Let $q>3$ be a rational number, such that $q^2-4$ is a perfect square of a rational number. The sequence $a_0, a_1, \ldots$ is defined by $a_0=2, a_1=q, a_{i+1}=qa_i-a_{i-1}$ for all $i \geq 1$. Is it true that there exist a positive integer $n$ and nonzero integers $b_0, b_1, \ldots, b_n$ with sum zero, such that if $\sum_{i=0}^{n} a_ib_i=\frac{A} {B}$ for $(A, B)=1$, then $A$ is squarefree?