Help required! Polynomials

Question

Let $D(p) = p^{20} - p^{18} - p^{16} - \dots - p^2 - 2$

Prove that the sum of fourth powers of all the real roots of $D(p) = 8.$

Please help.

Answer 1

You have already noted, that there are exactly $2$ real roots $\alpha$ and $-\alpha$. So you are done by checking $D(\sqrt 2)=0$.

That the product of the complex roots is $1$, is then a consequence of the fact, that the product of the two real roots is equal to the constant coefficient $-2$.

Here are some thoughts how you come up with the idea that $\sqrt 2$ is a root:

We substitute $p^2=z$ and get the equation

$$z^{10}-z^9-z^8 - \dotsc - z -2=0$$

or equivalent (for $z \neq 1$, but $z=1$ is obviously no solution)

$$z^{10}-1 = z^9+z^8 + \dotsc + z +1=\frac{z^{10}-1}{z-1} \Longleftrightarrow (z^{10}-1)(z-2)=0$$

So you get the solutions exactly: The nine non-trivial $10$th roots of unity and $z=2$.

The roots of $D$ are precisely the square roots of this solutions.