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Let $a,b,c$ be distinct nonzero real numbers. If the equations $E_1: ax^3+bx+c=0, E_2: bx^3+cx+a=0$ and $E_3: cx^3+ax+b=0$ have a common root, prove that at least one of these equations has three real roots(not necessarily distinct).

This question was a math olympiad problem.Below was my attempt but I've got confused:

I know that the discriminant of the equation $x^3+Ax+B=0$ is given by $-4A^3-27B^2$, so it suffices to prove that the discriminant of at one least of the equations is non-negative.

It's easy to prove that the common root $r$ must be real. Now assume on the contrary that all three equations had a pair of complex conjugate roots. Let $\alpha, \bar{\alpha}, \beta, \bar{\beta}, \gamma, \bar{\gamma}$ be the complex conjugate roots of $E_1, E_2$ and $E_3$ respectively.

Then by viete theorem, we have $$\alpha \bar{\alpha} r=|\alpha|^2r=\frac{-c}{a},$$ $$|\beta|^2r=\frac{-a}{b},$$ $$|\gamma|^2r=\frac{-b}{c}.$$ Multiplying the three equations gives $$|\alpha \beta \gamma|^2r^3=-1.$$ Hence $r<0.$ Hence $-c/a<0$ or $c/a>0.$ Similarly $a/b>0$ and $b/c>0.$ This implies that $a,b,c$ all all positive or or negative.

By symmetry, we can assume $a>b>c>0$, but then none of the discriminant could be non-negative. So what's wrong with the argument and how can I complete the proof? Thanks in advance!

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  • $\begingroup$ "none of the discriminant could be non-negative" is contradict to your assumption "$\alpha, \bar{\alpha}, \beta, \bar{\beta}, \gamma, \bar{\gamma}$be the complex conjugate roots of $E_1, E_2$ and $E_3$ respectively". $\endgroup$ – Zack Ni Jul 19 '16 at 12:11
  • $\begingroup$ Your proof is almost done! $\endgroup$ – Zack Ni Jul 19 '16 at 12:11
  • $\begingroup$ I am sorry, but I dont get it, we have $a>b>c>0$, hence the discriminant of $E_1$ is $-4(b/3)^3-27(c/a)^2$ must be negative. Similarly for $E_2$ and $E_3$, pleas helps... thanks. $\endgroup$ – Schurik Jul 19 '16 at 16:20
  • $\begingroup$ All your reasoning is correct but the definition of the discriminant is a bit problematic: it should be $A^3/27+B^2/4$ $\endgroup$ – Zack Ni Jul 20 '16 at 8:08
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Let $r$ be the common root, then

$$(a+b+c)(r^3+r+1)=0 \tag{$E_1+E_2+E_3$}$$

Case I:

If $r^3+r+1=0$, then $(ar^3+br+c)-a(r^3+r+1)=0$

$$\implies (a-b)r=(c-a)$$

By symmetry, $$r=\frac{c-a}{a-b}=\frac{a-b}{b-c}=\frac{b-c}{c-a}$$

$$c=\frac{a+b \pm i(a-b)\sqrt{3}}{2}$$

Or equivalently,

$$r=\frac{-1 \pm i\sqrt{3}}{2}$$

that contradicts with $r^3+r+1=0$.

Case II:

If $a+b+c=0$, then $r=1$.

For $x\neq 1$,

$$\frac{ax^3+bx+c}{x-1}=ax^2+ax-c=0$$

$$\Delta_{1}=a^2+4ac$$

Similarly,

$$\Delta_{2}=b^2+4ba$$

$$\Delta_{3}=c^2+4cb$$

Since $a+b+c=0$, either two of $a$, $b$ and $c$ are of equal sign.

That is either one of $ab$, $bc$ and $ca$ is positive.

At least one of the $\Delta_{1}$, $\Delta_{2}$ and $\Delta_{3}$ is positive.

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