34th Hellenic Mathematical Olympiad

(i) Since $\xi =\frac{-1+\sqrt{17}}{2}$ is irrational and the polynomial $Fx=Px-2017$ has rational coefficients and $\xi$ as a root, then it will have also the conjugate $\frac{-1-\sqrt{17}}{2}$ as a root, and therefore it is divided by the polynomial $\phi x=x^2+x-4$. It comes easily from the identity $$Fx=Px-2017=x^2+x-4Qx+\kappa x+\lambda ,$$ by putting $x=\xi$. Then $\kappa \xi +\lambda =0$ which gives $\kappa =\lambda =0$, taking in mind that $\xi$ is irrational. Therefore there exists a polynomial $Qx$ such that: $$\begin{array}{rl}Fx=Px-2017& =x^2+x-4\mathcal{Q}x\\ \iff a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0-2017& =x^2+x-4\mathcal{Q}x\end{array}(1)$$ From (1) for $x=1$ we get: $$\begin{array}{rl}& a_0+a_1+\cdots +a_n-2017=-2Q1\\ \Rightarrow a_0+a_1+\cdots +a_n=2017-2Q1\equiv 1\text{mod }2& \end{array}$$

(ii) We consider the set $a_0,a_1,...,a_n$ with elements nonnegative integers satisfying the following: (α) $a_n{\xi }^n+a_{n-1}{\xi }^{n-1}+...+a_1\xi +a_0=2017$ and (β) the sum $a_0+a_1+...+a_n$ is minimal. First we observe that: $0\le a_i\le 3$, for all $i=1,2,...,n-2$. In fact, if it was not true for someone $i=1,2,...,n-2$, then the elements of the set $a_0,...,a_{i-1},a_{i-4},a_{i-1}+1,a_{i+1}+1,a_{i+2}+1,a_{i+3},...,a_n$ would be nonnegative integers, it would satisfy relation (α), while the sum of its elements would be less than of $a_1+a_2+...+a_n$, which is absurd. Let now $Qx=b_{n-2}{x}^{n-2}+b_{n-1}{x}^{n-1}+...+b_{1}x+b_0$. Then from the identity $a_n{x}^n+a_{n-1}{x}^{n-1}+...+a_{1}x+a_0-2017=x^2+x-4b_{n-2}{x}^{n-2}+b_{n-3}{x}^{n-3}+...+b_{1}x+b_0$ we get the equations: $$\left\{\begin{array}{l}{a}_0-2017=-4b_0\\ a_1=-4b_1+b_0\\ a_2=-4b_2+b_1+b_0\\ a_3=-4b_3+b_2+b_1\\ \text{multicolumn}5c\text{........................................}\\ a_{n-2}=-4b_{n-2}+b_{n-3}+b_{n-4}\\ a_{n-1}=b_{n-2}+b_{n-3}\\ a_n=b_{n-2}\end{array}\right\}\iff \left\{\begin{array}{l}{a}_0-2017=-4b_0\\ a_1-b_0=-4b_1\\ a_2-b_1-b_0=-4b_2\\ a_3-b_2-b_1=-4b_3\\ \text{multicolumn}5c\text{........................................}\\ a_{n-2}-b_{n-3}-b_{n-4}=-4b_{n-2}\\ a_{n-1}-b_{n-2}=b_{n-3}\\ a_n=b_{n-2}\end{array}\right\}$$ In general we have: $a_{i+2}-b_{i+1}-b_i=-4b_{i+2}$, for all $i=0,1,...,n-4$ Since $0\le a_i\le 3$, for all $i=1,2,...,n-2$, from the first equation we have $a_0=1$ and $b_0=504$. From the second equation we get $a_1=0$ and $b_1=126$. From the third equation we get $a_2=2$ and $b_2=157$. Continuing in the same way we find the sets $b_0,b_1,b_2,...b_{14}=504,126,157,70,56,31,21,13,8,5,3,2,1,0,0$ $a_0,a_1,a_2,...a_{14}=1,0,2,3,3,2,3,0,2,1,1,0,1,3,1$ Therefore the least possible value of the sum $a_0+a_1+...+a_n$ is 23.