Hellenic Mathematical Olympiad ARCHIMEDES

(a) The sum of the coefficients of the polynomial $P(x)$ is equal to $0$. It means that $1$ is one of its roots and so $$P(x)=a{x}^3+(b-a)x^2-(c+b)x+c=(x-1)(a{x}^2+bx-c)$$ If $x_0=1$, from Vieta's formulas we have: $$x_1+x_2=-\frac{b}{a}\text{and}{x}_1{x}_2=-\frac{c}{a}\ne 0,(1)$$ Moreover, from the hypothesis $x_0,x_1,x_2$ are roots of the polynomial $$\begin{array}{rl}F(x)& =Q(x)-P(x)=x^4+(b-a-1)x^3+2(a-b)x^2+(b-a)x\\ & =x(x^3+(b-a-1)x^2+2(a-b)x+b-a)\\ & =x[x^3-x^2+(b-a)(x^2-x)+(a-b)(x-1)]\\ & =x(x-1)[x^2+(b-a)x+(a-b)].\end{array}$$ Since $F(x)=0\iff x=0$ or $x=1$ or $x^2+(b-a)x+(a-b)=0$ and $x_0,x_1,x_2\ne 0$, we conclude that: $$x_0=1,x_1+x_2=a-b,x_1{x}_2=a-b.(2)$$ From (1) and (2) we obtain: $a-b=-\frac{b}{a}=-\frac{c}{a}\Rightarrow$ $$b=c(3)\text{or}{a}^2-ab=-b(4)$$ From (3) and (4) we have: $$a^2=b(a-1)\Rightarrow a>1\text{ (since b>0) and }b=c=\frac{a^2}{a-1}.$$ We have that $$\begin{array}{rl}abc& =a{\left(\frac{a^2}{a-1}\right)}^2=\frac{a^5}{(a-1{)}^2}\stackrel{x=a-1\ge 0}{=}\frac{(x+1{)}^5}{x^2}=\frac{x^5+5x^4+10x^3+10x^2+5x+1}{x^2}\\ & \Rightarrow abc=x^3+\left(5x^2+\frac{1}{x^2}\right)+\left(10x+\frac{5}{x}\right)+10.\end{array}(5)$$ Now we observe that: $x^3>0$, $5x^2+\frac{1}{x^2}>4\iff 5x^4-4x^2+1>0\iff x^4+(2x^2-1{)}^2>0$, valid, $10x+\frac{5}{x}>14\stackrel{x\ge 0}{\iff }10x^2-14x+5>0$, is valid.

Finally, from (5) we get that: $abc>28$.

(b) As in the first question from relation $b=a+1+\frac{1}{a-1}$, since $b,a+1$ are integers, we obtain that $\frac{1}{a-1}\in \mathbb{Z}$ Therefore: $$a-1=\pm 1\iff a=0\text{ (it is rejected) or }a=2.\text{ Then }b=c=\frac{a^2}{a-1}=4.\text{ For these values the polynomial}$$ $$P(x)=a{x}^3+(b-a)x^2-(c+b)x+c=2x^3+2x^2-8x+4=2(x-1)(x^2+2x-2)$$ has roots $x_0=1$, $x_{1,2}=-1\pm \sqrt{3}$ which are also roots of the polynomial $$F(x)=Q(x)-P(x)=x(x-1)(x^2+(b-a)x+(a-b))=x(x^2+2x-2),$$ and therefore they are roots of the polynomial $Q(x)=F(x)+P(x)$.