Talk:Perfect number

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.

References

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

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Euler said all EPN are Euclid's PN. All even composite numbers can be expressed as E=2^kP, where P is an odd number. If E is perfect, then σ(E)=(2^k+1–1)σ(P)=2E=2^k+1P. Therefore, P=2^k+1–1 and σ(P)=2^k+1, which in turn yields σ(P)=P+1, implying that P has to be an odd prime. Thus, all EPN are indeed Euclid's PN. 209.167.89.139 19:50, 1 October 2006 (UTC)

Odd 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)

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 $N=q^{4n+1}p_{1}^{2e_{1}}\ldots p_{k}^{2e_{k}},$, 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)

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 $N=S^{4s+1}A_{1}^{2a_{1}}A_{2}^{2a_{2}}\cdots A_{k}^{2a_{k}}$ with S, A1, ..., Ak distinct primes, whereas there's no reason to assume that the same is true of $N=SA_{1}^{2a_{1}}A_{2}^{2a_{2}}\cdots A_{k}^{2a_{k}}$. (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)

new argument

JoshuaZ, please forgive me for cleaning the "board" for easy reading of the edited text. Thanks.

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)

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, X is a prime of 1 mod 16. Such primes can be expressed as m²+8n². If so, 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　$x^{3}+1$ 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 $(x^{3}+1)=N$ with N an even perfect number. So

$(x+1)(x^{2}-x+1)=(2^{p}-1)(2^{p-1})$ Now we have $x^{2}-x+1$ odd so $x^{2}-x+1=2^{p}-1$ 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)

• 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, lest they will share 3 as a common factor and the EPN will contain 3² as a factor. 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 can be 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)

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)

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)

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)

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)

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)

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 $\alpha$ we have the set of n less than x such that σ(n)/n =$\alpha$ is bounded by $e^{\frac {\log x}{\log logx}}$ 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)

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 • contribs)

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)

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 $\sigma (n)=2n$ (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 you.

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 bounds must be satisfied:

[(1+X+X²+…+X⁵)/X⁵]×[(1+Y+Y²)/Y²]×……<2<[X/(X−1)]×[Y/(Y−1)]×……

It does not seem that the bounds can be met simultaneously. 69.158.101.148 18:29, 31 December 2006 (UTC)

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)

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)

"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)

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 $N=q^{e}a^{2B_{1}}\ldots a^{2B_{n}}$, 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)

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)