4
$\begingroup$

Let $a$, $b$ and $c$ be roots of the equation $$x^3+15x^2-198x+1=0.$$ Prove that: $$\sqrt[5]a+\sqrt[5]b+\sqrt[5]c=0$$

I have a solution for this problem, but I want to see another solutions.

My solution is the following.

We'll take the equation $x^3-3x+1=0$,

which has three roots $a=2\cos40^{\circ}$, $b=2\cos80^{\circ}$ and $c=-2\cos20^{\circ}$.

Now, easy to show that

$a^5+b^5+c^5=-15$, $a^5b^5+a^5c^5+b^5c^5=-198$, $a^5b^5c^5=-1$ and since $a+b+c=0$,

we obtain the starting equation.

My question is how we can solve this equation without previous way?

Thank you very much!

  • 5
    It would be helpful if you showed your solution so others don't waste effort coming up with the same solution... – 2017-02-12
  • 3
    If $\alpha$ is a root of the polynomial $p(x)=x^3+15x^2-198x+1$ then $\alpha^{1/5}$ is a root of the polynomial $p(x^5)=(1-3x+x^3)\cdot q(x)$. If we manage to prove that $(1-3x+x^3)$ is the minimal polynomial of $\alpha^{1/5}$, the claim easily follows from Vieta's theorem. – 2017-02-12
  • 0
    @David Quinn If I'll post my solution so it not would interesting to solve. – 2017-02-12

1 Answers 1

6

If $\alpha$ is a root of the polynomial $p(x)=x^3+15x^2-198x+1$ then $\alpha^{1/5}$ is a root of the polynomial $p(x^5)=(1-3x+x^3)\cdot q(x)$. If we manage to prove that $(1-3x+x^3)$ is the minimal polynomial of $\alpha^{1/5}$, the claim easily follows from Vieta's theorem. For the last part, it is simpler to go in the opposite direction.

Claim: if $\theta$ is a root of $x^3-3x+1$, then $\theta^5$ is a root of $x^3+15x^2-198x+1$.

Let we work in $\mathbb{Q}[x]/(x^3-3x+1)$. A base of this ring as a vector space over $\mathbb{Q}$ is given by $\{1,x,x^2\}$, and by computing a few polynomial remainders we have: $$\begin{eqnarray*}1 &=& 1 \\ x^5 &=& -x^2+9x-3\\ x^{10} &=& 90x^2-109x+27\\ x^{15}&=&-1548 x^2+3417 x-1000 \end{eqnarray*} $$ so the previous Claim follows from gaussian elimination.

  • 0
    It was beautiful reasoning! – 2017-02-12
  • 1
    How did you come up with $(1-3x+x^3)$? – 2017-02-12
  • 0
    @i9Fn: I factored $p(x^5)$ in WA. – 2017-02-12
  • 0
    @ i9Fn $x^{15}+15x^{10}-198x^5+1=(x^3-3x+1)(x^{12}+3x^{10}-x^9+9x^8+9x^7+28x^6+18x^5+75x^4+26x^3+9x^2+3x+1)$, where the right polynomial is positive for all real $x$. – 2017-02-12
  • 0
    @MichaelRozenberg: I guess it is a bit redundant, since in the last part we are essentially just re-proving that $x^3-3x+1$ is a divisor of $p(x^5)$. Anyway, it works, thanks. – 2017-02-12
  • 0
    @Jack D'Aurizio I created this problem from $x^3-3x+1=0$ and you showed, how we can see it! Thank you! – 2017-02-12
  • 0
    @JackD'Aurizio How did you know to look at $p(x^5)$? – 2017-02-14
  • 0
    @Qudit: if $\alpha$ is a root of $p(x)$, then $\alpha^{1/5}$ is a root of $p(x^5)$. – 2017-02-14
  • 0
    @JackD'Aurizio Sure, but what I'm wondering is if there is some way to tell that $5$ is the right number to look at or if you just checked for the first few numbers. – 2017-02-14
  • 0
    @Qudit: the problem is about a sum of fifth roots. – 2017-02-14