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.
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.