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 between $-20$ and $20$ , inclusive. There is exactly one integer $m$ such that $P(m) = P(2)$ . How many possible values are there for the ordered triple $(a, b, c)$ ?

Solution

If $m$ is the only integral value that satisfies $P(m) = P(2)$ , we can show that $m$ is the only real value that satisfies $P(m) = P(2)$ .

Next, we have $P(m) = P(2)$ , so therefore

$m^3 + am^2 + bm + c - 8 - 4a - 2b - c = 0.$

We can now simplify:

$m^3 - 8 + am^2 - 4a + bm - 2b = 0$ $(m-2)(m^2 + 2m + 4) + a(m-2)(m+2) + b(m-2) = 0$ $(m-2)(m^2 + 2m + 4 + am + 2a + b) = 0.$

Since $m \neq 2$ ,

$m^2 + (a+2)m + 2a + b + 4 = 0.$

We can now apply the quadratic formula, yielding

$m = \frac{-a-2 \pm \sqrt{a^2 - 4a - 4b - 12}}{2}.$

For this to have exactly $1$ solution, we must have $a^2 - 4a - 4b - 12 = 0$ , and thus $(a-2)^2 = 4(b+2)$ . This means that $|a| \leq 9$ , yielding $9$ solutions for $a$ . For any solution of $a$ , $b$ can only attain one value. And, the value of $c$ doesn't matter. Our answer is thus $9 \cdot 41 = \boxed{369}$ .

~mathboy100

I believe this solution is wrong. The answer is $738$ . ~r00tsOfUnity

??? I checked with like 5 people ~mathboy100

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2023 AIME I Problems/Problem 9

Solution