How to solve this Diophantine equation? Can anyone say how one can find solutions to the Diophantine equation $$x^3+y^4=z^2$$ in General? Only a few triples of numbers have been found, and most likely this equation has infinitely many solutions.
Examples of triples: $(6,5,29),(2,1,3),(9,6,45)$...

"OP" enquired about integer coefficent's for the parametric 
solution for the equation $(x^2+y^4=z^2)$. "OP" just needs 
to substitute $k=(m/n)$ in the parametrization & the resulting 
solution after removing common factors is given below.
$x=6(u^3)(v^2)$
$y=(u^2)(v)(10m-27n)$
$z=(u^4)(v^2)(116m^2-540mn+621n^2)$
And $u=(4m^2-27n^2)$  & $v=(2m-5n)$
For $(m,n)=(13,5)$ we get:
$6^3+5^4=29^2$
