Which Prime Numbers Satisfy These Divisibility Conditions?

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Discussion Overview

The discussion revolves around identifying prime numbers (p, q, r) that satisfy specific divisibility conditions related to the expressions pq + pr + rq and p³ + q³ + r³ - 2pqr, with a focus on their divisibility by p + q + r. The scope includes mathematical reasoning and exploration of the problem's origins.

Discussion Character

  • Exploratory, Mathematical reasoning

Main Points Raised

  • One participant seeks to find all prime numbers (p, q, r) that meet the given divisibility conditions.
  • Another participant notes that this problem appears frequently on various math forums, suggesting it may have a broader context or origin.
  • A third participant inquires about the specific forums where the problem has been discussed, indicating a desire to explore the topic further.
  • One participant suggests that the problem evokes thoughts of cubing numbers, hinting at a potential approach to the solution.
  • A later reply mentions having seen the problem on a specific math forum, indicating that it may be a known question in certain circles.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the origins of the problem or its specific solutions, and multiple viewpoints regarding its prevalence on different forums are expressed.

Contextual Notes

The discussion does not clarify the assumptions behind the divisibility conditions or the implications of the expressions involved, leaving these aspects unresolved.

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Find all prime numbers [tex](p,q,r)[/tex], that numbers [tex]pq+pr+rq[/tex] and [tex]p^3+q^3+r^3-2pqr[/tex] are divided by [tex]p+q+r[/tex]
 
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I'm seeing this problem asked a lot on the various math forums I frequent. Where does it come from?
 
Which forums (fora) are those, CR?

(See it this way: if you tell me I'll go pester somewhere else.) :P
 
Doesn't looking at the problem just make you instantly think of cubing things?
 
I think I saw the problem, at the least, at
http://www.mymathforum.com/
 

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