Rational solutions of quadratic Diophantine equation $ax^2+by^2+cz^2+du^2=v^2$?

Question

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.

Answer 1

indefinite forms in at least five variables are isotropic. There is an integer solution, not all variables zero.

If you are not worried about common divisors, you can parametrize all rational solutions by stereographic projection around a fixed solution.

Answer 2

$ax^2+by^2+cz^2+du^2=v^2$
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}