VMO

a. We consider the following transformation $$\left(x^2-\frac{\sqrt{15}}{2}x+1\right)\left(x^2+\frac{\sqrt{15}}{2}x+1\right)=x^4-\frac{7}{4}{x}^2+1,$$ $$\left(x^4-\frac{7}{4}{x}^2+1\right)\left(x^4+\frac{7}{4}{x}^2+1\right)=x^8-\frac{17}{16}{x}^4+1,$$ $$\left(x^8-\frac{17}{16}{x}^4+1\right)\left(x^8+\frac{17}{16}{x}^4+1\right)=x^{16}+\frac{223}{256}{x}^8+1.$$ It follows that $f(x)$ is the quotient of $x^{16}+\frac{223}{256}{x}^8+1$ and $$\left(x^2+\frac{\sqrt{15}}{2}x+1\right)\left(x^4+\frac{7}{4}{x}^2+1\right)\left(x^8+\frac{17}{16}{x}^4+1\right).$$

b. Suppose $\frac{P(x)}{Q(x)}=x^2-\alpha x+1$ where $P,Q$ are polynomials with non-negative coefficients. Substituting $x=1$, we have $$2-\alpha =\frac{P(1)}{Q(1)}>0\text{ so }\alpha <2.$$ We will prove that every real number $\alpha <2$ satisfies the problem. Indeed, if $\alpha \le 0$ then the polynomial $f(x)$ itself is satisfied, so we can choose $P(x)=f(x)$, $Q(x)=1$. If $\alpha \in (0;2)$, let us consider the multiplication $$(x^2-\alpha x+1)(x^2+\alpha x+1)=x^4+(2-{\alpha }^2)x^2+1.$$ Continuing like that, we find that the coefficients of the first and last terms of the polynomial are always 1, and the middle coefficient is determined by the sequence $(u_n)$ as follows $$\{\begin{array}{l}{u}_0=\alpha ,\\ u_{n+1}=2-u_n^2,n\ge 0.\end{array}$$ We will prove that there exists a positive term in this sequence. Suppose that for every $n\ge 1$, $u_n<0$. Then, since $\alpha \in (0;2)$ so by induction, we can show that $-2<u_n<0,\forall n\ge 1$. Note that $$u_{n+1}-u_n=2-u_n-u_n^2=(2+u_n)(1-u_n)>0,$$ so $u_{n+1}-u_n>0,\forall n\ge 1$; shows that this sequence increases. Since the sequence is bounded by 0 so it has a limit $L\in (-2;0]$. By letting $n$ tend to infinity, we have $$L=2-L^2\text{ so }L\in \{1;-2\}.$$ This contradiction shows that there exists $n=N$ so that $u_N\ge 0$. Consider the polynomials sequence $$f_n(x)=x^{2n+1}+u_n{x}^{2n}+1$$ with $n=1,2,3,\dots ,N$ it is easy to see that $f(x)$ is the quotient of two polynomials $f_N(x)$ and $f_1(x)f_2(x)\dots f_{N-1}(x)$. Clearly, these polynomials have non-negative coefficients. $\square$