN is of the form N=X^4x+1Y^2yZ^2z…. Now, σ(N)=2N requires that X be a prime of 1 mod 4. Due to Fermat, such a prime is a sum of two squares. We know that any integer power of a sum of two distinct squares is also a sum of two squares. So an OPN is a sum of two squares.

However, I think X is a prime of 1 mod 16. Such primes can be expressed as m²+8n². And product or power of such primes is also of that form. That being the case, OPN=a²+8b². 69.158.101.148 18:36, 31 December 2006 (UTC)

For the record, here's the quote: "The only even perfect number of the form　x³ + 1 is 28 (Makowski 1962)."[1]

Is it trivial? How do you show it? WAREL 04:51, 29 March 2006 (UTC)

There probably quite a few proofs since the conditions are very restrictive, but this seems to work(it is possible that I am making some stupid error here, but I doubt it): We have (x³ + 1) = N with N an even perfect number. So

(x + 1)(x² − x + 1) = (2^p − 1)(2^{p − 1}) Now we have x² − x + 1 odd so x² − x + 1 = 2^p − 1 and x = 2^{p − 1} This gives us (2^{p − 1})² − (2^{p − 1}) + 1 = 2^p − 1 and the left hand side grows much faster than the right hand side so it is easy to see that p=x=3 is the only solution. JoshuaZ 06:11, 29 March 2006 (UTC) Small fixes. Septentrionalis 06:52, 29 March 2006 (UTC)

Joshua, I think x≠2^p−1, for x³+1=odd≠EPN. 69.158.101.116 14:08, 1 January 2007 (UTC)

• Another proof. First, x³+1=2^p(2^p+1−1) implies x is odd. Next, x+1 and x²−x+1 are relatively prime, or they share 3 as a common factor and the EPN will contain 3² as a factor (Recall 2^p+1−1 is a prime). Hence, x+1=2^p and x²−x+1=2^p+1−1. They in turn yield (x−1)²=2^p. So p=2k. Now x+1=2^2k and x−1=2^k leads to 2=2^k(2^k−1). It is seen, k=1 is the only admissible integer and x=3 is the only solution. Donkey2ft 13:30, 8 December 2006 (UTC)

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Is there a use or application for perfect numbers, or are they just kind of a "hey, this is cool" type of thing? --Ryan Salisbury 21:34, 18 October 2005 (UTC)

[edit] Is 1 a perfect Number?

1's divisable numbers are just 1 so 1=1

and if so is 0 one aswell

No, a number is perfect if it is the sum of all its positiv divisors less than the number. The sum of all such divisors of 1 is zero. An alternate definition is the number is twice the sum of all its positive divisors in which case we would need that sum to be 2 for 1 to be perfect. But the sum of all the positive divisors of 1 is 1. JoshuaZ 14:28, 30 January 2006 (UTC)

0 is divisible by all natural numbers (zero excluded). Their sum is not finite, while 0 is. Hence 0 cannot be perfect. Ocolon 21:12, 9 March 2007 (UTC)

[edit] Number of Distinct Prime Factors

Warel, do you have a citation for your modification? I haven't seen anything about the number being pushed up by Nielsen. JoshuaZ 13:28, 28 February 2006 (UTC)

Unfortunately, Warel reacts rarely to comments on the talk page. He also has a history of questionable edits. Therefore, I reverted his edit. Warel can always put it back in with a proper reference. -- Jitse Niesen (talk) 11:13, 1 March 2006 (UTC)

[edit] Pomerance Heuristic

As far as I am aware, Pomerance's heuristic has not been published in a journal, although it is discussed on www.oddperfect.org which is run by William Lipp and is generally reliable. Should we cite that or should we just remove the comment about Pomerance's heuristic? JoshuaZ 00:32, 5 March 2006 (UTC)

That will do. I think the site could be mentioned under external links, but I'm not sure about it since it is described as "preannounced", so perhaps William Lipp would prefer that it isn't mentioned yet. What do you think about this? -- Jitse Niesen (talk) 01:15, 5 March 2006 (UTC)

I talked to Lipp. He says that he is fine with it as an external link but thinks that his webpage is not ready to be discussed in the article beyond that. This seems reasonable to me, and I will therefore add it to the external link set. JoshuaZ 18:56, 5 March 2006 (UTC)

[edit] Undecidability

Warel, can you please explain why you think this statement about undecidiablity is noteworthy when it applies to a very large class of problems? JoshuaZ 04:18, 16 March 2006 (UTC)

For the record, I agree with Joshua. -- Jitse Niesen (talk) 06:29, 16 March 2006 (UTC)

Because, it really encounters our intuition. No one will think about the undecidability when they first learn about odd perfects as a high school student. And it's good to let everyone know about "undecidability". I don't immediately think there are thousands of problem like this. WAREL 19:58, 19 March 2006 (UTC)

While it may be good to let people know about undecidability and its implications for specific problems, the fact is that it is highly incidental to this topic. Note that there aren't just 1000s of problems like this, but in fact an infinite number in any consistent axiomatic system that models Z. Just take any diophantine equation and one gets an essentially identical statement. JoshuaZ 20:01, 19 March 2006 (UTC)

I don't believe that. If that's true, that explanation surely is what really needs to be included in the article.WAREL 20:12, 19 March 2006 (UTC)

Warel, please find yourself a book on theoretical computer science that treats Computability theory (computer science) (books and/or classes are still the way to go if you want to learn something new). The argument goes as follows. Consider the following algorithm for answering the question "Does an odd perfect number exist":

1. Start with k = 1
2. If k is an odd perfect number, then the answer is YES.
3. If k is not an odd perfect number, then increase k by one and go to step 2.

This algorithm terminates if an odd perfect number exists, so it that case the question is decidable. But this argument does not depend on any properties of odd perfect numbers; the same argument holds for Goldbach's conjecture (is there an even integer greater than 2 which can not be written as the sum of two primes?) and any similar question. -- Jitse Niesen (talk) 04:37, 20 March 2006 (UTC)

Thank you, Jitse, that's a much better explanation of the matter than the one I was giving. JoshuaZ 04:42, 20 March 2006 (UTC)

That's not saying anything when odd perfect doesn't exist.That's just saying that we can surely know in finite time if it exists. WAREL 07:05, 20 March 2006 (UTC)

Exactly. So there are three possibilities:

1. There is no odd perfect number, and this can be proven (within a certain system of axioms).
2. There is no odd perfect number, and there is also no proof for this. In this case, the question "Do odd perfect numbers exists?" is undecidable.
3. An odd perfect number exists. In this case, there is always a proof for it.

Did you want to add something else to the article? If so, what? -- Jitse Niesen (talk) 07:51, 20 March 2006 (UTC)

What makes second possibility possible? Could you explain it?WAREL 08:37, 20 March 2006 (UTC)

I find this hard to understand intuitively, but there are statements that are true and yet unprovable (in a given theory); this is Gödel's first incompleteness theorem. It might be the case that "There are no odd perfect numbers" is one of these true but unprovable statements. -- Jitse Niesen (talk) 09:16, 20 March 2006 (UTC)

Of course the concept of "true" is itself a rather technical one here. If I understand correctly, this is to be taken as "there is no model of the theory in which the statement is false". For example neither the axiom of choice nor the continuum hypothesis is true in Zermelo-Fraenkel set theory, as there are models of ZF in which they are false (both proved by Cohen in 1963). Perhaps a logician could clarify if, in general, the existence of a model for a theory in which a statement is true is equivalent to the existence of an extension to the theory in which the statement is true? Elroch 09:49, 20 March 2006 (UTC)

To avoid being accused of being deliberately confusing, of course neither the negation of the axiom of choice nor the negation of the continuum hypothesis is true in ZF either. Elroch 12:14, 20 March 2006 (UTC)

John Barkley Rosser later strengthened the theorem by removing the requirement for the theory to be omega-consistent. How did his result change the recognition to this problem (odd perfect number) ?WAREL 19:39, 20 March 2006 (UTC)

[edit] Ohno's result trivia vs. trivial

Warel, there is a difference between trivia and trivial. In this case it is non-trivial trivia. JoshuaZ 22:22, 30 March 2006 (UTC)

I would say this is a result of recreational mathematics (if it is a result at all), as concatenating the perfect numbers as decimal integers has no obvious mathematical motivation (as will be seen if you try to express it as a series). Few will be surprised that Ohno's paper on a lower bound for odd perfect numbers does not include this result about a strange decimal number. I hope WAREL will be able to demonstrate that Ohno did indeed amuse himself by proving the result that WAREL stated, so as to remove any suspicion of falsification. The lack of any hits on google is somewhat surprising for such an elementary result. Incidentally, there are 13 edits in the history to achieve exactly nothing. Perfectly ludicrous. Elroch 01:13, 31 March 2006 (UTC)

What is the citation for Ohno's paper? I can't find it on mathscinet. JoshuaZ 01:24, 31 March 2006 (UTC)

I concur; the only papers he did in 2005 are Sum relations for multiple zeta values and connection formulas for the Gauss hypergeometric functions; and Sum relations for multiple zeta values. Somehow this doesn't seem a place to find this. And I don't think the result is trivial unless we know more about the distribution of Mersenne primes than seems likely. Suppose it were rational and it just happened that no perfect number ever had length a multiple of the period; where's the contradiction? Septentrionalis 06:23, 31 March 2006 (UTC)

I would be surprised if there is any easy proof. Proofs dealing with the irrationality or transcendance of numbers defined by concatination are notoriously difficult. It isn't even trivial to show that .012345678910111213.... is transcendental. JoshuaZ 06:27, 31 March 2006 (UTC)

Warel has celebrated April Fool's day by providing a "source" for this. The paper exists; although Ohno is junior author to Michael Hoffmann. It contains neither "perfect" nor "Mersenne", nor even "irrational". This had better stop on 00:00 April 2. Septentrionalis 04:22, 1 April 2006 (UTC)

To be honest, I don't know where the source is. I just heard directly from his lecture. Please help finding with me.WAREL 04:27, 1 April 2006 (UTC)

If you just heard it in lecture, that means we can't check it. See WP:V --Trovatore 04:39, 1 April 2006 (UTC)

Are you sure he didn't say "the prime numbers"? That is true. Septentrionalis 04:46, 1 April 2006 (UTC)

For future reference, if you don't know where something is, it is unhelpful to give a citation that happens to be from the same person. JoshuaZ 04:51, 1 April 2006 (UTC)

[edit] Rare

While I believe the text Elroch has edited was intended to be colloquial, there are well-defined senses in which any infinite set can be rare. Even the set of numbers of the form x · y² where x is prime and y is a product of at least 37 odd prime factors, at least 8 of them distinct, excludes almost all integers. Septentrionalis 23:25, 19 April 2006 (UTC)

Well, there are, but which of them have been proved to hold of odd perfect numbers, but not of even perfect numbers? Both, for example, must be of asymptotic density zero, so you'd have to find a finer measure than that. --Trovatore 23:34, 19 April 2006 (UTC)

Proving an order would mean proving that the sets are actually infinite, but I'm sure there are a lot of cute results on odd perfect numbers which are largely proven from none less than 10³⁰⁰. Septentrionalis 23:49, 19 April 2006 (UTC)

The strongest rarity result is to due to Wirsing, from a series of papers in written in the late 50s. He showed a much more general result: There is a constant C such that for any α we have the set of n less than x such that σ(n)/n =α is bounded by JoshuaZ 03:45, 4 June 2006 (UTC)

Isn't there a better one for even perfect numbers? I vaguely recall that the best proven asymptotic for evens is more restrictive than that for odds, though evens are believed to be infinite and odds nonexistant. CRGreathouse (t | c) 01:08, 4 September 2006 (UTC)

Yes, if one just uses the Euclid-Euler characterization of even perfect numbers and then apply the Prime Number Theorem you get the much tighter bound - O(log x / log log x) - you can get an explicit constant in the Big-O if one is slightly more careful. JoshuaZ 15:57, 14 September 2006 (UTC)

[edit] Special term?

Isn't there a special term for numbers whereby both σ(n) and s(n) are perfect? —Preceding unsigned comment added by 66.222.230.97 (talk)

I've never seen a term for that, although I have seen numbers where σ(n) and tau(n) are both perfect being called "sublime" JoshuaZ 03:46, 4 June 2006 (UTC)

[edit] The first two lines in disagreement.

"In mathematics, a perfect number is defined as an integer which is the sum of its proper positive divisors, that is, the sum of the positive divisors not including the number. Equivalently, a perfect number is a number that is half the sum of all of its positive divisors, or σ(n) = 2 n."

Just wanted to point out that the first two lines do not agree. If a perfect number is "an integer which is the sum of its proper postitive divisors" then how is it also "a number that is half the sum of all of its positive divisors"? I am not sure where to go to point this error out and this looked like the only place. Sorry if I am incorrect. —Preceding unsigned comment added by Troubledone (talk • contribs)

A number always divides itself, but a number's proper divisors don't include itself. Thus if σ(n) = 2n (where sigma is the number of divisors) the proper divisors are equal to n (since 2n − n = n). Don't worry, it's easy to get confused here with the terminology, even though it's not actually difficult when you get past that. CRGreathouse (t | c) 01:04, 4 September 2006 (UTC)

It is said if an odd perfect number exists, it is of the form N=P⁴ⁿ⁺¹Q², where P=4m+1 is a prime and Q is a distinct odd prime. Suppose such N is perfect, then 2=[(1+P+P²+...+P⁴ⁿ⁺¹)(1+Q+Q²)]/(P⁴ⁿ⁺¹Q²)<[P/(P–1)]×[Q/(Q–1)]<(5/4)×(3/2)<2, a contradiction. 209.167.89.139 21:36, 29 September 2006 (UTC)

Almost, in fact Q is just known to be an odd number, not an odd prime. What you have above is in fact a valid proof that an odd perfect number needs to have at least three distinct prime factors (because 1 prime factor is sort of trivial, and if we have two then Q needs to be prime). JoshuaZ 21:39, 29 September 2006 (UTC)

• Happy New Year to all.

Suppose N=X(X²)^2xY^2yZ^2z is an OPN. Then

(1) 2X(X²)^2xY^2yZ^2z=(1+X)σ[(X²)^2xY^2yZ^2z]; and

(2) X(X²)^2xY^2yZ^2z=[(1+X)/2]σ[(X²)^2xY^2yZ^2z]

Suppose X is 1 mod 4. The LHS of (1) is 2 mod 4. Since 1+X on the RHS of (1) is 2 mod 4, σ[…] on the RHS od (1) is 1 mod 4. Now, the LHS of (2) is 1 mod 4. In order for (1+X)/2 on the RHS of (2) to be 1 mod 4, X is 1 mod 8.

With X being 1 mod 8, the LHS of (1) is 2 mod 8. Since 1+X on the RHS of (1) is 2 mod 8, σ[…] is 1 mod 8. Now, the LHS of (2) is 1 mod 8. X is 1 mod 16 so that (1+X)/2 is 1 mod 8.

That X is 1 mod 16 is more restrictive than X is 1 mod 4.

Also, if an OPN exists, the following necessary bounds must be satisfied, with X=16k+1:

[(1+X+X²+…+X⁵)/X⁵] [(1+Y+Y²)/Y²] [(1+Z+Z²)/Z²] …… < or = 2 < [X/(X−1)] [Y/(Y−1)] [Z/(Z−1)] ……

It does not seem that all the bounds can be met. 69.158.101.148 19:27, 31 December 2006 (UTC)

Would you stop

1. Deleting other's comments on your errors results. That is uncivil.
2. Using σ[(X²)^2x] to indicate 1 + X² + ... + X^2x. That's just wrong.
3. Trying to prove there are no odd perfect numbers in this forum. That would be [WP:OR]], even if you knew what you were doing.

— Arthur Rubin | (talk) 14:55, 1 January 2007 (UTC)

• You know I don't discuss with you. Your comments are not welcome. 69.158.101.116 21:47, 1 January 2007 (UTC)

Anon, please see Wikipedia:Etiquette and WP:NPA. Arthur Rubin made valid points, and you have yet to respond to them. Your notation is non-standard and any proof you present is OR -- and much as I'd like to see a proof, this isn't the place. Now if you put this in your userspace instead of here, it might be better-received, but unless you do that or get a paper published this just isn't the place. I do wish you luck with your proof, though. CRGreathouse (t | c) 22:35, 1 January 2007 (UTC)

[edit] Hare's result; WAREL's edits

WAREL/WATARU's version doesn't mean quite the same thing as the current explanation that the number must have at least 75 prime factors; one natural way of reading the latter is 75 distinct prime factors. Someone ought to check what Hare's result really says. If WAREL's version is more accurate then we should probably rephrase as "75 prime factors (counting repetition)" or some such. --Trovatore 22:23, 13 September 2006 (UTC)

I've modified the article to read: "N has at least 75 (not necessarily distinct) prime factors", which agrees with what is claimed in this PDF. Paul August ☎ 03:33, 14 September 2006 (UTC)

Since that's stated one line away from the result on distinct factors, I think this is a bit pedantic. Still, if it's the way you prefer it, it's fine I suppose. CRGreathouse (t | c) 15:36, 14 September 2006 (UTC)

[edit] Dead link

The link: Pace P. Nielsen, "An upper bound for odd perfect numbers," Integers, vol. 3, A14, 9 pp. (electronic), does not seem to work - in that the website is there, but none of the links to files actually seems to work. It may only be a temporary problem, so I have not removed it - yet. Madmath789 07:40, 14 September 2006 (UTC)

I just replaced the link with one directly to the INTEGERS site. CRGreathouse (t | c) 14:04, 14 September 2006 (UTC)

The above link seems to work for me. JoshuaZ 15:22, 14 September 2006 (UTC)

The Emis link worked for me, but the internal links (to the paper and its abstract) didn't. I don't see any disadvantage to changing to the actual INTEGERS site, though. CRGreathouse (t | c) 15:35, 14 September 2006 (UTC)

[edit] "not distinct prime factors" vs not 'necessarily distinct prime factors for odd perf #s

Hi I think the 2nd phrase above is correct. The first phrase would mean an odd perfect number must be divisible by p^75 for some prime p, which I doubt is the case. Regards,Rich 00:12, 31 October 2006 (UTC)

No, it doesn't. But if it can be so misunderstood, it is undesirable. But "not necessesarily distinct" is wrong; no odd perfect number can be the product of distinct primes. Septentrionalis 00:25, 31 October 2006 (UTC)

Are you saying that if an odd perf number exists, it cannot be squarefree? I agree with you, but although "not necessarily distinct" does leave open in itself the remote possibility of squarefreeness, it's not strongly indicated. For example, the square of the product of the first 79 odd primes is a product of at least 75 distinct primes, but it's not squarefree.But I may be completely misunderstanding your point. Thanks in advance for your patience.Regards,Rich 00:41, 31 October 2006 (UTC)

Of course it can't. Only one of its prime factors can occur an odd number of times; the rest must occur at least twice. Septentrionalis 01:17, 31 October 2006 (UTC)

I would interpret the "not distinct prime factors" version to mean that, although there must be 75 or more primes in the factorization, there should be fewer than 75 distinct primes. But that's not a known fact, and what is actually intended is that we don't know how the number of distinct primes relates to 75. Therefore I would prefer the "not necessarily distinct prime factors" wording — it's not maximally pedantic, as we know that some factors are certainly non-distinct, but it's less open to misinterpretation. But I think even better would be to combine the two bullets: "N has at least 75 prime factors, and at least 9 distinct prime factors. If 3 is not one of the factors of N, then N has at least 12 distinct prime factors." That way we avoid all problems of how "not" or "not necessarily" is supposed to modify "distinct", and the juxtaposition makes it obvious (or even more obvious) that the 75 factors are not assumed to be distinct. —David Eppstein 01:22, 31 October 2006 (UTC)

That sounds much better.Rich 02:10, 31 October 2006 (UTC)

[edit] McDaniel claim

I think the claim is actually in McDaniel's paper. I'll dig it up later today and let people know. JoshuaZ 17:01, 31 October 2006 (UTC)

WAREL claims it's here; I don't see it. Septentrionalis 17:27, 31 October 2006 (UTC)

Here is the relevant passage from the PDF mentioned above (It's on page 3, number 2, middle of the page):

Proving that no OP [odd perfect] numbers can exist in the form , with all of the B_i’s in the same congruence class. McDaniel proved ([MCD2]) that having all of the B_i’s ≡ 1 (mod 3) is sufficient for N not to be OP. Iannucci added onto this result ([IAN]) by proving that if each B_i’ is either ≡ 1 (mod 3) or ≡ 2 (mod 5) then N is not OP. The proofs of these results are quite extensive, and thus are beyond the scope of this paper. However, a new proof of a special case of Iannucci’s result is provided below (Theorem 2).

Paul August ☎ 18:04, 31 October 2006 (UTC)

[edit] Perisastri and Grun

According to MathSciNet, Grun 1952 proved that the smallest prime factor is < (2n/3)+2, where n is the number of distinct prime factors (MR0053123; requires subscription), and Perisastri 1958 published the same result (MR0115963). So let's stick to Grun. -- Jitse Niesen (talk) 02:40, 4 November 2006 (UTC)

[edit] Nielsen's result

• N is less than . N is less than if q = 5 and α = 1. N is less than if q = 5 and α = 1 and p₁ = 3 and e₁ = 1, assuming p₁ < p₂ < ... < p_k (Nielsen 2003).

Working off memory, this seems to be exactly what Nielsen's paper shows (well, it actually shows more, but these three are special cases). Does anyone actually have objections to the material, regardless of its source? CRGreathouse (t | c) 04:21, 26 January 2007 (UTC)

We already have some results listed that seem to rule out only unlikely-looking special cases (I feel the ones attributed to McDaniel and Yamada are of this type), and this seems like another such. I worry that continuing to add all possible published results of this type will make an even longer and less readable list of properties, only a few of which are actually interesting. But if someone other than WAREL adds the results to the list, I won't revert it. —David Eppstein 05:06, 26 January 2007 (UTC)

[edit] I'm confused

It says that a perfect number is the sum of its divisors...

but 28 is clearly not a multiple of 5, and 5 is listed as though it were one. If anyone can clear this up for me, that'd be grand.68.93.99.142 01:59, 10 March 2007 (UTC)

DOH! I got two things mixed up. 28 is 1+2+3+4+5+6+7 but also 1+2+4+7+14. disregard! 68.93.99.142 02:03, 10 March 2007 (UTC)

[edit] June 28

I saw that June 28 has been added to and removed from the See also section a several times now. I think discussing this and finding a common agreement here would be helpful. Why is it added? Because the date June 28 (06/28) consists of two perfect numbers. Why is it removed? I guess because it's rather arbitrary and has no mathematical relevance and is not the only date consisting of perfect numbers (→ June 6) anyway. What do you think about it? — Ocolon 19:49, 14 March 2007 (UTC)

It seems rather arbitrary to me, and quite unrelated to this article. CRGreathouse (t | c) 03:10, 15 March 2007 (UTC)

I agree that this is connected only in a trivial way to perfect numbers. There is no mathematics in the connection, and very little relevance to timekeeping. If one were making a mathematical desk calendar with cute mathematical factoids connected to every date, I'd include it, but that's not what Wikipedia is. So my preference would be to leave it out, though I don't feel strongly enough about it to make edits removing it myself. —David Eppstein 18:52, 17 March 2007 (UTC)

[edit] Odd perfect numbers

Are there some mathematicians who believe that there are odd perfect numbers; we just haven't found them?? Georgia guy 18:10, 19 March 2007 (UTC)

oddperfect.org does not believe the arguments they don't exist; which is as close as anybody is likely to come in public. Septentrionalis PMAnderson 01:28, 13 April 2007 (UTC)

I see above that William Lipp, who runs it, would prefer that this be only an external link. Septentrionalis PMAnderson 01:30, 13 April 2007 (UTC)

The text quotes "q congruent 1 mod 4". I believe q is congruent 1 mod 16.Zymogen (talk) 19:09, 12 January 2008 (UTC)

[edit] Yamada's statement

What about the statement:

• If e₁ = e₂ = ... = e_k = β, then k ≤ 4β² + 2β + 2 (Yamada 2005), and β does not equal 1, 2, 3, 5, 6, 8, 11, 12, 17, 24 or 62 (Steuerwald, McDaniel, Kanold, Hagis, Cohen, Williams)^{[citation needed]} or any 3m+1 (special case of McDaniel's above result).

The review in MathSciNet says that the betas in Yamada's paper are defined differently from our betas. An IP editor, probably User:WAREL, says otherwise. Additionally, it's not clear to me how important the result is. It's a very strong restriction to assume that all the e_i are equal. Then there is the fact tag which has been around for a few months. So, I moved the statement here. -- Jitse Niesen (talk) 13:39, 18 May 2007 (UTC)

The matter is very simple. Just go to the library and look at Yamada's original paper. The importance of the result is a different question. It's a necessary condition for the OPN to exist. So, the result is meaningless if the OPN doesn't exist at all. Nevertheless, no one can assure yet the nonexistence.

218.133.184.93 14:07, 18 May 2007 (UTC)

I checked and MathSciNet is indeed wrong. However, the result is not just a necessary condition for the existence of an odd perfect number; it's only a condition for the existence of an odd perfect number of a very special form. Since odd perfect numbers are extremely rare, if they exist at all, the chance that odd perfect numbers of a very special form exist is even smaller. That's why the result does not seem that important to me. -- Jitse Niesen (talk) 04:39, 22 May 2007 (UTC)

It still is a necessary condition for the existence of an odd perfect number.218.133.184.93 15:09, 24 May 2007 (UTC)

Wrong. It is a necessary condition for the existance of an odd perfect number with all exponents equal. The first case may be of some interest, but the general case is not. — Arthur Rubin | (talk) 19:37, 24 May 2007 (UTC)

An odd perfect number with all exponents equal is an odd perfect number.218.133.184.93 03:53, 25 May 2007 (UTC)

[edit] Curtiss result

Since the Curtiss result is massively superceded by the results of Heath-Brown et al. should we just remove it? JoshuaZ 07:53, 24 May 2007 (UTC)

Do we have the Heath-Brown result in the article?

In any case the Curtiss result is weak, so I have no strong feelings about it (although I have always liked that paper for some reason). On a similar note: the Kühnel reference¹ is rather weak, simple enough that it could probably be used as an undergraduate exercise. Should this be demoted to 'minor references'?

¹ I haven't read it, though I read a similar paper by Touchard with the same result, as well as a simplifying paper referencing Touchard's.

CRGreathouse (t | c) 12:31, 24 May 2007 (UTC)

Heath-Brown is included (in that we give the slightly tighter version due to Nielsen) which is a vast improvement on Curtiss. I have no strong impression about Kuhnel, although the 105m matter really is just a specific case of any multiple of an abundant number being abundant together with Euler's restriction. Furthermore, I think the 105m thing shows up in a paper of Sylvester's much earlier but I'll need to track down the reference. The Kuhnel thing is only interesting in so far as that the results of that form are normally attributed to Touchard and Kuhnel seems to have been slightly earlier. JoshuaZ 18:42, 24 May 2007 (UTC)

That's a fair point on Kühnel; I hadn't read about it before this article. On those grounds alone we should keep it in this article. (Of course it could still be demoted to minor results or not.)

You're quite right on Heath-Brown. I couldn't place the reference at first but now that I think on it I believe I have read the paper.

CRGreathouse (t | c) 20:08, 24 May 2007 (UTC)

[edit] What do you mean unsupported?

There was an addition to the article (changing the OPN lower bound to 10^400) that was reverted; on re-adding it, the comment What do you mean unsupported? [2] was used.

The reference added was

• Buxton, M. and Elmore, S. "An Extension of Lower Bounds for Odd Perfect Numbers." Not. Amer. Math. Soc. 22, A-55

This paper is from 1976. As far as I can tell it ostensibly proves a 10^200 lower bound but Brent & Cohen 1989 don't believe that it even manages that ("Details of these expansions [to 10^200] have never been published…").

As a result, I'm going to revert the change again unless there's excellent justification.

CRGreathouse (t | c) 04:04, 27 May 2007 (UTC)

Thanks for dealing with that. Yeah, my understanding of that paper is that, at best, it claims proves a 10^200 bound, not 10^300. Cheers, Doctormatt 05:25, 27 May 2007 (UTC)

[edit] Application?

Are there any applications of perfect numbers (as primes are used in cryptography, for instance), or are they simply a mathematical curiosity? I think a note either way would be nice. --67.102.218.34 17:41, 10 July 2007 (UTC)

I second that. The article is notably lacking details of why perfect numbers are important, besides giving theoretical mathematicians something to write papers about :) Brianski 09:12, 28 October 2007 (UTC)

Having practical applications is not relevant to whether something is notable in the mathematical community, which is all that is necessary for it to be notable enough to be included in a general encyclopedia, especially one that purports to include all human knowledge. Xihr 10:24, 28 October 2007 (UTC)

I didn't see anyone suggesting that there was anything wrong with having the article in WP. 67 and Brianski are quite right that if there were any practical applications, it would be good to mention them in the article. Unfortunately I don't actually know of any. --Trovatore (talk) 02:28, 9 January 2008 (UTC)

[edit] These are all proven

• N has at least 9 distinct prime factors. If 3 is not one of the factors of N, then N has at least 12 distinct prime factors. (Nielsen 2006). If 3 and 5 are not one of the factors of N, then N has at least 15 distinct prime factors.If 3 and 5 and 7 are not one of the factors of N, then N has at least 27 distinct prime factors. (Norton 1960,1961)

• N is less than . N is less than if q = 5 and α = 1. N is less than if q = 5 and α = 1 and p₁ = 3 and e₁ = 1, assuming p₁ < p₂ < ... < p_k (Nielsen 2003). —Preceding unsigned comment added by 218.133.184.93 (talk) 22:05, 15 September 2007 (UTC)

• The (i+1)th smallest prime factor is less than for (Masao Kishore 1981).

• If e₁ = e₂ = ... = e_k = β, then k ≤ 4β² + 2β + 2 (Yamada 2005), and β does not equal 1, 2, 3, 5, 6, 8, 11, 12, 17, 24 or 62 (Steuerwald, McDaniel, Kanold, Hagis, Cohen, Williams)^{[citation needed]} or any 3m+1 (special case of McDaniel's above result).

When e_i ≤ 2 for every i

• α ≡ 1 (mod 12) or α ≡ 9 (mod 12) (McDaniel 1970).
• 　The smallest prime factor of N is greater than 2500000 and less than exp(4.97401 * 10¹⁰) (Yamada 2005).

—Preceding unsigned comment added by 218.133.184.93 (talk)

Beyond any issue of mathematics, I find it fascinating that this anoymous editor continues failing to communicate in useful ways. Here they make a statement in a header, and list a bunch of claims. No sentences, no explanations, no greetings, no challenges, no discussion. This is just fascinating to me. Why don't they communicate? And the revert wars! Oh! Anyway...

I've casually read through Yamada 2005 several times today, and I fail to see how anything in that paper applies to yield this bound claimed by this anonymous editor. The pdf (apparently of slides from a talk) that this editor has crammed into the bottom of the reference section doesn't have a proof either, as quite a lot of detail is lacking (and the pdf mentions methods that are definitely not in Yamada 2005). In any case, it is unpublished, so not useful as a citation, and this editor should know this by now. If Yamada publishes the result, and we can cite it, perhaps the result could go into the article. What else is there to say? Cheers, Doctormatt 19:35, 15 September 2007 (UTC)

"This anonymous editor" is almost certainly WAREL. Not just by pattern of edits but also similarity of IP address to one already listed in suspected Wikipedia sockpuppets of WAREL. —David Eppstein 20:00, 15 September 2007 (UTC)

Hey Mat, try to read this one.

• Tomohiro Yamada, On the divisibility of odd perfect numbers by a high power of a prime,arXiv:math.NT/0511410.

—Preceding unsigned comment added by 218.133.184.93 (talk)

"Try"? Oy. Doctormatt 20:07, 15 September 2007 (UTC)

The pdf looks fine to me, and although it doesn't contain more than a sketch of the proof. I imagine I could come up with a proof, having read it. It's quite elementary by all appearances, nothing like slogging through Ianucci's excellent papers on the subject. On the other hand, though, limiting the smallest prime factor of N subject to the assumption that all non-special exponents are either 2 or 4 strikes me as an extremely weak result -- much weaker than the McDaniel result to be sure. Now if he can bring down the astronomical exp(4.97401×10¹⁰) bound and actually test all primes in the range, we'd have something barely worth mentioning on the minor results page (if it gets published). CRGreathouse (t | c) 00:55, 16 September 2007 (UTC)

Do you guys realize you're all editing the talk page? I don't understand all the edits, re-edits, and reverts going on here. Xihr 09:11, 16 September 2007 (UTC)

[edit] Minor results ?

I'm a bit confused about this statement in the "Minor Results" section:

The number of divisors of a perfect number (whether even or odd) must be even, since N cannot be a perfect square.

The example of the first perfect number is 6 = 3 + 2 + 1. Isn't that an odd number of divisors? What am I missing?

Thanks —Preceding unsigned comment added by 149.173.6.51 (talk) 20:30, 22 October 2007 (UTC)

6 has four divisors: 6, 3, 2, and 1. Xihr 20:33, 22 October 2007 (UTC)

If they exist, OPN=A^4a+1P^2pQ^2qR^2r……, a≥0, can be written as OPN=BX^2x, which in turn leads to an alternative expression for OPN: OPN=CX^2xσ(X^2x), and an alternative condition for perfection: 2CX^2x=σ[Cσ(X^2x)]. The expressions are applicable to: N=A^4a+1P^2p and N=A^4a+1P^2pQ^2q. With A congruent 1 mod. 16, one can show that these two N's violate the necessary bound: 2<[A/(A–1)]×[P/(P–1)]×[Q/Q–1)]×…… Based on imperfection of these two N's, one is led to surmise that given any C and x, no odd prime X can be found to satisfy 2CX^2x=σ[Cσ(X^2x)]. Conversely, for whatever odd prime X is chosen, no C and x can be found to satisfy 2CX^2x=σ[Cσ(X^2x)]. Incidentally, 2CX^2x=σ[Cσ(X^2x)] is satisfied by C=1, X=2, and a prime 2^2x+1–1. Do I make sense?? Zymogen (talk) 16:13, 8 January 2008 (UTC)

No, you do not make sense. X^2x is not what you've got. You can write OPN=A X² (where A is prime, but A and X are not necessarily relatively prime), or OPN = A^4a+1X², where A is prime and A and X are relatively prime, but I don't see where you get OPN =CX^2xσ(X^2x) out of any of these. You may be able to get results where all p=q=r=..., but I don't see where else you can go with this. — Arthur Rubin | (talk) 22:40, 8 January 2008 (UTC)

OPN=(A^4a+1P^2p…) X^2x (…Z^2z…)=BX^2x. We group all other terms into B. B is distinct from X. σ(BX^2x)=σ(B)σ(X^2x)=2BX^2x. Because σ(X^2x) is odd and not divided by X, σ(B) is even and divided by 2 and X. So σ(B)/(2X^2x)=C, C is distinct from X, B=Cσ(X^2x), and OPN=CX^2xσ(X^2x). Now, σ[CX^2xσ(X^2x)]=σ(X^2x)σ[Cσ(X^2x)]=2CX^2xσ(X^2x) gives σ[Cσ(X^2x)]=2CX^2x. In the derivation, X really need not be odd. So the result is also satisfied by all Euclid's perfect numbers except N=6, for 6 cannot be written as BX^2x. But all OPN ought to be expressible as such. I hope I had clarified my points and made sense to you. Zymogen (talk) 02:13, 9 January 2008 (UTC)

Do you have any comments or any thought to share with me? Zymogen (talk) 20:13, 10 January 2008 (UTC)

I still don't see the point of your analysis, as you haven't indicated which of the numbers are primes, prime powers, or relatively prime. (And it's clearly not X^2x, it's just X².) — Arthur Rubin | (talk) 22:27, 10 January 2008 (UTC)

For the discussion of OPN, A, P, Q, …, X, Y, Z, … all are distinct odd primes; P^2p, X^2x, …, etc. are each a power of an odd prime; small letters p, q, …, x, y, … are positive integers; B and C are positive odd numbers. It is simpler to deal with OPN=CX^2xσ(X^2x) and perfection as expressed by σ[Cσ(X^2x)]=2CX^2x. Both are true for any OPN. Now, we use N=A^4a+1P^2p and N=A^4a+1P^2pQ^2q, a≥0, to contradict σ[Cσ(X^2x)]=2CX^2x. These N can be expressed in the form of CX^2xσ(X^2x) if they are perfect. But they are imperfect by virtue of the necessary bound. This suggests that for whatever odd prime X is chosen, no integers C and x can be found to satisfy σ[Cσ(X^2x)]=2CX^2x. Zymogen (talk) 22:37, 10 January 2008 (UTC)

Arthur, do I make sense? I would like to think so! Zymogen (talk) 01:33, 22 January 2008 (UTC)

I'd like to think so, also, but it doesn't seem to help unless the number of distinct prime factors is small. — Arthur Rubin | (talk) 01:44, 22 January 2008 (UTC)

We can write an OPN, if it exists, as OPN=BX^2x to single out a power of a prime X and dump all other prime powers in B. So however many prime powers there are in an OPN does not matter. Zymogen (talk) 13:50, 23 January 2008 (UTC)

You can write any number as bX^2x trivially with X = 1, but how does that help? The special thing about odd perfect numbers (or even perfect numbers greater than 6) is that they can be written bX² with b a prime power and gcd(b, X) = 1. CRGreathouse (t | c) 14:23, 23 January 2008 (UTC)

Sorry, X=1 is not a prime. I require that X be an odd prime number. You lost me. Zymogen (talk) 23:04, 23 January 2008 (UTC)

[edit] OPN proofs

Well, it's true that an odd perfect number can be written as pn² with p prime. In fact that's true for any perfect number greater than 6. But more is true by the Euler form -- we know that p is 1 mod 4 and that for some k ≥ 0, where iff and . I guess I just don't see where you're going. CRGreathouse (t | c) 14:16, 24 January 2008 (UTC)

First, it is important that OPN=CX^2xσ(X^2x) and perfection stated as 2CX^2x=σ[Cσ(X^2x)], where C is an odd number, X is an odd prime number, and x≥1 is an integer, be true for any OPN.

Next, consider N=A^4a+1X^2x and N=A^4a+1X^2xY^2y, a≥0, where A, X and Y are distinct odd prime numbers. If these two N's were OPN, then we ought to be able to express them in the form as shown in the previous paragraph.

By means of the necessary bound, we know such N's are not perfect. Then we conclude that whatever odd number C and integer x≥1 we choose, we cannot find an odd prime number X to satisfy CX^2xσ(X^2x) and 2CX^2x=σ[Cσ(X^2x)]. Conversely, for whatever odd prime number X we choose, we cannot find an odd number C and an integer x≥1 to satisfy CX^2xσ(X^2x) and 2CX^2x=σ[Cσ(X^2x)].

Suppose we could find an odd prime number X, an odd number C, and an integer x≥1 to satisfy CX^2xσ(X^2x) and 2CX^2x=σ[Cσ(X^2x)]. Then we contradict the fact that N=A^4a+1X^2x and N=A^4a+1X^2xY^2y, a≥0, are not perfect.

Therefore, I am inclined to say that OPN does not exist. Zymogen (talk) 16:07, 24 January 2008 (UTC)

I think we both agree that OPN=A^4a+1P^2pQ^2q……X^2x…Z^2z, a≥0. Euler viewed it as OPN=A[(A²)^aP^pQ^q……Z^z]²=AM². So M is composite while A is a prime, as you said. I view it as OPN=X^2x[A^4a+1P^2pQ^2q……Z^2z]=X^2xB. So B is composite but X is a prime. Euler singled out A, whereas I single out X^2x. Really the views are of the same OPN but from different angles. It seems you have cast my representation in the form of Euler's representation. Perhaps that's why you do not see what I see. My view allows me to arrive at the two new expressions, which I believe are useful. Zymogen (talk) 19:02, 24 January 2008 (UTC)

For N = A^{4a + 1}X^2xY^2y, where A, X and Y are distinct odd prime numbers, as you suggested, N cannot be an odd eprfect number because it has only three distinct prime factors and OPNs must have at least 9.

Your form, where you single out an arbitrary prime power rather than the special prime, seems less useful to me because facts are known about the special prime but nothing is known about your arbitrary prime, just what applies to all nonspecial primes. CRGreathouse (t | c) 03:38, 25 January 2008 (UTC)

My representation of OPN=CX^2x, where C is an odd number and X is an odd prime number, leads to a condition for perfection as:

2CX^2x=σ[Cσ(X^2x)].

An observation of the condition suggests that an odd number of the form N=CX^2x is not perfect. If it were, it ought to satisfy 2CX^2x=σ(CX^2x) instead of 2CX^2x=σ[Cσ(X^2x)]. My view is able to reveal this dilemma. Zymogen (talk) 21:19, 27 January 2008 (UTC)

[edit] Interesting Product

I was attempting to prove that there were no odd primes, and although unsuccessful I did come up with something that might help either those who are looking for one, or perhaps just as a nail to say it is even less likely...

I'm not sure if the same idea is already listed and I missed it?

I've taken this right back to basics, so please excuse me if I'm not using the appropriate notation.

Let N be a perfect number, with its prime factorisation expressed as....

Then it turns out that...

Equation 1:

To me what is interesting about this is that each factor on the RHS... needs to render a certain number of prime factors on the LHS. Exactly one of them must be even, and the total count of their prime factors must be = .

Where it gets interesting is that for a lot of these factors their prime factorisation produces fewer factors than x_i, and so there needs to then be other factors that can then take up the slack.

Another result derived is that...

Equation 2:

By applying x_i = 1 (only allowed once) and , we see that each factor in this equation falls within a range...

Minimum = (p_i + 1) / p_i

Maximum = p_i / (p_i − 1)

So for all values of x_i, the value > 1.

This should be useful in trying to construct a perfect number, because as you add each new factor, then if the product of terms in Equation 2 exceeds 2 then that subset is no longer valid. And if it equals 2 you have found the first ever odd perfect number!

I hope this all makes sense to someone. I really think that it is a fairly strong limiting factor on any possible odd prime, and perhaps someone with better maths skills than I have can make use of it.

Xtempore (talk) 07:42, 11 December 2007 (UTC)

Yes, this equation 2 of yours is used implicitly in all sigma-proofs: the idea that multiplying a number by a prime always increases the value, and that the contribution of a given prime power with an unknown exponent is bounded. Although practically small primes must be treated separately since the contribution of, say, a second 5 is significant. CRGreathouse (t | c) 13:17, 12 December 2007 (UTC)

[edit] Why Even Perfect Numbers Work (and Odd Perfect Numbers may not!)

Each element in the RHS of Equation 1 (as in my previous post) contributes a certain number of prime factors for the LHS, and as noted the total on the right must equal the sum of the x's + 1 (to account for the 2 on the LHS).

For any pairing of p_i and x_i the following equation gives what I like to call the generosity of the element...

The interesting thing about this is that for even Perfect numbers we have...

N = 2^n-1(2ⁿ-1)

When p₀ = 2ⁿ - 1, x₀ = 1, then g₀=n

When p₁ = 2, x₁=n - 1, then g₁ = 1-n

Adding these together we find the total generosity, G = n + 1-n = 1 ... As is required for a perfect number.

So Mersenne primes are great for generating Even Perfect Numbers, because even with x=1 they are very generous.

BUT Mersenne primes are not so good for odd perfect numbers.

For ODD values of x, they have a high generosity, but the main thing they contribute is a whole lot of 2's, in other words they can only produce even numbers.

For EVEN values of x, their generosity seems to slump into negative numbers (exceptions/proof?).

In other words, using the form as per the main page...

q CANNOT be a Mersenne Prime (because alpha must be odd).

Mersenne Primes are also poor candidates for p_i because of their negative generosity.

Xtempore (talk) 03:53, 13 December 2007 (UTC)

[edit] Nitpick!

The article says "It is straightforward to show that the last digit of any even perfect number must be 6 or 8." Is it really? Then tell me how this can be shown. Of course it follows trivially from the theorem that all even perfect numbers fit the Mersenne prime formula, but since this theorem is far from "straightforward" to prove, that can't be what's meant. 91.105.63.239 (talk) 19:48, 16 April 2008 (UTC)