, 2015

Assume, for the sake of contradiction, that there exists an integer $k \geq 2$ such that $a_{k}$ and $a_{1}$ are relatively prime and that $a_{1}$ does not divide $a_{0}$. We deduce that there exists a prime number $p$ such that $0 \leq v_{p}\left(a_{0}\right)<v_{p}\left(a_{1}\right)$, where $v_{p}(a)$ is the $p$-adic valuation of the integer $a$, that is the greatest integer $m$ such that $p^{m}$ divides $a$.

Assume that there exists an integer $m \geq 0$, for which we have $v_{p}\left(a_{m}\right)< v_{p}\left(a_{m+1}\right)$. Because $a_{m+1}^{2}$ divides $a_{m} a_{m+2}$, we have $2 v_{p}\left(a_{m+1}\right) \leq v_{p}\left(a_{m}\right)+ v_{p}\left(a_{m+2}\right)$, and therefore $v_{p}\left(a_{m+1}\right)<v_{p}\left(a_{m+2}\right)$. This proves by induction that the sequence $v_{p}\left(a_{n}\right)$ is increasing, and therefore $0<v_{p}\left(a_{1}\right)<v_{p}\left(a_{k}\right)$. This implies that $p$ divides $\operatorname{gcd}\left(a_{1}, a_{k}\right)=1$, which is a contradiction.