show that every odd prime divisor of $3^{(2^n)}-1$ is of form $4k+1$

Question

ganeshie8 · Answer

I have a really nice proof, but it uses many luxury theorems.
Looking for an elementary proof if it is available... :)

ganeshie8 · Answer

Haha! I see what you have discovered... 
It seems the sequence is spitting out a "new" prime in the prime factorization of each subsequent term...

ganeshie8 · Answer

It doesn't break at 7 :

2^3, 
2^4×5, 
2^5×5×41, 
2^6×5×17×41×193, 
2^7×5×17×41×193×21523361, 
2^8×5×17×41×193×21523361×926510094425921
2^9×5×17×41×193×21523361×926510094425921×1716841910146256242328924544641

parthkohli · Answer

Oh, never mind.

ganeshie8 · Answer

I have checked first 11 terms, each has a new prime

parthkohli · Answer

Yeah, makes sense. And that prime is of course, of the form, $4k+1$. 

I was actually thinking of proving $3^{2^n}+1$ has the same property as mentioned in the question, so that if we multiply this and $3^{2^n}-1$, we inductively have the next term.

ganeshie8 · Answer

Looks you're onto something !

ganeshie8 · Answer

Let $F_n = 3^{2^n}-1$

basically you're saying $F_{n+1} =(F_n+2)F_n $

kainui · Answer

I found this quite strange too, that it seems like we're proving that $3^{2^n} \pm 1$ all have odd prime divisors of the form 4k+1 which seems like a kind of big deal.

parthkohli · Answer

Exactly. I brought it into that form and all, but couldn't follow through.

ganeshie8 · Answer

Yeah it is amazing there are no online references about this
I'm not sure where @imqwerty found this problem...

imqwerty · Answer

(:  try this i think u guys can do it 
i'll post the solution soon

ganeshie8 · Answer

@ParthKohli that factorization also gives a trivial result :

$p\mid F_n \implies p\mid F_{n+1}$

that means $F_k\mid F_{n}$ for all $k\le n$

parthkohli · Answer

yeah but it's the $F_n + 2$ term that's troubling

parthkohli · Answer

@imqwerty were you able to solve this

kainui · Answer

I don't think what I'm about to say will be particularly important to this problem, however it seems interesting. I noticed a peculiar thing with exponents m > 2 (where have I heard this kind of thing before...?) that took me off track, this inequality here:

$$\sum_{k=0}^{m^n-1}a^k > \prod_{k=0}^{n-1} (a^{m^k}+1)$$

However it is an equality when m = 2 because since the right side will then have every permutation of sums of powers of 2 (whew) which in binary works out to be all the numbers less than 1's for n-1 times... You can't do this in base 3 or higher, since we need to multiply a power of 3 by 2 to get more than just the 0 and 1 digits if this makes sense.... lol.

imqwerty · Answer

yes pk :) my method is lil long 
m trying to shorten it
i think u guys can get it too

parthkohli · Answer

ok i'm going to go to a corner and cry myself to sleep