Bulgarian Winter Tournament

We will prove that such numbers do not exist.

The quadratic equation $x^2-qx+1=0$ has two rational roots $t$ and $\frac{1}{t}$ for which $t+\frac{1}{t}=q$. It easily follows by induction that $a_m=t^m+\frac{1}{t^m}$. Let's assume that there exist numbers $b_0,b_1,\dots ,b_n$ satisfying the condition of the problem.

Lemma. Let $f(x)=c_n{x}^n+c_{n-1}{x}^{n-1}+\cdots +c_{1}x+c_0$ be a polynomial with nonzero integer coefficients for which $c_{n-k}=c_k$ for each $k=0,1,\dots ,n$ and ${\sum }_{i=0}^n{c}_i=0$. Then $f(x)=(x-1{)}^{2}g(x)$, where $g(x)$ is a polynomial with integer coefficients.

Proof: From the condition we have that $f(1)=0$. As $$f^{\prime }(x)=n{c}_n{x}^{n-1}+(n-1)c_{n-1}{x}^{n-2}+\cdots +c_1,$$ it follows that $$2f^{\prime }(1)=(n{c}_n+(n-1)c_{n-1}+\cdots +c_1)+(n{c}_0+(n-1)c_1+\cdots +c_{n-1})=n(c_n+c_{n-1}+\cdots +c_1+c_0)=0$$ and therefore $x=1$ is a double root. The lemma is proved.

The polynomial $f(x)=b_n{x}^{2n}+b_{n-1}{x}^{2n-1}+\cdots +b_1{x}^{n+1}+2b_0{x}^n+b_1{x}^{n-1}+b_2{x}^{n-2}+\cdots +b_{n-1}x+b_n$ satisfies the conditions of the lemma. It's not hard to see that $$t^n\left(b_n\left(t^n+\frac{1}{t^n}\right)+b_{n-1}\left(t^{n-1}+\frac{1}{t^{n-1}}\right)+\cdots +2b_0\right)=f(t)=(t-1{)}^{2}g(t).$$ Moreover, if $t=\frac{r}{s}$, $(r,s)=1$, then from $\frac{r}{s}+\frac{s}{r}=q>3$ it easily follows that $r\ge s+2$, i.e. $r-s\ge 2$. Then $g\left(\frac{r}{s}\right)$ is a number of the form $\frac{l}{s^{2n-2}}$.

Finally $b_0{a}_0+b_1{a}_1+\cdots +b_n{a}_n$ is presented in the form $\frac{(r-s{)}^{2}l}{r^n{s}^n}$ and because the number $r-s\ge 2$ and $(r-s,r)=(r-s,s)=1$, then the numerator will always square a prime number. $\square$