<s>[INST] Rational solutions of quadratic Diophantine equation $ax^2+by^2+cz^2+du^2=v^2$? What do we know about the rational solutions of quadratic Diophantine equation  $ax^2+by^2+cz^2+du^2=v^2$ in five variables $x,y,z,u,v$?
I am looking for references/papers related to this equation.
 [/INST] $ax^2+by^2+cz^2+du^2=v^2$
Let assume $a+b+c+d=r^2.$
$p,q$ are arbitrary.
Substitute  $x=pt+1, y=qt+1, z=pt-1, u=qt-1, v=t+r$  to above equation, then we get  $$t = \frac{2(ap-qr^2+qa+2bq+qc-cp-r)}{(-ap^2-q^2r^2+q^2a+q^2c-cp^2+1)}.$$
Thus, we get a parametric solution below.
\begin{eqnarray}
  &x& = (a-3c)p^2+(2qc-2qr^2+2qa+4bq-2r)p+q^2a+q^2c+1-q^2r^2 \\ 
  &y& = (-a-c)p^2+(2qa-2qc)p+3q^2c-3q^2r^2+3q^2a+4bq^2+1-2qr \\ 
  &z& = (3a-c)p^2+(2qc-2qr^2+2qa+4bq-2r)p-q^2a-q^2c-1+q^2r^2 \\ 
  &u& = (c+a)p^2+(2qa-2qc)p+q^2c-q^2r^2+q^2a+4bq^2-1-2qr \\ 
  &v& = (-ra-rc)p^2+(2a-2c)p+2qc-2qr^2+2qa+4bq-q^2r^3+rq^2a-r+rq^2c.  
\end{eqnarray}
Example for $(a,b,c,d,r)=(1,2,3,3,3).$
\begin{eqnarray}
&x& = -8p^2+(-2q-6)p+1-5q^2\\
&y& = -4p^2-4qp-7q^2-6q+1\\
&z& = (-2q-6)p+5q^2-1\\
&u& = 4p^2-4qp+3q^2-6q-1\\
&v& = -12p^2-4p-2q-3-15q^2
\end{eqnarray}
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