Is there an ordered 4-tuple of rational numbers $(a,b,c,d)$ such that $(b,d)\ne(0,0)$ and $2a^2+3b^2+30c^2+45d^2=2$?


The former (deleted) question was just about cases $(a,b,c,d)\ne(1,0,0,0)$ but it was quite silly :( I apologize and I think now it makes sense. I guess there is a canonical proof of nonexistence or an algorithm for examples.



sum of four squares with some coefficients
Actually infinitely many, and a parametrization of all rational solutions is e.g.


$a:=\frac{3B^2+30C^2+45D^2-2E^2}{3B^2+30C^2+45D^2+2E^2}$


$b:=\frac{4BE}{3B^2+30C^2+45D^2+2E^2}$


$c:=\frac{4CE}{3B^2+30C^2+45D^2+2E^2}$


$d:=\frac{4DE}{3B^2+30C^2+45D^2+2E^2}$.


