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If the roots of the equation $x^4 + ax^3 + bx^2 + cx + d = 0$ are in geometric progression then,

$a) b^2 = ac$

$b) a^2 = b$

$c) c^2 = a^2d$

Using Vieta's relations, finding values of coefficients in terms of the assumed roots is what first comes to mind. That process works out well to give the answer ($c)$) But, as it can be seen, this method is quite lengthy. Is there a shorter method to solve this question or a trick?

mathx
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2 Answers2

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Here is the OP's original method. Assume the roots are $k, kr, kr^2, kr^3$. Then:

$$k(1+r+r^2+r^3) = -a$$ $$k^2r+k^2r^3+k^2r^4+k^2r^3+k^2r^4+k^3r^5 = b$$ $$k^3r^3 + k^3r^4+k^3r^5+k^3r^6 = k^3 r^3(1+r+r^2+r^3) = -c$$ $$k^4r^6 = d$$

By comparing the maximum powers of $r$, a) is not correct ($r^{10}$ vs $r^9$) and so is b). But $c/a = k^2r^3$ or $c^2/a^2 = k^4r^6 = d$, hence c) is correct.

This relationship between $a$ and $c$ can be extended to any power $n$ in general. $-c$ can also be written as $k^4 r^6 (1/k + 1/(kr) + 1/(kr^2) + 1/(kr^3))$ as you miss out one term in the product of three (or $n$) roots, and in general this is $k^n r^{n(n-1)/2} (1/k + \cdots + \frac{1}{kr^{n-1}}) = k^{n-1} r^{(n-1)(n-2)/2} (1 + \cdots + r^{n-1})$.

Toby Mak
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Hint: let the roots be $t,u,v,w$, then a sufficient condition for a GP is $tw=uv$ $\implies tw=uv=\sqrt{tuvw}=\sqrt{d}$.

This requires $d\gt 0$, and in that case let $x=x'\sqrt[4]{d}$ then the polynomial

$$dx'^4+a\sqrt[4]{d^3} x'^3+b\sqrt{d} x'^2+c\sqrt[4]{d} x' +d $$

has roots $t'=t/\sqrt[4]{d}, u'=u/\sqrt[4]{d},\dots$ with $t'w'=u'v'=1$ i.e. it is reciprocal, so $a\sqrt[4]{d^3}=c\sqrt[4]{d}\,$.

dxiv
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  • That was elegant! – mathx Jul 13 '21 at 07:44
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    @mathx This does not even require the full premise of a GP, but just that the two pairs of roots have the same ratio $u/t=w/v$. – dxiv Jul 13 '21 at 07:48
  • Oh Yes. Nice observation. – mathx Jul 13 '21 at 07:51
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    For an example, $x^4 - 24 x^3 + 163 x^2 - 336 x + 196$ has roots $1,2,7,14$ which are not in a GP, but $2/1 = 14/7$ and indeed $336^2 = 24^2 * 196$. – dxiv Jul 13 '21 at 08:19
  • OK. Before posting the question, I was thinking if some coefficient substitution could make the solution shorter. How did you come up with that equation? It seems difficult to create something like that. – mathx Jul 13 '21 at 08:55
  • @mathx A GP with geometric mean $1$ is invariant to the transformation $,x \mapsto \frac{1}{x},$, so the first step "normalizes" the GP such that it has geometric mean $,1,$. Then, a polynomial invariant to the transformation $,x \mapsto \frac{1}{x},$ must have the symmetric coefficients equal, so the result follows. – dxiv Jul 13 '21 at 15:40
  • dxiv OK. Got it. – mathx Jul 13 '21 at 16:12
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    @mathx Loosely related, see an example here of additive symmetry, where the pairwise sums are equal (instead of their products). In that case, what simplifies the problem is a translation $,x \mapsto x-x_0,$ which brings the mean to zero. – dxiv Jul 13 '21 at 16:49
  • That one is a clever solution too. Immensely appreciate it. Really nice and elegant! – mathx Jul 13 '21 at 17:03