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How many ordered triples of rational numbers $(a,b,c)$ are there such that the cubic polynomial $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$?

The polynomial is allowed to have repeated roots.

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1 Answer 1

[There was a mistake in the first version of this answer that caused one solution to go missing.]

The only monic polynomial with roots $a,b,c$ is $(x-a)(x-b)(x-c)$, so we must have

$$ \begin{align} a&=-a-b-c\;,\\ b&=ab+bc+ca\;,\\ c&=-abc\;. \end{align} $$

If $c=0$, this becomes

$$ \begin{align} a&=-a-b\;,\\ b&=ab\;,\\ \end{align} $$

and thus either $b=0$ and $a=0$ or $a=1$ and $b=-2$.

If $c\ne0$, the third equation becomes $ab=-1$; substituting $b=-1/a$ into the first equation yields

$$ c=-2a+\frac1a\;, $$

and then the second equation becomes

$$ -\frac1a=-1+2-\frac1{a^2}-2a^2+1\;. $$

Multiplying through by $a^2$ yields

$$ 2a^4-2a^2-a+1=0\;. $$

The solution $a=1$ is readily guessed, and dividing through by $a-1$ yields

$$ 2a^3+2a^2-1=0\;, $$

which has one irrational and two complex roots (computation). The solution $a=1$ leads to $b=c=-1$.

Thus the only ordered triples are $(0,0,0)$, $(1,-2,0)$ and $(1,-1,-1)$, with corresponding polynomials $x^3$, $x^3+x^2-2x$ and $x^3+x^2-x-1$, respectively.

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Not much familiar with solutions of biquadratic equations..."no real solutions since the first term is non-negative and the quadratic equation obtained by dropping it has no real solutions" Please explain... and I hope you need to arrange the possibilities of the second case (1,-2,0) in 3! ways because a,b and c can be permuted in 3! ways... Please tell if I'm wrong... –  Raj Dec 7 '12 at 14:38
    
@Raj: No, there's no symmetry among $a$, $b$ and $c$. They are coefficients of different terms of the polynomial. You can also see from the three equations in the beginning that they play different roles. Perhaps more convincingly, just try it out; $0$ isn't a root of $x^3+x^2+0\cdot x-2$. On your first question: This is not a biquadratic equation. What are the roots of $2a^2+a+1$? The fact that they're not real implies that $2a^2+a+1$ never changes sign; since it's positive at $a=0$, it's always positive. Adding the non-negative term $2a^4$ doesn't change that. –  joriki Dec 7 '12 at 14:55
    
Abc 123 points out that the third equation should be $c=-abc$ –  Ross Millikan Dec 7 '12 at 17:49
    
@Ross: Thanks; correcting that mistake yields one further solution. –  joriki Dec 7 '12 at 18:08
    
@Raj: Please note that the first version (to which your comment refers) contained a mistake that caused one solution go missing; sorry about that. –  joriki Dec 7 '12 at 18:10
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