Seventeenth ROMANIAN MASTER OF MATHEMATICS

The answer is in the negative. Notice first that, if $a_n>1$, then $2^{a_n}-1\equiv 3(\mathrm{mod}4)$; since $(2^{a_n}-1)a_{n+1}$ is a perfect square, we should have $a_{n+1}\equiv 0(\mathrm{mod}4)$ or $a_{n+1}\equiv 3(\mathrm{mod}4)$, so in particular $a_{n+1}>1$. As $a_1>1$, we conclude that all terms of the sequence are greater than 1.

Denote the largest prime divisor of an integer $k>1$ by $g(k)$. We will show that $g(a_{n+1})>g(a_n)$ for all $n$, which yields the desired result. To this end, usage is made of the lemma below.

Lemma: For any prime $p$, each prime divisor of $2^p-1$ is greater than $p$.

Proof. Let $q$ be a prime factor of $2^p-1$; then $q$ is odd. The multiplicative order $d$ of $2$ modulo $q$ divides $p$ and is larger than $1$, so $d=p$. On the other hand, by Fermat's little theorem, $2^{q-1}\equiv 1(\mathrm{mod}q)$, so $p=d\mid q-1$ and the lemma follows.

Choose now any positive integer $n$, and denote, for convenience, $k=a_n$ and $\ell =a_{n+1}$. Let $p=g(k)$; then $2^p-1\mid 2^k-1$. Since $2^p-1\equiv 3(\mathrm{mod}4)$, this number is not a square, so there exists a prime $q$ such that $v_q(2^p-1)$ is odd. By the Lemma, $q>p$, so in particular $q\nmid k$. Therefore, by the Lifting Exponent Lemma, $$v_q(2^k-1)=v_q(2^p-1)+v_q(k/p)=v_q(2^p-1)+0,$$ so $v_q(2^k-1)$ is odd as well. Since $(2^k-1)\ell$ is a perfect square, we should then have $q\mid \ell$, so $g(\ell )\ge q>p=g(k)$, as desired.