<s>[INST] 4-dim. generalization of $ab+ac+bc=0$ 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.
 [/INST] 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.
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