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Prove that $2r^4 +20r^2 = 15r^3 + 15r - 6$ has no rational solutions without solving for $r$.

My first thought was using remainders upon division, but I'm not sure how to apply this with variables.

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    $\begingroup$ What tools are you allowed to use? The rational roots theorem? Modular arithmetic? There are a number of ways to proceed, so some bounds on the acceptable method are needed. $\endgroup$ – Eric Towers Oct 20 at 22:21
  • $\begingroup$ The Rational Root Theorem would give you a finite set of possible roots, which you could plug in one by one. $\endgroup$ – David Diaz Oct 20 at 22:22
  • $\begingroup$ No rational root theorem. I believe modular arithmetic is fine. $\endgroup$ – LetmeKnow Oct 20 at 22:22
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Hint: There are no solutions mod $5$.

Indeed, if $r=x/y$ then $2x^4 \equiv -6 y^4$ or $x^4 \equiv -3 y^4$. Now use Fermat's theorem.

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  • $\begingroup$ How do I mod this equation? $\endgroup$ – LetmeKnow Oct 20 at 22:25
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    $\begingroup$ @LetmeKnow : If $5 \mid x$ then $5 \mid y$ and $x/y$ is not in lowest terms. $\endgroup$ – Eric Towers Oct 20 at 23:01

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