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

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SUMMARY

The discussion focuses on the divisibility of the expression 2^(12n+9)+1 by the linear polynomial (24n+19). Participants confirmed that for specific values of n, such as 0, 1, 2, 5, 6, and 8, the expression is divisible, while for others like 3, 4, 7, 9, and 10, it is not. The introduction of the variable N=4n+3 and the analysis of prime factors of the form 4k+3 are critical to understanding the divisibility pattern. A counterexample was identified at n=38, where the divisibility does not hold, prompting further investigation into the conditions under which the pattern applies.

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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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