Possible factor(s) of 2^(12n+9)+1

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Discussion Overview

The discussion revolves around the factorization of the expression \(2^{12n+9}+1\) and its relationship with the polynomial \(24n+19\). Participants explore potential factors, counterexamples, and specific cases related to prime factors of certain forms.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that \(24n+19\) divides \(2(12n+9)+1\) for certain values of \(n\), providing specific examples where this holds true.
  • Another participant rewrites the problem using \(N=4n+3\) and expresses the original equation in terms of \(N\), indicating a potential reformulation of the problem.
  • A different participant notes that \(24n+19\) is of the form \(4k+3\) and discusses the implications for the number of prime factors, particularly those of the form \(4k+3\) and \(4k+1\).
  • One participant identifies \(n=38\) as a case where there are three prime factors of the form \(4k+3\) and presents a counterexample regarding divisibility.
  • Another participant reflects on their previous examples, noting that all instances where \(2(12n+9)+1\) divides \(2(12n+9)+1\) involved prime divisors, and expresses intent to find more counterexamples.
  • A later reply revisits the case of \(n=38\) and questions whether the observed pattern holds for distinct prime factors of the form \(P4k+3\).

Areas of Agreement / Disagreement

Participants express various viewpoints and hypotheses regarding the factorization and divisibility of the expressions, with no consensus reached on the validity of the patterns or the existence of counterexamples.

Contextual Notes

Some participants' claims depend on specific values of \(n\) and the nature of prime factors, which may not be universally applicable. The discussion includes unresolved mathematical steps and assumptions about the forms of prime factors.

Janosh89
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TL;DR
If the expression 2 x (12n+9) +1 has no prime factors of the form
4k+1 where k>0 , then 2^(12n+9)+1 is necessarily divisible by 24n+19.
Is this provable ?
It seems to hold true ..

nfactor(s) of (24n+19)(24n+19) divides 2(12n+9)+1 ?
019YES
143YES
267YES
37, 13NO
423, 5NO
5139YES
6163YES
711, 17NO
8211YES
947, 5 NO
107, 37NO
11283YES
 
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I have no idea on solution but rewrited the problem introducing
N=4n+3
24n+19= 6N+1
2^{12n+9}+1=8^{N}+1
 
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24n+19 is of the form 4k+3, so it must have an odd number of prime factors of the form 4k+3. In particular, if it does not have a prime factor of the form 4k+1 it must have an odd number of prime factors.

n=38 is the first time that we get three prime factors of the form 4k+3 as 2*(12*38+9) +1 = 72*19.
212*38+9+1 is not divisible by 2*(12*38+9)+1, so we have a counterexample.

Does this pattern only apply if 2*(12*n+9) +1 is a prime?
 
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I noticed that all my examples where 2*(12*n+9)+1 divides 2(12n+9)+1 were
prime divisors. I will endeavor to find further counterexamples. Thanks .
 
I should have noticed :-
for n=38

2(12*38+9)+1 is divisible by 7*19 . Does the pattern hold for the distinct
prime factors , P4k+3 , of 2*(12n +9)+1 ?
 

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