condition for tuple of non-zero integers $(a,b,c,d)$ such that $ad+bc=ac-bd=ab+cd=a^2-b^2+c^2-d^2=0$ As the title says, what would be the condition for tuple of non-zero integers $(a,b,c,d)$ such that $ad+bc=ac-bd=ab+cd=a^2-b^2+c^2-d^2=0$? Would there be infinitely many tuples that satisfy the condition above?
 A: Suppose that there is a set of non-zero integers $(a,b,c,d)$.
First, since $b\not=0$, we have
$$ac=bd\Rightarrow d=\frac{ac}{b}\tag1$$
Moreover, we have
$$ad+bc=ac-bd\Rightarrow (a+b)d=c(a-b).$$
If $a+b=0$, then we have $a-b=0$ because $c\not=0$. This leads $a=b=0$. This is a contradiction. Hence, we have $a+b\not =0$.
Hence, we have $$d=\frac{a-b}{a+b}c\tag2$$
Hence, since $c\not=0$, from $(1),(2)$ we have
$$\frac{ac}{b}=\frac{a-b}{a+b}c\Rightarrow ac(a+b)=bc(a-b)\Rightarrow a^2+b^2=0\Rightarrow a=b=0.$$
This is a contradiction. Hence, we know that there is no sets of non-zero integers $(a,b,c,d)$.
A: Let $i = \sqrt{-1}$. Then, $(a+bi)(c+di) = (ac-bd)+(ad+bc)i = 0+0i = 0$, so either $a+bi = 0$ or $c+di = 0$, i.e. either $a = b = 0$ or $c = d = 0$. 
If $a = b = 0$, then we are left with $cd = c^2-d^2 = 0$, from which we get $c = d = 0$. 
If $c = d = 0$, then we are left with $ab = a^2-b^2 = 0$, from which we get $a = b = 0$. 
Therefore, the only solution to $ad+bc=ac-bd=ab+cd=a^2-b^2+c^2-d^2=0$ is the trivial solution $(a,b,c,d) = (0,0,0,0)$. 
