2023 AIME I Problems/Problem 9

Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are integers in ${−20, −10, −18, . . . , 18, 19, 20}$ (Error compiling LaTeX. Unknown error_msg), such that there is a unique integer $m \neq 2$ with $p(m) = p(2)$ .

Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are integers in ${-20, -10, -18, . . . , 18, 19, 20}$ , such that there is a unique integer $m \neq 2$ with $p(m) = p(2)$ .

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.

Because $c$ can take any value from $\left\{ -20, -19, -18, \cdots , 18, 19, 20 \right\}$ , the number of feasible $c$ is 41.

Therefore, the number of $\left( a, b, c \right)$ is $8 \cdot 41 = \boxed{\textbf{(328) }}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution

It can be easily noticed that $c$ is independent of the condition $P(m) = P(2)$ , and can thus safely take all $41$ possible values between $-20$ and $20$ .

There are two possible ways for $m\ne2$ to be the only integer satisfying $P(m) = P(2)$ : $P$ has a double root at $2$ or a double root at $m$ .

Case 1: $P$ has a double root at $2$ :

In this case, $\frac{dP}{dx}(2) = 0$ , or $12 + 4a + b = 0$ . Thus $a$ ranges from $-8$ to $2$ . One of these values, $(a,b) = (-6,-12)$ corresponds to a triple root at $2$ , which means $m=2$ . Thus there are $10$ possible values of $(a,b)$ . (It can be verified that $m$ is an integer).