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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.

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    $\begingroup$ There is a lot known. On this site you'll find almost $100$ posts related to it, with many special cases, e.g., for integers with arbitrary many squares here, or for $2$ square, $3$ squares and $4$ squares separately (e.g., here). $\endgroup$ – Dietrich Burde Dec 21 '19 at 19:48
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    $\begingroup$ Special cases as "consecutive integers" here, or here, etc. $\endgroup$ – Dietrich Burde Dec 21 '19 at 19:56
  • $\begingroup$ Thanks for the references but they are very specialized cases. $\endgroup$ – mathisgood Dec 21 '19 at 20:02
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    $\begingroup$ They are all valid cases of your equation because you allow $a=0$ etc. $\endgroup$ – Dietrich Burde Dec 21 '19 at 20:24
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$ax^2+by^2+cz^2+du^2=v^2$
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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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.

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  • $\begingroup$ How should I start such a parametrization? Should it look like a rational parametrization of sphere given here $\endgroup$ – mathisgood Dec 22 '19 at 1:35

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