Question 111
Given a rational number $q \gt 3$ such that $q^2 - 4$ is the square of a rational number. The sequence $\{a_i\}_{i=0}^{\infty}$ is defined as follows:
$$
a_0 = 2, \ a_1 = q, \ a_{i+1} = q a_i - a_{i-1}, \text{ for each } i = 1, 2, \dots
$$
Do there exist a natural number $n$ and nonzero integers $b_0, b_1, \dots, b_n$ such that $\sum_{i=0}^n b_i = 0$ and, if we write the number $b_0 a_0 + b_1 a_1 + \dots + b_n a_n$ in the form $\frac{A}{B}$, where $A$ and $B$ are coprime integers, the number $A$ is square-free?
$$
a_0 = 2, \ a_1 = q, \ a_{i+1} = q a_i - a_{i-1}, \text{ for each } i = 1, 2, \dots
$$
Do there exist a natural number $n$ and nonzero integers $b_0, b_1, \dots, b_n$ such that $\sum_{i=0}^n b_i = 0$ and, if we write the number $b_0 a_0 + b_1 a_1 + \dots + b_n a_n$ in the form $\frac{A}{B}$, where $A$ and $B$ are coprime integers, the number $A$ is square-free?