IMC 2005 · Day 1 · P4

Note that $P(x)$ does not have any positive root because $P(x) > 0$ for every $x > 0$. Thus, we can represent them in the form $-\alpha_i$, $i = 1, 2, \dots, n$, where $\alpha_i \ge 0$. If $a_0 \ne 0$ then there is a $k \in \mathbb{N}$, $1 \le k \le n-1$, with $a_k = 0$, so using Viete's formulae we get $$\alpha_1 \alpha_2 \dots \alpha_{n-k-1} \alpha_{n-k} + \alpha_1 \alpha_2 \dots \alpha_{n-k-1} \alpha_{n-k+1} + \dots + \alpha_{k+1} \alpha_{k+2} \dots \alpha_{n-1} \alpha_n = \frac{a_k}{a_n} = 0,$$ which is impossible because the left side of the equality is positive. Therefore $a_0 = 0$ and one of the roots of the polynomial, say $\alpha_n$, must be equal to zero. Consider the polynomial $Q(x) = a_n x^{n-1} + a_{n-1} x^{n-2} + \dots + a_1$. It has zeros $-\alpha_i$, $i = 1, 2, \dots, n-1$. Again, Viete's formulae, for $n \ge 3$, yield: $$\tag{1} \alpha_1 \alpha_2 \dots \alpha_{n-1} = \frac{a_1}{a_n}$$ $$\tag{2} \alpha_1 \alpha_2 \dots \alpha_{n-2} + \alpha_1 \alpha_2 \dots \alpha_{n-3} \alpha_{n-1} + \dots + \alpha_2 \alpha_3 \dots \alpha_{n-1} = \frac{a_2}{a_n}$$ $$\tag{3} \alpha_1 + \alpha_2 + \dots + \alpha_{n-1} = \frac{a_{n-1}}{a_n}.$$ Dividing (2) by (1) we get $$\tag{4} \frac{1}{\alpha_1} + \frac{1}{\alpha_2} + \dots + \frac{1}{\alpha_{n-1}} = \frac{a_2}{a_1}.$$ From (3) and (4), applying the AM-HM inequality we obtain $$\frac{a_{n-1}}{(n-1) a_n} = \frac{\alpha_1 + \alpha_2 + \dots + \alpha_{n-1}}{n-1} \ge \frac{n-1}{\frac{1}{\alpha_1} + \frac{1}{\alpha_2} + \dots + \frac{1}{\alpha_{n-1}}} = \frac{(n-1) a_1}{a_2},$$ therefore $\frac{a_2 a_{n-1}}{a_1 a_n} \ge (n-1)^2$. Hence $n^2 \ge \frac{a_2 a_{n-1}}{a_1 a_n} \ge (n-1)^2$, implying $n \le 3$. So, the only polynomials possibly satisfying (i) and (ii) are those of degree at most three. These polynomials can easily be found and they are $P(x) = x$, $P(x) = x^2 + 2x$, $P(x) = 2x^2 + x$, $P(x) = x^3 + 3x^2 + 2x$ and $P(x) = 2x^3 + 3x^2 + x$.

Consider the prime factorization of $P$ in the ring $\mathbb{Z}[x]$. Since all roots of $P$ are rational, $P$ can be written as a product of $n$ linear polynomials with rational coefficients. Therefore, all prime factor of $P$ are linear and $P$ can be written as $$P(x) = \prod_{k=1}^{n} (b_k x + c_k)$$ where the coefficients $b_k, c_k$ are integers. Since the leading coefficient of $P$ is positive, we can assume $b_k > 0$ for all $k$. The coefficients of $P$ are nonnegative, so $P$ cannot have a positive root. This implies $c_k \ge 0$. It is not possible that $c_k = 0$ for two different values of $k$, because it would imply $a_0 = a_1 = 0$. So $c_k > 0$ in at least $n-1$ cases.

Now substitute $x = 1$. $$P(1) = a_n + \dots + a_0 = 0 + 1 + \dots + n = \frac{n(n+1)}{2} = \prod_{k=1}^{n} (b_k + c_k) \ge 2^{n-1};$$ therefore it is necessary that $2^{n-1} \le \frac{n(n+1)}{2}$, therefore $n \le 4$. Moreover, the number $\frac{n(n+1)}{2}$ can be written as a product of $n-1$ integers greater than 1.

If $n = 1$, the only solution is $P(x) = 1x + 0$.

If $n = 2$, we have $P(1) = 3 = 1 \cdot 3$, so one factor must be $x$, the other one is $x + 2$ or $2x + 1$. Both $x(x+2) = 1x^2 + 2x + 0$ and $x(2x+1) = 2x^2 + 1x + 0$ are solutions.

If $n = 3$, then $P(1) = 6 = 1 \cdot 2 \cdot 3$, so one factor must be $x$, another one is $x + 1$, the third one is again $x + 2$ or $2x + 1$. The two polynomials are $x(x+1)(x+2) = 1x^3 + 3x^2 + 2x + 0$ and $x(x+1)(2x+1) = 2x^3 + 3x^2 + 1x + 0$, both have the proper set of coefficients.

In the case $n = 4$, there is no solution because $\frac{n(n+1)}{2} = 10$ cannot be written as a product of 3 integers greater than 1.

Altogether we found 5 solutions: $1x + 0$, $1x^2 + 2x + 0$, $2x^2 + 1x + 0$, $1x^3 + 3x^2 + 2x + 0$ and $2x^3 + 3x^2 + 1x + 0$.