Yes, consider the sequence $\{ a_n\}_{n\in \mathbb{Z}^+}$ defined by $a_1=1$ and $a_{i+1}=(3a_i+100)!+1$ for all $i\in \mathbb{Z}^+$. The proof is easy.

Can anyone write a full solution please? Thank you.

LGKOm1 wrote: Can anyone write a full solution please? Thank you. First and foremost, for these types of problems, it helps to consider factorials since they handle relatively primeness via the euclidean algorithm quite well. [There were a couple usamo problems of this sort, refer to: https://artofproblemsolving.com/community/c6h55343p343870 for an example] To deal with infiniteness, it suffices to generate the sequence starting from a base case recursively. For this problem specifically, if your i+1th term is of the form M!+1 where M encompasses all of the previous elements in your set, the conclusion follows quite rapidly from the euclidean algorithm.

In @above, There is a small problem that arises if $\max(m,n,k) = \max(i,j) = L$. That actually has a easy fix: in that case too, $a_L$ can be changed to $1$. For example, $$ \gcd(a_1 + a_4+a_5,a_2 + a_5) = \gcd(a_1 + a_4 - a_2,a_2 + a_5) = \gcd(a_1 + a_4 - a_2,a_2 + 1) = \gcd(a_1 + a_4 + 1,a_2 + 1) $$Remark: $2,3$ can be replaced by any positive integers $p,q$ satisfying $\gcd(p,q) = 1$. Conversely, if $\gcd(p,q) = d > 1$, then problem isn't even true as infinitely many numbers are same mod $d$, then $\gcd$ can be forced to divisible by $d$