[Number Theory] Challenge Problem 1.

[Steffen Schuler]

Are there a finite set I of odd primes and a family (n_p: p in I)
of integers larger than one such that the product

Prod(p in I: (1 - (1/p)^(n_p)) / (1 - 1/p))

is equal to two?

Either give an example or prove that this is impossible or reduce an
open problem to this problem!

--
Steffen Schuler <schuler....@gmail.com>

[Steffen Schuler]

Charlie-Boo wrote:
>
> Impossible since each product is >1 and <2 so if there is one
> element then the sum is <2 and if two or more it is >2.

The limit of the product for ``I --> |P \ {2},'' and for the
exponents n_p --> oo is

Prod(p in |P \ {2}: 1/(1 - 1/p)) = (1/2) zeta(1) = oo.

Thus, for infinitely many instances the original product is larger
than or equal to two.

From which "sum" do you write? If you have denoted inadvertently the
original product with the term "sum", then note that the product of
two given factors can clearly be less than or equal to two. Consider
for example I = {3, 5} and n_3 = n_5 = 2. Then the given product
is equal to

(1 + 1/3)(1 + 1/5) = (4/3)(6/5) = 8/5 < 2.

Currently, I don't want to say anything about the possibility or
impossibility of the product being equal to two, since I want to give
also other people a chance to solve the problem.

--
Steffen Schuler <schuler....@gmail.com>

[quasi]

Charlie-Boo wrote:

>Steffen Schuler wrote:
>>
>>Are there a finite set I of odd primes and a family
>>(n_p: p in I) of integers larger than one such that
>>the product
>>
>> Prod(p in I: (1 - (1/p)^(n_p)) / (1 - 1/p))
>>
>>is equal to two?
>>
>>Either give an example or prove that this is impossible
>>or reduce an open problem to this problem!
>

>Impossible since each product is >1 and <2

You mean each factor.

>so if there is one element then the sum is <2

Of course.

>and if two or more it is >2.

No, that doesn't follow.

quasi

[Steffen Schuler]

I post here a solution to the task I set in the original root article
of this thread. I'm grateful for reviews and constructive criticism.
Please, tell me eventual mathematical or syntactical errors or
flaws. Since nobody (besides me) has posted a correct and complete
solution, I'm fearing that the task either didn't have the right
difficulty level or was not interesting enough. Please, tell me
your opinion. Shall I continue to post mathematical problems?

----8----------------------------------------------

The Equivalence of Challenge Problem 1 to some Open Problem

Version 1.0

http://schuler.bplaced.net/chall01.txt

Steffen Schuler <schuler....@gmail.com>

May 6, 2017


Dedicated to my sister Eva Maria Schuler.

Abstract. We state a difficult arithmetical problem and show that
it is equivalent to some open problem.


1. Introduction

The difficult arithmetical problem is Challenge Problem 1. It was
posted originally by the author at April 30, 2017 in the Usenet
newsgroup "sci.math". In this posting, the author set the explicit
task that either the problem should be solved in a positive or a
negative way, or that some open problem should be reduced to this
problem.

Challenge Problem 1. Are there some finite set M of odd primes and
some family (e_p: p in M) of integers larger than 1 such that

2 = Prod(p in M: [1 - (1/p)^(e_p)] / [1 - 1/p])?

We will show that Challenge Problem 1 is equivalent to the well-known
open problem whether there exists an odd perfect number.


2. Preliminaries

Let x in M mean that the object x is an element of the set M. Let
i = a..b mean that i is an integer such that a <= i <= b. Let
(x_i: i in M) stand for the family of all objects x_i where i is
an element of M. Let Sum(i in I: x_i) be defined as the sum of all
numbers x_i for elements i of I. Let Sum(i = a..b: x_i) be defined
as the sum of all numbers x_i for all integers i such that
a <= i <= b. Let Prod(i in I: x_i) be defined as the product of
all numbers x_i for elements i of I.

An integer x is said to be divisible by an integer y if there exists
an integer z such that x = yz. If x is divisible by y, we say also
that y divides x, and we call y a divisor or factor of x, and x a
multiple of y. An integer x is said to be even if 2 divides x, and
x is said to be odd if x is not even. An integer p is said to be
prime if p > 1 and for every positive divisor d of p, either d = 1
or d = p.

For any positive integer n, we define sigma(n) to be the sum of all
positive divisors of n. A positive integer n is said to be a perfect
number if sigma(n) = 2n.

We phrase a fact of Harold Davenport as Lemma L1 (see [1: p. 14]).

Lemma L1. Let M be a finite set of primes, let (e_p: p in M) be a
family of positive integers, and let n = Prod(p in M: p^(e_p)).
Then

sigma(n) = Prod(p in M: Sum(i = 0..e_p: p^i)).

Without proof.


3. Main Part

First we make the claimed equivalence of the two problems precise.
For this purpose, we formulate two statements A and B whose truth
values are currently unknown.

Statement A. There exist some finite set M of odd primes and
some family (e_p: p in M) of integers larger than 1 such that

2 = Prod(p in M: [1 - (1/p)^(e_p)] / [1 - 1/p]).

Statement B. There exists some odd perfect number.

To demonstrate the equivalence of Challenge Problem 1 to the problem
whether there exists an odd perfect number, we will prove that the
two statements A and B are equivalent.

Theorem T1. Statement A is equivalent to Statement B.

Proof. A -> B. Suppose Statement A is true. Then we can choose
some set M of odd primes and some family (e_p: p in M) of integers
larger than 1 such that

(1) 2 = Prod(p in M: [1 - (1/p)^(e_p)] / [1 - 1/p]).

For every element p of M, let m_p = e_p - 1. Let p be an arbitrary
element of M. Since e_p is an integer larger than 1, it follows
that m_p is a positive integer and e_p = m_p + 1. Then

[1 - (1/p)^(e_p)] / [1 - 1/p]

=

[1 - (1/p)^(m_p + 1)] / [1 - 1/p]

= (geometric sum formula)

Sum(i = 0..m_p: (1/p)^i)

=

Sum(i = 0..m_p: p^(m_p - i)) / p^(m_p)

= (substitute j for m_p - i)

Sum(j = 0..m_p: p^j) / p^(m_p)

= (rename index variable j into variable i)

Sum(i = 0..m_p: p^i) / p^(m_p).

Since p was arbitrary, we can conclude that for every element p of
M, equation (2) is true:

(2) [1 - (1/p)^(e_p)] / [1 - 1/p]

= Sum(i = 0..m_p: p^i) / p^(m_p).

Then we can infer

2

= (equation (1))

Prod(p in M: [1 - (1/p)^(e_p)] / [1 - 1/p])

= (equation (2))

Prod(p in M: Sum(i = 0..m_p: p^i) / p^(m_p))

=

Prod(p in M: Sum(i = 0..m_p: p^i)) / Prod(p in M: p^(m_p)).

Thus, we can conclude that

(3) 2 Prod(p in M: p^(m_p))

= Prod(p in M: Sum(i = 0..m_p: p^i)).

Let

(4) n = Prod(p in M: p^(m_p)).

Then since for every element p of M, both p is an odd positive
integer and m_p is an positive integer, it follows that n is
an odd positive integer. Whence

2n

= (definition (4))

2 Prod(p in M: p^(m_p))

= (equation (3))

Prod(p in M: Sum(i = 0..m_p: p_i))

= (Lemma L1 with definition (4))

sigma(n).

Thus, since n is an odd positive integer, we can conclude that n is
an odd perfect number, so Statement B is true.

B -> A. Suppose Statement B is true. Then we can choose some odd
perfect number n. By the fundamental theorem of arithmetic (see
[1: pp. 9-11]), we can let M be the finite set of primes and
(m_p: p in M) be the family of positive integers such that

(5) n = Prod(p in M: p^(m_p)).

Since n is odd and every exponent m_p is a positive integer for
elements p in M, every element p of M must also be odd. For every
element p of M, let e_p = m_p + 1. Then M is a finite set of odd
primes and (e_p: p in M) is a family of integers larger than 1.
Whence

2 Prod(p in M: p^(m_p))

= (equation (5))

2n

= (since n is a perfect number)

sigma(n)

= (Lemma L1 with equation (5))

Prod(p in M: Sum(i = 0..m_p: p^i)).

Thus, we can infer that

(6) 2 = Prod(p in M: Sum(i = 0..m_p: p^i)) / Prod(p in M: p^(m_p)).

Then

2

= (equation (6))

Prod(p in M: Sum(i = 0..m_p: p^i)) / Prod(p in M: p^(m_p)).

=

Prod(p in M: Sum(i = 0..m_p: p^i) / p^(m_p)).

Thus, it follows that

(7) 2 = Prod(p in M: Sum(i = 0..m_p: p^i) / p^(m_p)).

Let p be an arbitrary element of M. Then

Sum(i = 0..m_p: p^i) / p^(m_p)

=

Sum(i = 0..m_p: 1/(p^(m_p - i)))

= (substitute j for m_p - i)

Sum(j = 0..m_p: 1/(p^j))

= (rename the index variable j into the variable i)

Sum(i = 0..m_p: 1/(p^i))

=

Sum(i = 0..m_p: (1/p)^i)

= (geometric sum formula)

[1 - (1/p)^(m_p + 1)] / [1 - 1/p]

=

[1 - (1/p)^(e_p)] / [1 - 1/p].

Since p was arbitrary, we can conclude that for every element p of
M, equation (8) is true:

(8) Sum(i = 0..m_p: p^i) / p^(m_p)

= [1 - (1/p)^(e_p)] / [1 - 1/p].

Then we can infer that

2

= (equation (7))

Prod(p in M: Sum(i = 0..m_p: p^i) / p^(m_p))

= (equation (8))

Prod(p in M: [1 - (1/p)^(e_p)] / [1 - 1/p]).

Thus, since M is a finite set of odd primes and (e_p: p in M) is a
family of integers larger than 1, it follows that Statement A is
true.

QED


4. Conclusion

With Theorem T1 we have demonstrated that Challenge Problem 1 is
equivalent to the open problem whether there exists an odd perfect
number. This open problem is known as notoriously difficult.
Therefore, we have good reasons to anticipate that it is really
difficult to solve Challenge Problem 1 in a positive or negative
way.


5. References

[1]: Davenport, Harold: The Higher Arithmetic, 8th ed.; Cambridge
University Press; (c) the estate of Harold Davenport 2008;
ISBN 978-0-521-72236-0 (pbk.).


--
Steffen Schuler <schuler....@gmail.com>