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

[Janosh89]

TL;DR Summary

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 ..

n	factor(s) of (24n+19)	(24n+19) divides 2(12n+9)+1 ?
0	19	YES
1	43	YES
2	67	YES
3	7, 13	NO
4	23, 5	NO
5	139	YES
6	163	YES
7	11, 17	NO
8	211	YES
9	47, 5	NO
10	7, 37	NO
11	283	YES

 

[anuttarasammyak]

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

 

[mfb]

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.
2^12*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?

 

[Janosh89]

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 .

 

[Janosh89]

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 , P_4k+3 , of 2*(12n +9)+1 ?