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
$$P(x) = a x^3 + (b - a) x^2 - (c + b) x + c = (x - 1)(a x^2 + b x - c)$$
If $x_0 = 1$, from Vieta's formulas we have:
$$x_1 + x_2 = -\frac{b}{a} \quad \text{and} \quad x_1 x_2 = -\frac{c}{a} \neq 0, \qquad (1)$$
Moreover, from the hypothesis $x_0, x_1, x_2$ are roots of the polynomial
$$\begin{aligned} F(x) &= Q(x) - P(x) = x^4 + (b - a - 1) x^3 + 2(a - b) x^2 + (b - a) x \\ &= x(x^3 + (b - a - 1) x^2 + 2(a - b) x + b - a) \\ &= x[x^3 - x^2 + (b - a)(x^2 - x) + (a - b)(x - 1)] \\ &= x(x - 1)[x^2 + (b - a)x + (a - b)]. \end{aligned}$$
Since $F(x) = 0 \Leftrightarrow x = 0$ or $x = 1$ or $x^2 + (b - a)x + (a - b) = 0$ and $x_0, x_1, x_2 \neq 0$, we conclude that:
$$x_0 = 1, \quad x_1 + x_2 = a - b, \quad x_1 x_2 = a - b. \qquad (2)$$
From (1) and (2) we obtain: $a - b = -\frac{b}{a} = -\frac{c}{a} \Rightarrow$
$$b = c \qquad (3) \quad \text{or} \quad a^2 - a b = -b \qquad (4)$$
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
$$\begin{aligned} abc &= a \left( \frac{a^2}{a - 1} \right)^2 = \frac{a^5}{(a - 1)^2} \stackrel{x = a - 1 \ge 0}{=} \frac{(x + 1)^5}{x^2} = \frac{x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1}{x^2} \\ &\Rightarrow abc = x^3 + \left( 5x^2 + \frac{1}{x^2} \right) + \left( 10x + \frac{5}{x} \right) + 10. \end{aligned} \qquad (5)$$
Now we observe that:
* $x^3 > 0$, $5x^2 + \frac{1}{x^2} > 4 \Leftrightarrow 5x^4 - 4x^2 + 1 > 0 \Leftrightarrow x^4 + (2x^2 - 1)^2 > 0$, valid,
* $10x + \frac{5}{x} > 14 \stackrel{x \ge 0}{\Leftrightarrow} 10x^2 - 14x + 5 > 0$, 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:
$$a - 1 = \pm 1 \Leftrightarrow a = 0 \text{ (it is rejected) or } a = 2. \text{ Then } b = c = \frac{a^2}{a - 1} = 4. \text{ For these values the polynomial}$$
$$P(x) = a x^3 + (b - a) x^2 - (c + b) x + c = 2x^3 + 2x^2 - 8x + 4 = 2(x - 1)(x^2 + 2x - 2)$$
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 $Q(x) = F(x) + P(x)$.