2025 SSMO Team Round Problems/Problem 4

Problem

Let $P(x) = x^2 - ax + b$ be a quadratic with nonzero real coefficients. Given that $P(a)$ and $P(-b)$ are roots of $P(x),$ there exists a value of $c$ such that $P(c)$ is constant for all possible $P(x)$ . Find $c$ .

Solution

Note that $P(a) = b$ . Since $b$ is always a root of $P(x)$ , by Vieta, $1$ is always a root of $P(x)$ as well. Therefore, we have $P(1)=0$ over all possible $P(x)$ .

We now show that there is no value of $c$ other than $1$ such that $P(c)$ constant over all possible $P(x)$ . If there exist at least two such quadratics $P(x)$ , we are done because any two distinct monic quadratics cannot be equal on more than one input. It is straightforward to check that the three quadratics $(x-1)(x + \tfrac{1}{2})$ , $(x-1)(x-\tfrac{-1+\sqrt{3}}{2})$ , and $(x-1)(x-\tfrac{-1-\sqrt{3}}{2})$ satisfy the conditions imposed on $P(x)$ . (In fact, these are the only possible $P(x)$ ; finding them is only a matter of showing that $P(P(a))=0$ implies $a = b+1$ , and then determining the possible values of $b$ from the equation $P(P(-b))= 0$ .) Thus, the answer is $\boxed{1}$ .

~Sedro

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2025 SSMO Team Round Problems/Problem 4

Problem

Solution