2023 AIME I Problems/Problem 9

Problem (Unofficial, please update when official one comes out):

$P(x) = x^3 + ax^2 + bx + c$ is a polynomial with integer coefficients in the range $[-20, -19, -18\cdots 18, 19, 20]$ , inclusive. There is exactly one integer $m \neq 2$ such that $P(m) = P(2)$ . How many possible values are there for the ordered triple $(a, b, c)$ ?

Solution

Solution 1

Plugging $2$ into $P(x)$ , we get $8+4a+2b+c = m^3+am^2+bm+c$ . We can rewrite into $(2-m)(m^2+2m+4+a(2+m)+b)=0$ , where $c$ can be any value in the range. Since $m\neq2, m^2+2m+4+a(2+m)+b$ must be $0$ . The problem also asks for unique integers, meaning $m$ can only be one value for each polynomial, so the discriminant must be $0$ . $m^2+2m+4+a(2+m)+b = m^2+m(2+a)+(2a+b+4)= 0$ , and $(2+a)^2-4(2a+b+4)=0$ . Rewrite to be $a(a-4)=4(b+3)$ . $a$ must be even for $4(b+3)$ to be an integer. $-6<=a<=10$ because $4(20+3) = 92$ . However, plugging in $a=-6, b=12$ result in $m=2$ . There are 8 pairs of $(a,b)$ and 41 integers for $c$ , giving $41\cdot8 = \boxed{328}$

~chem1kall

Solution

Define $q \left( x \right) = p \left( x \right) - p \left( 2 \right)$ . Hence, for $q \left( x \right)$ , beyond having a root 2, it has a unique integer root that is not equal to 2.

We have $\begin{align*} q \left( x \right) & = p \left( x \right) - p \left( 2 \right) \\ & = \left( x - 2 \right) \left( x^2 + \left( 2 + a \right) x + 4 + 2a + b \right) . \end{align*}$

Thus, the polynomial $x^2 + \left( 2 + a \right) x + 4 + 2a + b$ has a unique integer root and it is not equal to 2.

Following from Vieta' formula, the sum of two roots of this polynomial is $- 2 - a$ . Because $a$ is an integer, if a root is an integer, the other root is also an integer. Therefore, the only way to have a unique integer root is that the determinant of this polynomial is 0. Thus, $\left( 2 + a \right)^2 = 4 \left( 4 + 2a + b \right) . \hspace{1cm} (1)$

In addition, because two identical roots are not 2, we have $2 + a \neq - 4 .$

Equation (1) can be reorganized as $4 b = \left( a - 2 \right)^2 - 16 . \hspace{1cm} (2)$

Thus, $2 | a$ . Denote $d = \frac{a-2}{2}$ . Thus, (2) can be written as $b = d^2 - 4 . \hspace{1cm} (3)$

Because $a \in \left\{ -20, -19, -18, \cdots , 18, 19, 20 \right\}$ , $2 | a$ , and $2 + a \neq -4$ , we have $d \in \left\{ - 11, - 10, \cdots, 9 \right\} \backslash \left\{ 4 \right\}$ .

Therefore, we have the following feasible solutions for $\left( b, d \right)$ : $\left( -4 , 0 \right)$ , $\left( -3 , \pm 1 \right)$ , $\left( 0 , \pm 2 \right)$ , $\left( 5, \pm 3 \right)$ , $\left( 12 , 4 \right)$ . Thus, the total number of $\left( b, d \right)$ is 8.