4. A4 (BUL 2) Determine all real values of the parameter a for which the equation
$$
16 x^{4}-a x^{3}+(2 a+17) x^{2}-a x+16=0
$$
has exactly four distinct real roots that form a geometric progression.
4. Suppose that $a$ satisfies the requirements of the problem and that $x, q x$, $q^{2} x, q^{3} x$ are the roots of the given equation. Then $x \neq 0$ and we may assume that $|q|>1$, so that $|x|<|q x|<\left|q^{2} x\right|<\left|q^{3} x\right|$. Since the equation is symmetric, $1 / x$ is also a root and therefore $1 / x=q^{3} x$, i.e., $q=x^{-2 / 3}$. It follows that the roots are $x, x^{1 / 3}, x^{-1 / 3}, x^{-1}$. Now by Vieta's formula we have $x+x^{1 / 3}+x^{-1 / 3}+x^{-1}=a / 16$ and $x^{4 / 3}+x^{2 / 3}+2+x^{-2 / 3}+x^{-4 / 3}=$ $(2 a+17) / 16$. On setting $z=x^{1 / 3}+x^{-1 / 3}$ these equations become
$$
\begin{aligned}
z^{3}-2 z & =a / 16, \\
\left(z^{2}-2\right)^{2}+z^{2}-2 & =(2 a+17) / 16 .
\end{aligned}
$$

Substituting $a=16\left(z^{3}-2 z\right)$ in the second equation leads to $z^{4}-2 z^{3}-$ $3 z^{2}+4 z+15 / 16=0$. We observe that this polynomial factors as $(z+$ $3 / 2)(z-5 / 2)\left(z^{2}-z-1 / 4\right)$. Since $|z|=\left|x^{1 / 3}+x^{-1 / 3}\right| \geq 2$, the only viable value is $z=5 / 2$. Consequently $a=170$ and the roots are $1 / 8,1 / 2,2,8$.