48th Austrian Mathematical Olympiad

We look at this sequence modulo $5$. This is possible as long as $a_n\ne 0(\mathrm{mod}5)$ so that the next element is defined modulo $5$. If we start to compute the elements modulo $5$ we obtain $$a_0\equiv 1(\mathrm{mod}5),$$ $$a_1\equiv 1+2\equiv 3(\mathrm{mod}5),$$ $$a_2\equiv 3+2\cdot 3^{-1}\equiv 3+2\cdot 2\equiv 2(\mathrm{mod}5),$$ $$a_3\equiv 3(\mathrm{mod}5),$$ $$a_4\equiv 2(\mathrm{mod}5),$$ So we see that after $a_0$, the sequence just alternates between the values $2$ and $3$ modulo $5$. These are not quadratic residues modulo $5$. Since $a_0=2016$ is also not the square of a rational number, there is indeed no square of a rational number in this sequence.