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I have a polynomial

$$ \gamma_k(x) = x^{2k+2} - m x^{2k} - n x^k - 1 $$

where $m$ and $n$ are positive real numbers and $k$ is a positive even integer. How can I prove $\gamma_k(x)$ has distinct roots for any $m$,$n$ and $k$ ?

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    $\begingroup$ Have you tried finding $\gcd(\gamma_k,\gamma_k')$? That would be my first approach, as a root is multiple iff it also is a root of the derivative. Also, i think you should swap $k$ for $2\ell$ so you don't have to go around remembering all the time that $k$ is even. $\endgroup$
    – Arthur
    Dec 14, 2017 at 18:04
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    $\begingroup$ Does "roots" here refer only to real roots, or to complex roots, too? $\endgroup$ Dec 14, 2017 at 18:05
  • $\begingroup$ @Arthur of course I tried to find $gcd(\gamma_k,\gamma_k ')$ but couldn't achieve. If you prove it with your way, could you please share it here? $\endgroup$
    – drxy
    Dec 14, 2017 at 18:25
  • $\begingroup$ @Travis both of them. $\endgroup$
    – drxy
    Dec 14, 2017 at 18:25
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    $\begingroup$ That's not "of course" (as evidenced by my comment getting three up votes). It's cool that you did, but you should've said so in your post. $\endgroup$
    – Arthur
    Dec 14, 2017 at 18:30

1 Answer 1

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This claim is false over $\Bbb C$. For example, for $k = 2$ we can generate polynomials of the form $x^6 - m x^4 - n x^2 - 1$ with repeated roots by taking $$\left(x - \frac{1}{\lambda}\right) \left(x + \frac{1}{\lambda}\right)(x^2 + \lambda)^2 = x^6 - \left(\frac{1}{\lambda^2} - 2 \lambda\right) x^4 - \left(\frac{2}{\lambda^2} - \lambda\right) x^2 - 1 .$$ It can be checked that that the coefficients of $x^4, x^2$ are negative as required for $0 < \lambda < \frac{1}{\sqrt[3]{2}}$. For these values, the repeated roots are (nonreal) imaginary: $\pm i \sqrt\lambda$.

(On the other hand, a deleted answer shows that for any even $k$ the polynomial has a single positive real root $\mu$ as a straightforward application of Descartes' Rule of Signs, and since the polynomial is even, a single negative real root, $-\mu$.)

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  • $\begingroup$ That's it @Travis. Thank you. $\endgroup$
    – drxy
    Dec 14, 2017 at 20:03

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