Nice! Let $2^{a_i} - 3$, with $0\leq i \leq n$, numbers with every two relative primes. to find our '$a_{n+1}$' the idea is see that iff $2^k - 1$ is multiple of all of this numbers, $2^k - 3$ is relative prime with all of this! to find this k, is very easy

Peter wrote: Prove that the set of integers of the form $ 2^{k} - 3$ ($ k = 2,3,\cdots$) contains an infinite subset in which every two members are relatively prime. We assume to the contrary that all such subsets are finite. Now we take a subset of maximal size. Let it be, $ A = \{ 2^{a_1} - 3,2^{a_2} - 3,\cdots 2^{a_m} - 3 \}$ Now, we shall prove that, there exists a $ m > a_m$ such that $ (2^m - 3,2^{a_l} - 3) = 1$; $ 1\leq l\leq m$. We take $ q = \prod ( 2^{a_l} - 3 )$. We have that among, $ 2^0 - 3,\cdots, 2^{q} - 3$, two of the integers will give same residue, when divided by $ q$. That is $ q | 2^i - 2^j$; where $ 0\leq j\leq i\leq q$. i.e. $ q | 2^{(i - j)} - 1$. As $ 2^{(i - j)} - 1$ is odd, $ 2^{(i - j)} - 3$ is relatively prime to $ 2^{(i - j)} - 1$, and apparently so to the elements of $ A$. Also, $ 2^{a_l} - 3 < q\leq 2^{(i - j)} - 1$. (That is we may take $ m = (i - j)$) Which contradicts our assumption, so $ A$ is infinite.

use euler lemmar add induction

Moonmathpi496 wrote: Peter wrote: Prove that the set of integers of the form $ 2^{k} - 3$ ($ k = 2,3,\cdots$) contains an infinite subset in which every two members are relatively prime. We assume to the contrary that all such subsets are finite. Now we take a subset of maximal size. Let it be, $ A = \{ 2^{a_1} - 3,2^{a_2} - 3,\cdots 2^{a_m} - 3 \}$ Now, we shall prove that, there exists a $ m > a_m$ such that $ (2^m - 3,2^{a_l} - 3) = 1$; $ 1\leq l\leq m$. We take $ q = \prod ( 2^{a_l} - 3 )$. We have that among, $ 2^0 - 3,\cdots, 2^{q} - 3$, two of the integers will give same residue, when divided by $ q$. That is $ q | 2^i - 2^j$; where $ 0\leq j\leq i\leq q$. i.e. $ q | 2^{(i - j)} - 1$. As $ 2^{(i - j)} - 1$ is odd, $ 2^{(i - j)} - 3$ is relatively prime to $ 2^{(i - j)} - 1$, and apparently so to the elements of $ A$. Also, $ 2^{a_l} - 3 < q\leq 2^{(i - j)} - 1$. (That is we may take $ m = (i - j)$) Which contradicts our assumption, so $ A$ is infinite. Can we not just choose $m=t\phi(q)$ where $t$ is sufficiently large such that $m>q$?