4-dim. generalization of $ab+ac+bc=0$

Question

The equation $ab+ac+bc=0$ can be parameterized by $(a,b,c)=\lambda(-pq, p(p+q), q(p+q))$.

Is there a (similar) parameterization for $ab+ac+ad+bc+bd+cd=0$?

What about the 5-dimensional case?

Edit: Proof of the parametrization in the 3-dim. case:

Write $x=a+b, y=a-b$. Then $ab+ac+bc=0 <=> (x-y)(x+y) +2(x+y)c+2(x-y)c=0 <=> (x+2c)^2-y^2-4c^2=0$. Then use the pythogorean triple parametrization to get $x+2c=\lambda(m^2+n^2), y=\lambda(m^2-n^2), c=\lambda mn$.
Now $a=(x+y)/2= \lambda(m^2+n^2-2mn+m^2-n^2)/2=m(m-n), b=(x-y)/2=\lambda n(n-m), c=\lambda mn$. Now set $m=p$ and $n=q+p$ and the result follows (with $b$ and $c$ switched)

Edit2: yes, I am looking for a parameterization in the integers. Sorry, that I didn't make myself clear.

Answer 1

This kind of equations are solved everything. And even what some coefficients. equation

$ab+ac+ad+bc+bd+cd=0$

Has the following solutions:

$a=(t+k)ps-(t^2+tk+k^2)s^2$

$b=(t-2k)ps+(k^2+tk-t^2)s^2$

$c=-3p^2+4(t+k)ps-(t^2+3tk+k^2)s^2$

$d=(k-2t)ps+(t^2+tk-k^2)s^2$

And more.

$a=3p^2-(t+k)ps$

$b=3p^2-(t+4k)ps+(t+k)ks^2$

$c=-3p^2+2(t+k)ps-tks^2$

$d=3p^2-(4t+k)ps+(t+k)ts^2$

Although I like this equation:

$a^2+b^2+c^2+d^2=ab+ac+ad+bc+bd+cd$

And solutions:

$a=p^2-2tps+(3k^2+t^2)s^2$

$b=p^2+3(k+t)^2s^2$

$c=4p^2-4tps+4t^2s^2$

$d=3p^2+6kps+(3k^2+t^2)s^2$

And more:

$a=7p^2+2(3k-t)ps+(3k^2+6kt+7t^2)s^2$

$b=7p^2+6(k-t)ps+3(k-t)^2s^2$

$c=p^2+2(3k+t)ps+(3k+t)^2s^2$

$d=3p^2-6(k+t)ps+(3k^2+6kt+7t^2)s^2$

number $t,k,p,s$ integers and set us.

Answer 2

Solution of above inquiry regarding parametric form for equations (given below) for four and five variables is shown:

$ab+ac+ad+bc+bd+cd=0$

$ab+ac+bc+bd+ad+cd+ea+eb+ec+ed=0$

For $ab+ac+ad+bc+bd+cd=0$

$(a,b,c,d)=((p^2+2p+1),(2p+2),-(2p+2),(4))$

$p=2$ we get $(a,b,c,d)=(9,6,-6,4)$

For $ab+ac+bc+bd+ad+cd+ea+eb+ec+ed=0$

$(a,b,c,d,e)=((3p+7),-3(p-3),(3p-5),-(3p+5),(3p^2-p+12))$

For $p=0$ we get: $(a,b,c,d,e)=(7,9,-5,-5,12)$