Prove that for every positive integer $d > 1$ and $m$ the sequence $a_n=2^{2^n}+d$ contains two terms $a_k$ and $a_l$ ($k \neq l$) such that their greatest common divisor is greater than $m$. Proposed by T. Hakobyan
Problem
Source: Tuymaada 2018 Senior League/Problem 4, Junior League/Problem 4
Tags: greatest common divisor, Sequence, number theory
dangit
20.07.2018 20:57
This is also China TST 4/2017 problem 4 (day 2), but i don't know how to make a shortcut to it )
ThE-dArK-lOrD
20.07.2018 20:59
@above here
MatBoy-123
11.07.2021 06:09
By Kobayashi , the sequence has infinitely many primes dividing it, so choose a prime $p$ $>>>>$ $m$ and select $k \equiv l$ $mod$ $ord_{ord_{p} 2} 2$ , from here we can ensure that $a_k$ and $a_l$ have atleast $p$ as a common factor which is $>>>>$ $m$. and we Win !!
lvym
19.07.2021 11:40
MatBoy-123 wrote: By Kobayashi , the sequence has infinitely many primes dividing it, so choose a prime $p$ $>>>>$ $m$ and select $k \equiv l$ $mod$ $ord_{ord_{p} 2} 2$ , from here we can ensure that $a_k$ and $a_l$ have atleast $p$ as a common factor which is $>>>>$ $m$. and we Win !! I guess there's something wrong with your solution. Actually we need (p-1) isn't divisible by 2^(l+1), or we can't solve it by Euler's Theorem.
MatBoy-123
19.07.2021 12:01
lvym wrote: MatBoy-123 wrote: By Kobayashi , the sequence has infinitely many primes dividing it, so choose a prime $p$ $>>>>$ $m$ and select $k \equiv l$ $mod$ $ord_{ord_{p} 2} 2$ , from here we can ensure that $a_k$ and $a_l$ have atleast $p$ as a common factor which is $>>>>$ $m$. and we Win !! I guess there's something wrong with your solution. Actually we need (p-1) isn't divisible by 2^(l+1), or we can't solve it by Euler's Theorem. We have choosen a large prime $p$ in common which is much larger than $m$ , and then we have constructed $k$ and $l$ such that $ p |a_k, a_l$