Crossposted from [MSE](https://math.stackexchange.com/questions/1401214/what-is-the-type-of-the-surfaces-x5-y5-z2-x-0-and-x5-y5-z2).


I am interested what is the type of the surfaces over the
rationals
$$ x^5 - y^5 + z^2 + x=0$$


and


$$ x^5 - y^5 + z^2 + x+1=0$$


Magma's `KodairaEnriquesType(S : CheckADE:=true);` fails
to compute it.


According to Magma they are not rational.


Partial answers (e.g. it is not $X$) or approaches
how to compute with a CAS are welcome.



What is the type of the surfaces $x^5 - y^5 + z^2 + x=0$ and $x^5 - y^5 + z^2 + x+1=0$?
Both are $K3$: they are double planes ramified at sextics (the order $5$ curves in $x$, $y$ and the line at infinity). One should only check carefully that these sextics have simple singulrities. I use the coordinates $x=z\_1/z\_0$, $y=z\_2/z\_0$ and multiply by $z\_0$ (the line at infinity).


The former: $z\_0z\_1^5-z\_0z\_2^5+z\_0^5z\_1=0$. Obviously, there are no singular points in the affine part ($z\_0\ne0$), and the residual quintic intersects the component $z\_0=0$ at five distinct points, thus forming five $A\_1$, which are all simple.


The latter: $z\_0z\_1^5-z\_0z\_2^5+z\_0^5z\_1+z\_0^6=0$. The same conclusion applies literally.


