<s>[INST] Show that $3x^4+4y^4=19z^4$ has no integer solution If $x,y,z$ are integers such that $xyz\neq 0$, prove or disprove that
$$3x^4+4y^4=19z^4$$
has no solution.
Maybe this is can use quadratic residue to solve it, maybe this equation seem is famous? because I have solve paper Introduced this.
 [/INST] COMMENT.- Just for fun. Do you have for the equation $3x^4+7y^4=19z^4$ the following irrational parameterization
$$\begin{cases}x=\left(\sqrt[4]{\dfrac{19}{3}}\sqrt{\dfrac{1-t^2}{1+t^2}}\right)z\\y=\left(\sqrt[4]{\dfrac{19}{7}}\sqrt{\dfrac{2t}{1+t^2}}\right)z\end{cases}$$ There is a value of $t$ such that both  $\sqrt[4]{\dfrac{19}{3}}\sqrt{\dfrac{1-t^2}{1+t^2}}$ and $\sqrt[4]{\dfrac{19}{7}}\sqrt{\dfrac{2t}{1+t^2}}$ be rational $\ne0$? 
If the equation $3x^4+7y^4=19z^4$ is not solvable, certainly NOT. But if the equation is solvable, certainly YES.  For example for the equation $10x^4+6y^4=285z^4$ we have the parameterization $$\begin{cases}x=\left(\sqrt[4]{\dfrac{285}{10}}\sqrt{\dfrac{1-t^2}{1+t^2}}\right)z\\y=\left(\sqrt[4]{\dfrac{285}{6}}\sqrt{\dfrac{2t}{1+t^2}}\right)z\end{cases}$$ and for $t=\pm\dfrac15\sqrt{\dfrac{179\pm12\sqrt{114}}{5}}$ (which is root of $125t^4-358t^2+125=0$) we get the solution $(x,y,z)=(3,5,2)$.
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