amparvardi wrote: Determine all real values of the parameter $a$ for which the equation \[16x^4 -ax^3 + (2a + 17)x^2 -ax + 16 = 0\] has exactly four distinct real roots that form a geometric progression. Let $u^3,u^3v,u^3v^2,u^3v^3$ be the four roots (with $u\ne 0$, $v\ne 0$ and $|v|\ne 1$ in order to have distinct roots). $x$ root $\implies$ $\frac 1x$ root but : $u^3\cdot u^3v=1$ and $u^3v^2\cdot u^3v^3=1$ is impossible with $|v|\ne 1$ $u^3\cdot u^3v^2=1$ and $u^3v\cdot u^3v^3=1$ is impossible with $|v|\ne 1$ So $u^6v^3=1$ and the four roots are $u^3,u,\frac 1u,\frac 1{u^3}$ Equating the coefficients implies then : $u^3+u+\frac 1u+\frac 1{u^3}=\frac a{16}$ $u^4+u^2+\frac 1{u^2}+\frac 1{u^4}+2=\frac{2a+17}{16}$ Setting $t=u+\frac 1u$, this can be written : $t^3-2t=\frac a{16}$ and $t^4-3t^2+2=\frac{2a+17}{16}$ which implies $t^4-3t^2+2=\frac{17}{16}+2(t^3-2t)$ $\iff$ $t^4-2t^3-3t^2+4t+\frac{15}{16}=0$ $\iff$ $(t+\frac 32)(t-\frac 52)(t-\frac{1-\sqrt 2}2)(t-\frac{1+\sqrt 2}2)=0$ And , since $|t|\ge 2$, we get $t=\frac 52$ and so $\boxed{a=170}$ And it is easy to check that this value fits the requirements (the roots are $8,2,\frac 12,\frac 18$)

My solution from WOOT: Note that the product of these roots is $1$ by Vieta's, and that $\frac1x$ is a root iff $x$ is. Then set the roots as $s^{-3},s^{-1},s,s^3$. Let $r=s+\frac1s$ and note that we have: $$\begin{cases}s^2+\frac1{s^2}=r^2-2\\s^3+\frac1{s^3}=r^3-3r\\s^4+\frac1{s^4}=r^4-4r^2+2\end{cases}$$Then Vieta's gives us the two equations: $$\frac a{16}=r^3-2r\text{ and }\frac{2a+17}{16}=r^4-3r^2+2,$$so we have $(2r+3)(2r-5)(4r^2-4r-1)=0$. If $s>0$, then $r=s+\frac1s\ge2$ by AM-GM. If $s<0$, then $-r=-s+\frac1{-s}\ge2$, so $r\le-2$. We conclude that $r\notin(-2,2)$. This immediately eliminates $r=-\frac32$, and since $4r^2-4r-1$ implies $r\in\left\{\frac{1-\sqrt2}2,\frac{1+\sqrt2}2\right\}\subset(-2,2)$, we must have $r=52$. Then a simple calculation yields $a=\boxed{170}$.

Let the roots be $\frac d{r^3}, \frac dr, dr, dr^3$ for some $d, r$. Observe that from the constant term, $d = \pm 1$. Let $y = x+\frac 1x$. With some work, the roots to the given polynomial are also roots of $$16y^2 - ay + (2a-15) = 0.$$Note that the $d=1$ case gives the same answers as the $d = -1$ case, because 1 and 3 are both odd; thus, WLOG $d=1$. Then, $r, \frac 1r$ are roots to $x+\frac 1x = y_1$, and $r^3, \frac 1{r^3}$ are roots to $x+\frac 1x = y_2$, where $y_1, y_2$ are the roots of the first display. Set $r + \frac 1r = m$, so that $r^3+\frac 1{r^3} = m(m^2-3)$. By Vieta's Formulas, \begin{align*} m(m^2-3) \cdot m = \frac{2a-15}{16} \\ m(m^2-3)+m = \frac a{16}. \end{align*}Eliminating $a$, $$m(m^2-3) \cdot m - 2[m(m^2-3) + m] = -\frac{15}{16} \iff 16m^4-32m^3-48m^2+64m+15=0.$$Observe the following claim: Claim. $|m| \geq 2$. Proof. AM-GM. By RRT the only root with $|m|\geq 2$ is $m=\frac 52$, which gives $a = \boxed{170}$.

HamstPan38825 wrote: By RRT ... Why are roots rational (the first "R" of RRT) ?

pco wrote: HamstPan38825 wrote: By RRT ... Why are roots rational (the first "R" of RRT) ? I made a mistake in typing out the specific quartic (which should have been $16m^4-32m^3-48m^2+64m+15=0$) but the exhaustive list of roots of the correct quartic ends up being $-\frac 32, \frac 52, \frac{1 \pm \sqrt 2}2$, the first two of which are extractable by RRT.