Infinite subset with pairwise comprime elements I'm looking for solution for this problem > Let $a$ be an integer greater then 1. Then, the set $A\mathrel{:=}\\{a^n(a+1)-1\mid n\in\mathbb{Z}_{+}\\}$ a has an infinite subset b such that it's elements are pairwise coprime. I found similar question here but proposed solutions are wrong: Author of solution claims that " Then $d$ is also a multiple of $m$ and using the well-known fact that $x−y$ divides $x^k−y^k$, we see that $a^m−1$ is a multiple of $a^d−1$ " Here we can find a small but critical mistake: from $m\mid d$ follows that $a^m-1 \mid a^d-1$ but no $a^d-1\mid a^m-1$. Also author states that if $m\mid n$ that $\gcd(X_m,X_n)=1$. It's also false because we can take $X_1=a(a+1)-1$ and it's prime factor $p$. But then $X_p=a^p(a+1)-1$ is also divided by $p$ because of Fermat's little theorem.

**Construction by induction** :

As a basis step note that any two consecutive terms in the sequence $X_n=a^{n+1}+a^n-1$ are pairwise coprime.

Now assume that we constructed a set $S$ of cardinal $n$ constituted by the terms of the sequence $X_n$, let $P$ be the product of all elements of $S$ so (Euler's theorem) : $$a^{\varphi(P)+1}+a^{\varphi(P)}-1\equiv a\mod P $$ thus $X_{\varphi(P)}$ is coprime with $P$ (coprime with all elements of $S$) so we can add $X_{\varphi(P)}$ to $S$ and construct a set of cardinal $n+1$.

The constructed sequence looks like: $$X_1,X_2,X_{\varphi(X_1X_2)},X_{\varphi(X_1X_2X_{\varphi(X_1X_2)})},\cdots \cdots$$