HMMT-Feb 2015 guts, Problem 30

Find the sum of squares of all distinct complex numbers $x$ satisfying the equation \[ 0=4 x^{10}-7 x^{9}+5 x^{8}-8 x^{7}+12 x^{6}-12 x^{5}+12 x^{4}-8 x^{3}+5 x^{2}-7 x+4 \]

Solution: For convenience denote the polynomial by $P(x)$. Notice $4+8=7+5=12$ and that the consecutive terms $12 x^{6}-12 x^{5}+12 x^{4}$ are the leading terms of $12 \Phi_{14}(x)$, which is suggestive. Indeed, consider $\omega$ a primitive 14 -th root of unity; since $\omega^{7}=-1$, we have $4 \omega^{10}=-4 \omega^{3},-7 \omega^{9}=7 \omega^{2}$, and so on, so that \[ P(\omega)=12\left(\omega^{6}-\omega^{5}+\cdots+1\right)=12 \Phi_{14}(\omega)=0 \] Dividing, we find \[ P(x)=\Phi_{14}(x)\left(4 x^{4}-3 x^{3}-2 x^{2}-3 x+4\right) \] This second polynomial is symmetric; since 0 is clearly not a root, we have \[ 4 x^{4}-3 x^{3}-2 x^{2}-3 x+4=0 \Longleftrightarrow 4\left(x+\frac{1}{x}\right)^{2}-3\left(x+\frac{1}{x}\right)-10=0 \] Setting $y=x+1 / x$ and solving the quadratic gives $y=2$ and $y=-5 / 4$ as solutions; replacing $y$ with $x+1 / x$ and solving the two resulting quadratics give the double root $x=1$ and the roots $(-5 \pm i \sqrt{39}) / 8$ respectively. Together with the primitive fourteenth roots of unity, these are all the roots of our polynomial. Explicitly, the roots are \[ e^{\pi i / 7}, e^{3 \pi i / 7}, e^{5 \pi i / 7}, e^{9 \pi i / 7}, e^{11 \pi i / 7}, e^{13 \pi i / 7}, 1,(-5 \pm i \sqrt{39}) / 8 \] The sum of squares of the roots of unity (including 1) is just 0 by symmetry (or a number of other methods). The sum of the squares of the final conjugate pair is $\frac{2\left(5^{2}-39\right)}{8^{2}}=-\frac{14}{32}=-\frac{7}{16}$. $\fbox{-\frac{7}{16}}$.

Answer: -\frac{7}{16}