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 $\{ -20, -19, -18, \dots , 18, 19, 20 \}$ , such that there is a unique integer $m \neq 2$ with $p(m) = p(2).$

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).