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Algebra Difficulty 4.6 AIME Prove it India

Let $f(x) = \sum_{k=1}^{n} a_k x^k$ and $g(x) = \sum_{k=1}^{n} \frac{a_k}{2^k - 1} x^k$ be two polynomials with real coefficients, where $n \ge 3$ . Suppose $1, 2^{n+1}$ are roots of $g(x) = 0$ . Prove that $f(x) = 0$ has a positive root smaller than $2^n$ .

Solution

It is easy to see that $g(2x) - g(x) = f(x)$ . Thus
$\sum_{k=0}^{n} f(2^k) = g(2^{n+1}) - g(1) = 0.$
Consider the relation $f(1) + f(2) + \dots + f(2^n) = 0$ . If $f(2^n) \neq 0$ , then $f(1)$ and $f(2^n)$ have opposite signs. Hence there exists $\alpha$ in $(1, 2^n)$ such that $f(\alpha) = 0$ . If $f(2^n) = 0$ , omit this and the argument applies to $f(1) + f(2) + \dots + f(2^{n-1}) = 0$ .

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