# 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.

• 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). – Dietrich Burde Dec 21 '19 at 19:48
• Special cases as "consecutive integers" here, or here, etc. – Dietrich Burde Dec 21 '19 at 19:56
• Thanks for the references but they are very specialized cases. – mathisgood Dec 21 '19 at 20:02
• They are all valid cases of your equation because you allow $a=0$ etc. – Dietrich Burde Dec 21 '19 at 20:24

$$ax^2+by^2+cz^2+du^2=v^2$$
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 $$(a,b,c,d,r)=(1,2,3,3,3).$$ $$\begin{eqnarray} &x& = -8p^2+(-2q-6)p+1-5q^2\\ &y& = -4p^2-4qp-7q^2-6q+1\\ &z& = (-2q-6)p+5q^2-1\\ &u& = 4p^2-4qp+3q^2-6q-1\\ &v& = -12p^2-4p-2q-3-15q^2 \end{eqnarray}$$