How can I prove that this polynomial has distinct roots?

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 $k$ ?

• 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. – Arthur Dec 14 '17 at 18:04
• Does "roots" here refer only to real roots, or to complex roots, too? – Travis Willse Dec 14 '17 at 18:05
• @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? – drxy Dec 14 '17 at 18:25
• @Travis both of them. – drxy Dec 14 '17 at 18:25
• 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. – Arthur Dec 14 '17 at 18:30

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, $-\mu$.)