Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.6 AIME, harder Find the answer

The polynomial $f(x)=x^{2007}+17 x^{2006}+1$ has distinct zeroes $r_{1}, \ldots, r_{2007}$ . A polynomial $P$ of degree 2007 has the property that $P\left(r_{j}+\frac{1}{r_{j}}\right)=0$ for $j=1, \ldots, 2007$ . Determine the value of $P(1) / P(-1)$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

For some constant $k$ , we have $P(z)=k \prod_{j=1}^{2007}\left(z-\left(r_{j}+\frac{1}{r_{j}}\right)\right)$ . Now writing $\omega^{3}=1$ with $\omega \neq 1$ , we have $\omega^{2}+\omega=-1$ . Then $P(1) / P(-1)=\frac{k \prod_{j=1}^{2007}\left(1-\left(r_{j}+\frac{1}{r_{j}}\right)\right)}{k \prod_{j=1}^{2007}\left(-1-\left(r_{j}+\frac{1}{r_{j}}\right)\right)}=\prod_{j=1}^{2007} \frac{r_{j}^{2}-r_{j}+1}{r_{j}^{2}+r_{j}+1}=\prod_{j=1}^{2007} \frac{\left(-\omega-r_{j}\right)\left(-\omega^{2}-r_{j}\right)}{\left(\omega-r_{j}\right)\left(\omega^{2}-r_{j}\right)} =\frac{f(-\omega) f\left(-\omega^{2}\right)}{f(\omega) f\left(\omega^{2}\right)}=\frac{\left(-\omega^{2007}+17 \omega^{2006}+1\right)\left(-\left(\omega^{2}\right)^{2007}+17\left(\omega^{2}\right)^{2006}+1\right)}{\left(\omega^{2007}+17 \omega^{2006}+1\right)\left(\left(\omega^{2}\right)^{2007}+17\left(\omega^{2}\right)^{2006}+1\right)}=\frac{\left(17 \omega^{2}\right)(17 \omega)}{\left(2+17 \omega^{2}\right)(2+17 \omega)} =\frac{289}{4+34\left(\omega+\omega^{2}\right)+289}=\frac{289}{259}$ .

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.