As far as I know this is appear in Imo 1982 and 23rd Vietnam Olympiad (1985).
Find all real values of parameter "a" for which the equation in $ 16x^4 −ax^3 +(2a+17)x^2 −ax+16=0 $ has four solutions which form an geometric progression( the original was arithmetic )
For me, I let r as the first root so the other roots are $ rq, rq^2, rq^3 $ then I put them in the equation, but I don't know where I should I go next. I try the solution from the web too but it is too hard to understand. Please kindly solve this problem for me. Thanks
The roots have product $1$, so are of the form $k^{\pm1},\, k^{\pm3}$ (where without loss of generality $|k|\ge1$) so $\frac{a}{16}=k+k^{-1}+k^3+k^{-3}$. Write $t:=k+k^{-1}$ so $\frac{a}{16}=t(t^2-2)$. The $x^2$ coefficient gives
$$2t^3-4t+\frac{17}{16}=\frac{a}{8}+\frac{17}{16}=(k+k^{-1})(k^3+k^{-3})+2=t^2(t^2-3)+2,$$which rearranges to $$t^4-2t^3-3t^2+4t+\frac{15}{16}=0\implies t\in\left\{-\frac{3}{2},\,\frac{1}{2}-\frac{1}{\sqrt{2}},\,\frac{1}{2}+\frac{1}{\sqrt{2}},\,\frac{5}{2}\right\}.$$For real progressions $|k+k^{-1}|\ge2$, so$$t=\frac52\implies k=2\implies a=16\left(\frac18+\frac12+2+8\right)=170.$$Other values of $t$ give complex $k$, with $a=16t(t^2-2)$:$$t=-\frac32\implies a=-6,\,t=\frac12\pm\frac{1}{\sqrt{2}}\implies a=-2\mp6\sqrt{2}.$$