Determine all solutions in positive integers a, b, and c to this equation.

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Discussion Overview

The discussion revolves around finding all solutions in positive integers \(a\), \(b\), and \(c\) to the equation \(\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 5\), with the condition that \(a < b < c\). The scope includes mathematical reasoning and exploration of potential solutions.

Discussion Character

  • Exploratory, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant lists several potential solutions, including \(1,2,4\), \(2,4,8\), and \(3,6,12\), suggesting a pattern of solutions of the form \(n, 2n, 4n\).
  • Another participant asserts that any multiple of a solution is also a solution, implying that the solutions can be reduced to a single fundamental solution.
  • A question is raised about the existence of solutions that are not multiples of \(1, 2, 4\), with a claim that none exist for \(c < 1000\).
  • Concerns are expressed regarding the clarity and completeness of responses, particularly about the need for thorough explanations and the presentation of partial answers.
  • A participant defends their response, indicating that they provided an answer to the original poster and questioning the sufficiency of the search limit.

Areas of Agreement / Disagreement

Participants express differing views on the completeness and clarity of the solutions presented. There is no consensus on whether additional solutions exist beyond the multiples of the identified patterns.

Contextual Notes

There are limitations regarding the search for solutions, particularly the upper bound of \(c < 1000\) and the implications of using multiples of identified solutions. The discussion includes unresolved issues about the sufficiency of the search method employed.

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Determine all solutions for \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \ = \ 5, where \ \ a, \ b, \ and \ \ c \ \ are \ \ positive \ \ integers, \ \ and \ \ a &lt;b &lt; c.
 
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1,2,4
2,4,8
3,6,12
...
n,2n,4n
 
Wilmer said:
1,2,4
2,4,8
3,6,12
...
n,2n,4n

It is quite obvious that any multiple of a solution will also be a solution, so in essence you have a single solution here. Are there any solutions that are not a multiple of 1,2,4?

CB
 
CaptainBlack said:
> It is quite obvious that any multiple of a solution will also be a solution,
> so in essence you have a single solution here.
Thank you, Sir.

> Are there any solutions that are not a multiple of 1,2,4?
None with c < 1000
...
 
Wilmer said:
...

There is something wrong with your quoting, you are attributing to me an answer rather than the question. Also if that was your answer to the question that I did ask you need to give some explanation, like exhaustive search up to some limit, ... Also partial answers should not be presented as if they are complete answers.

You might be a man of few words but there is a point at which brevity stops conveying meaning.

CB
 
I do not visit this site often.
I noticed in this case that there had been no answer to the OP's post in over a week;
(plus I notice now that the OP is banned.)
I simply put up a quick reply TO THE OP, not to you.

What are you complaining to me about exactly?
I answered "within" the quote; I did specify a search up to c < 1000;
are you saying that's not "exhaustive" enough?

Or are you complaining about the "..."?
If so, that was because of the "minimum of 3 characters" required by this site.

Anyway, you have the capabilities of banning me, so just do so if I've
sinned appropriately; fine with me.
 

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