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Annual Harvard-MIT Mathematics Tournament

Algebra Difficulty 5.0 AIME, harder Prove it United States

Problem:

Let $f(x) = x^{3} + x + 1$ . Suppose $g$ is a cubic polynomial such that $g(0) = -1$ , and the roots of $g$ are the squares of the roots of $f$ . Find $g(9)$ .

Solution

Solution:

Let $a, b, c$ be the zeros of $f$ . Then $f(x) = (x - a)(x - b)(x - c)$ . Then, the roots of $g$ are $a^{2}, b^{2}, c^{2}$ , so $g(x) = k(x - a^{2})(x - b^{2})(x - c^{2})$ for some constant $k$ . Since $a b c = -f(0) = -1$ , we have $k = k a^{2} b^{2} c^{2} = -g(0) = 1$ . Thus,
$g(x^{2}) = (x^{2} - a^{2})(x^{2} - b^{2})(x^{2} - c^{2}) = (x - a)(x - b)(x - c)(x + a)(x + b)(x + c) = -f(x) f(-x)$
Setting $x = 3$ gives $g(9) = -f(3) f(-3) = -(31)(-29) = 899$ .

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