Find a solution of the polynomial Given $1<\beta<2$ and positive integer $n\geq2013$, then can we find a non-zero vector $(a_n\,,a_{n-1}\cdots\,a_1\,a_0 )$ where all $a_i\in\{-1\,,0\,,1\}$, such that $\sum\limits_{k=0}^{n-1}a_k\beta^k=0$ 
 A: If $\beta\in\mathbb{Q}$ then doesn't exists.
Indeed, suppose $(a_n,\cdots,a_0)\ne0$ such that
$$
a_n\beta^n+\cdots+a_1\beta+a_0=0
$$
Let $m=\max\{i:a_i\ne0\}$. Then, we have
$$
a_m\beta^m+\cdots+a_0=0
$$
and we can suppose $a_0\ne0$, otherwise, if $r=\min\{i:a_i\ne0\}$, then $a_m\beta^m+\cdots+a_0=a_m\beta^m+\cdots+a_r\beta^r=\beta^r(a_m\beta^{m-r}\cdots+a_r)=0$, and since $\beta^r\ne0$, then $a_m\beta^{m-r}+\cdots+a_r=0$.
Multiplying by $-1$, if necessary, we can suppose $a_m=1$. Now, writing $\beta=\frac ab$, with $a,b\in\mathbb{Z}$ and $\gcd(a,b)=1$, and multiplying the equation by $b^m$, we obtain
$$
a^m+ba_{m-1}\left(\frac ab\right)^{m-1} + \cdots + b^{m}a_m=0
$$
letting $b$ in evidence, we obtain
$$
a^m=bq
$$
hence, since $\gcd(a,b)=1$, we have that $b$ divides $a$, but this implies $\beta=\frac ab\in\mathbb{Z}$, contradiction.
A: Even if $\beta \in \mathbb{Q}$, there are infinitely many rational numbers $\in (1, 2)$, and set of all $\beta$s are finite (there are $3^{n+1}$ such polynomials and maximum # of solutions is much less than $(n-1)3^n$).
