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Let $a,b,c>0$ be pairwise relatively prime and $n>2$ be odd. Can the equation, $a^n\cdot x^2+b^n\cdot x+c^n=0$, have rational roots $x$?

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Start with finding those roots, with: $$x_{1,2}=\frac{-b^n\pm\sqrt{b^{2n}-4a^nc^n}}{2a^n}$$ –  Salech Alhasov Mar 4 '12 at 18:55
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Alternatively, I started to think of this problem as applying en.wikipedia.org/wiki/Rational_root_theorem –  user2468 Mar 4 '12 at 19:18

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The roots are $$ \frac{-b^n + \sqrt{b^{2n}-4a^n c^n}}{2a^n}, \frac{-b^n - \sqrt{b^{2n}-4a^n c^n}}{2a^n} $$ Let's rule out the trivial case $b^{2n} = 4a^n c^n$. If $\sqrt{b^{2n}-4a^n c^n}$ is rational then so do the roots. The question boils down to showing the existence of $r \in \mathbb{Q}$ such that $r^2 = b^{2n}-4a^n c^n$ under the conditions your gave for $a,b,c,n$. Not sure yet how to answer this.

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So we have $b^{2n}-r^2=4(ac)^n$ or $(b^n-r)(b^n+r)=4(ac)^n$, where $r$ is rational. Is this possible? –  Craig Feinstein Mar 4 '12 at 19:54
    
@CraigFeinstein Exactly. But I can't go anywhere from there. –  user2468 Mar 4 '12 at 20:08
    
@CraigFeinstein At least this implies that $r$ is an integer. For suppose $r=p/q$ is a reduced fraction and a prime $m$ divides $q$ then $(q b^n-p)(q b^n+p) = 4q^2(ac)^n$ and $m$ divides the RHS but not the LHS. –  WimC Mar 4 '12 at 20:48

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