In Heath-Brown's 2002 paper, "Rational points on curves and surfaces", he states 


"We may observe that if $d \geq 3$, the surface $$x\_1^d + x\_2^d - x\_2^{d-2} x\_3 x\_4 = 0$$
is absolutely irreducible, and contains no lines other than those in the planes $x\_2 = 0, x\_3 = 0$, and $x\_4 = 0$."


I am wondering how he is able to be so conclusive as to state no other lines lie on the surface. Can anyone explain why this is 'obvious'?



How does one know the following surface contains no other lines?
A line is given by a pair of equations:
\begin{equation\*}
a\_1 x\_1 +a\_2 x\_2+a\_3 x\_3 + a\_4 x\_4=0, \qquad 
b\_1 x\_1 + b\_2 x\_2 + b\_3 x\_3 + b\_4 x\_4=0.
\end{equation\*}
Suppose this line is on $X$.
If the minor $a\_3 b\_4-a\_4 b\_3$ is non-zero, then we may rewrite the equations of the line as
\begin{equation\*}
x\_3=a x\_1+ b x\_2, \qquad x\_4=c x\_1 + e x\_2.
\end{equation\*}
Substituting into the equation of the surface $X$ we see that the expression
\begin{equation\*}
x\_1^d+x\_2^d-x\_2^{d-2}(ax\_1 +b x\_2)(c x\_1 + e x\_2)
\end{equation\*}
vanishes as polynomial in x\_1 and x\_2. This is clearly impossible by considering the coefficient of $x\_1^d$. Hence the minor $a\_3 b\_4-a\_4 b\_3=0$. So we can suppose that one of the equations of the line is of the form $a x\_1 + b x\_2=0$. Assume that the line does not lie on either of the planes $x\_1=0$ or $x\_2=0$. Thus neither of $a$ or $b$ is zero and we may rewrite this equation as $x\_1=c x\_2$. Substituting in the equation for $X$ we see that the line lies on the conic
\begin{equation\*}
(1+c^d) x\_2^2 - x\_3 x\_4=0.
\end{equation\*} 
If $1+c^d \ne 0$, then the conic is irreducible and so does not contain a line. Hence $1+c^d=0$ and so the line is on one of the planes $x\_3=0$ or $x\_4=0$. 


