# Prove that there are no rational solutions

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.

• 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. – Eric Towers Oct 20 at 22:21
• The Rational Root Theorem would give you a finite set of possible roots, which you could plug in one by one. – David Diaz Oct 20 at 22:22
• No rational root theorem. I believe modular arithmetic is fine. – LetmeKnow Oct 20 at 22:22

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.
• @LetmeKnow : If $5 \mid x$ then $5 \mid y$ and $x/y$ is not in lowest terms. – Eric Towers Oct 20 at 23:01