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Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO

Algebra Difficulty 4.1 AIME Find the answer Italy

Problem:

What is the sum of the sixth powers of the solutions of the equation x6−16x4+16x2−1=0x^{6}-16 x^{4}+16 x^{2}-1=0?

Pick one

Solution

Solution:

The answer is (E)\mathbf{(E)}. We have x6−16x4+16x2−1=(x2−1)(x4−15x2+1)x^{6}-16 x^{4}+16 x^{2}-1=\left(x^{2}-1\right)\left(x^{4}-15 x^{2}+1\right) (all the roots are real), so if zz is a root different from ±1\pm 1, we have z4=15z2−1z^{4}=15 z^{2}-1, z6=15z4−z2=224z2−15z^{6}=15 z^{4}-z^{2}=224 z^{2}-15. Denoting by z1,…,z4z_{1}, \ldots, z_{4} these latter roots, we have:
∑j=14zj=0,∑1≤i<j≤4zizj=−15 \sum_{j=1}^{4} z_{j}=0, \quad \sum_{1 \leq i<j \leq 4} z_{i} z_{j}=-15
∑i=14zi2=(∑j=14zj)2−2∑1≤i<j≤4zizj=30,∑i=14zi6=224∑i=14zi2−60=6660, \begin{gathered} \sum_{i=1}^{4} z_{i}^{2}=\left(\sum_{j=1}^{4} z_{j}\right)^{2}-2 \sum_{1 \leq i<j \leq 4} z_{i} z_{j}=30, \\ \sum_{i=1}^{4} z_{i}^{6}=224 \sum_{i=1}^{4} z_{i}^{2}-60=6660, \end{gathered}
now reintroducing into the count the roots ±1\pm 1 we find that the correct answer is 6662. (D'Aurizio)

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.