Let $m$ be a fixed rational number. Are there any rational solutions to this equation

$$x^{8}-m^{2}x - m = 0$$ ?

My attempt: The equation can be rewritten as $$x^{9} = mx(mx+1)$$

Hence we should have $mx = a^{9}$ and $mx+1 = b^{9}$ for some rational $ab$, so that $b^9 - a^9 = 1$. By Fermat's Last Theorem, this yields $ab=0$ as the only rational solutions.

However, i'm not sure if $mx$ and $mx+1$ should necessarily be of the form that i stated.

  • $\begingroup$ Why the downvote please ? $\endgroup$ – Knowhow Oct 3 '16 at 10:10
  • $\begingroup$ Please would the downvoter explain his/her action ? $\endgroup$ – Knowhow Oct 3 '16 at 10:28
  • 2
    $\begingroup$ Downvote is not mine but let me guess the reason - you did not provide any motivating background, why should anybody be interested in this equation $\endgroup$ – მამუკა ჯიბლაძე Oct 3 '16 at 10:41

Considering your equation as a quadratic equation in $m$, it is equivalent to $$y^2 = 4 x^9 + 1$$ (with $y = 2xm + 1$). A solution will in particular give a rational point on the elliptic curve $E \colon y^2 = 4x^3 + 1$. But the only rational points on $E$ are the point at infinity and the two points with $x = 0$. ($E$ is isomorphic to the curve 27a3 in the Cremona database, where you can check that its group of rational points has order 3.) So $x = 0$ and hence $m = 0$ in your equation.

(This is basically equivalent to your attempt: the equation $y^2 = 4 x^n + 1$ has nontrivial rational solutions if and only if the $n$th Fermat equation has nontrivial solutions.)

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