Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.2 AIME Find the answer Italy

Problem:

Let $P(X) = a X^{2} + b X + c$ be a second-degree polynomial with real coefficients (that is, $a, b, c$ are real numbers and $a \neq 0$ ). If $P(2000) = 2000$ and $P(2001) = 2001$ , then $P(2002)$ cannot be equal to:

Pick one

Solution

Solution:

The answer is $(\mathbf{C})$ . For simplicity, let us set $Q(X) = P(X + 2000) - 2000$ : note that $Q(X)$ is always a polynomial of the same degree as $P(X)$ . Let $Q(X) = \alpha X^{2} + \beta X + \gamma$ . The given conditions become $Q(0) = 0$ and $Q(1) = 1$ , that is, $\gamma = 0$ and $\alpha + \beta = 1$ . The condition $P(2002) = 2002$ becomes $Q(2) = 2$ , that is, $4 \alpha + 2 \beta = 2$ , which, together with the previous ones, gives $\alpha = 0, \beta = 1$ ; it follows that $Q(X)$ , and hence also $P(X)$ , must be a first-degree polynomial.

For the other values, on the other hand, one easily finds the following examples:
$Q(X) = -X^{2} + 2 X, \quad \frac{-X^{2} + 3 X}{2}, \quad \frac{X^{2} + X}{2}, \quad X^{2}$

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.