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Hi all.

I've had this idea - a conjecture in the field of Number Theory - for a few years now. The conjecture is rather simple, as were the logical steps that I made in order to infer it, so I would have assumed that it had already been suggested in the past. Nevertheless, I have not been able to find any piece of evidence that it had (possibly due to the difficulty of "phrasing" it into Google's search engine).

I would appreciate your opinion of the following:

  • Are you familiar with this theorem in any way?
  • Is it eligible to be stated as an open conjecture in Number Theory?

My conjecture can be stated in any of the following ways:

  1. No set S ⊂ { (3n+2)/(2n+1) │ n∈N } exists such that ∏Si is a power of 2
  2. No set S ⊂ { (3n+2)/(2n+1) │ n∈N } exists such that ∏Si is integer
  3. No multi-set S ⊂ { (3n+2)/(2n+1) │ n∈N } exists such that ∏Si is integer

Reminder:

  • In a set, no element can appear more than once
  • In a multi-set, any element may appear more than once

In simple words: Take any group of numbers from the series {5/3, 8/5, 11/7, 14/9, 17/11, 20/13, ...}. Calculate the product (multiplication) of these numbers - the result will never be an integer number.

Note:

In its weakest form (#1), my conjecture is sufficient for proving that there are no cyclic sequences in the '3n+1' conjecture (the proof for that is pretty simple, but I am not including it here because it is not the main purpose of my question). I believe that my conjecture also holds in its strongest form (#3).

Thank you very much for your time.

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  • $\begingroup$ There is a trivial counterexample to your conjectures; you should rephrase to remove it. $\endgroup$
    – Charles
    Commented Oct 22, 2012 at 14:27
  • $\begingroup$ Charles, you mean the empty set ?? $\endgroup$
    – Joël
    Commented Oct 22, 2012 at 14:44
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    $\begingroup$ Just a remark : going far enough, these rational numbers will not be multiplicatively independent, because they are supported on primes $\leq 3n+2$, and there are not enough such primes. $\endgroup$
    – François Brunault
    Commented Oct 22, 2012 at 14:49

1 Answer 1

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Even conjecture 1. is false: For $S=\{(3n+2)/(2n+1)\;|\;n=9, 12, 14, 27, 41\}$ your product is $8$.

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  • $\begingroup$ Nice! Computer search? $\endgroup$
    – Daniel Litt
    Commented Oct 22, 2012 at 17:14
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    $\begingroup$ @Daniel: Yes, prime factorizations of the terms $(3i+2)/(2i+1)$ yields a system of linear equations in non-negative integral unknowns, which the integer linear program tools of Sage (www.sagemath.org) can solve. $\endgroup$
    – Peter Mueller
    Commented Oct 22, 2012 at 17:18

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