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Let $a, b, c$ be real numbers and assume that all roots of $x^3 + ax^2 +b x+c=0$ have the same absolute value. Show that, $a=0 $ if and only if $ b=0$.

Math137
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2 Answers2

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At least one of the roots $x_1,x_2,x_3$ is real, let $x_1$ be that root. Then from $0=-a=x_1+x_2+x_3$ you get $x_1=-(x_2+x_3)$ and $b=x_1(x_2+x_3)+x_2x_3=-x_1^2+x_2x_3$.

Now consider the cases $x_3=\pm x_2$, which can be excluded since with 3 real roots $a=0$ is not possible. It remains to analyse the case of a complex conjugate pair $x_3=\overline{x_2}$.

One can treat this more directly by analysing the factorization into linear factors, exploiting that the coefficients are real:

  • triple root at $x=0$
  • all roots real, nonzero gives the structure $(x-r)^2(x+r)$ or $(x-r)^3$
  • one real, one conjugate complex pair gives the structure $(x-r)(x+2cr+r^2)$, $|c|\le1$.

In each case the inner coefficients of the expanded polynomials are either both zero or both non-zero.

Lutz Lehmann
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  • Can you please elaborate upon the assertion " Now consider the cases $x_3=\pm x_2$ which can be excluded" ? I dont understand what you mean by excluded? Or rather why is it so ? – Arthur Apr 24 '23 at 09:01
  • Because that would imply that all roots are real and in either case $a=0$ is not possible. – Lutz Lehmann Apr 24 '23 at 09:28
  • So that means the question is incorrect? I mean the way, the question is presented it seems that if $b=0$ then $a$ is always equal to zero ...? – Arthur Apr 24 '23 at 12:05
  • No, the claim in the question is correct. The only solution for the given configuration is $x^3-r^3$ where $a=b=0$, in all other cases both inner coefficients are non-zero. – Lutz Lehmann Apr 24 '23 at 12:39
  • Ok ok ok,now this highly absurd. Let me put the situation clearly: First when we calculate $b=-x_1^2+x_2x_3$, we have many cases. One such case, is, if, $x_2,x_3$ be real, then $x_2=\pm x_3$. If $x_2=x_3$ then, $b=-x_1^2+x_2^2=0.$ But now, if, $x_2=-x_3$ then, $b=-x_1^2+x_2x_3=-x_1^2-x_2^2=-2x_1^2,$ but then again, $-a=0=x_1+x_2+x_3=x_1-x_3+x_3=x_1$ so, $x_1=0$ this implies $x_2=x_3=0$ or rather more specifically, $b=-2x_1^2=0$. Is this what you meant ? – Arthur Apr 25 '23 at 05:27
  • Yes, that was the idea. But this "running in (near) circles" can be reduced by first concentrating on what configurations are possible with the roots on a circle and the coefficients real, like I wrote in the recent add-on. The original approach of taking just one real root and jumping into the algebra of the Viete equations is not wrong but appears to be less systematic, splitting in many sub-cases. – Lutz Lehmann Apr 25 '23 at 06:22
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Let the roots be $x_{1,2,3}$ with $|x_1|=|x_2|=|x_3|=r \ge 0$. It can be assumed WLOG that $c \ge 0$, otherwise apply the following argument to the equation $x^3-ax^2+bx-c=0$ which has roots $-x_{1,2,3}$ of the same magnitude. By Vieta's relation $r^3 = |x_1x_2x_3|=|-c|=c \iff r = \sqrt[3]{c}$.

$P(x)=x^3+ax^2+bx+c$ is non-negative at $x=0$ since $p(0)=c\ge0$, so it must have a real root $x_1 \le 0$ with $|x_1| = \sqrt[3]{c} \ge 0$, which means $x_1=-\sqrt[3]{c}$. Substituting in the equation:

$$ \require{cancel} \cancel{\left(-\sqrt[3]{c}\right)^3} + a\,\left(-\sqrt[3]{c}\right)^2 + b\,\left(-\sqrt[3]{c}\right) + \cancel{c} = 0 \quad\iff\quad \sqrt[3]{c^2}\,a = \sqrt[3]{c}\, b $$

If $c=0$ then all roots are $0$ so $a=b=0$, else if $c \ne 0$ the above implies $a=0 \iff b=0$.

dxiv
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