<s>[INST] $a^2+b^2=c^2+d^2=1\ \ \text{and}\ \ ac+bd=0$, compute $ab+cd$ 
Consider that $\{a,b,c,d\}\subset \Bbb R$ and it is known that
  $$a^2+b^2=c^2+d^2=1\ \ \text{and}\ \ ac+bd=0.$$
Compute the value of $ab+cd$. (Edit: originally written with a typo as $ab+bd$), 

This is from a list of training problems used within the preparation for the "Cone-Sul" math olympics. I tried to solve using newton identities, but without success. Any help will be appreciated. Sorry if this is a duplicate.
 [/INST] Since 
$$ac+bd=0\implies a=-\frac{bd}{c}; c\neq0$$
Then
$$1=a^2+b^2=b^2[\frac{d^2}{c^2}+1]=b^2\cdot\frac{c^2+d^2}{c^2}=\frac{b^2}{c^2}$$
$$\implies c=\pm b\implies d=\mp a$$
The solution is given by
$$(a,b,c,d)=(a,b,\pm b,\mp a)$$
So
$$ab+cd=ab+(\pm b)(\mp a)=ab-ab=0$$
(For $c=0$, the case is trivial.)

Another solution
Let $a=mb$, $c=nd$.
$$ac+bd=0\implies mbnd+bd=0\implies bd=0 \text{ or }mn=-1$$
For $mn=-1$
$$a^2+b^2=b^2(m^2+1)=1\implies b^2=\frac{1}{m^2+1}$$
Similarly, $\displaystyle d^2=\frac{1}{n^2+1}$
Then
$$ab+cd=mb^2+nd^2=\frac{m}{m^2+1}+\frac{n}{n^2+1}=\frac{(m+n)(mn+1)}{(m^2+1)(n^2+1)}=0$$
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