Consider the quartic system in four variables $a,b,c,d\in\mathbb R$:
$$-(c^2-d^2)(a^2-b^2)=2(ad-bc)(bd+ac).$$



> 
> Does this system admit rational solution with $$abcd(c^2-d^2)(a^2-b^2)(a^2-c^2)(b^2-d^2)\neq0?$$
> 
> 
> 



> 
> Is there any easy way to compute for rational solution?
> 
> 
> 



Non-trivial solutions for $-(c^2-d^2)(a^2-b^2)=2(ad-bc)(bd+ac)$?
There is no such solution. Let
$$
Q(a,b,c,d) = 2(ad-bc)(bd+ac) + (a^2-b^2)(c^2-d^2)
$$
be the difference between the two sides of the equation,
so we seek to solve $Q(a,b,c,d) = 0$. This is a quadratic equation
in each variable, so in a rational solution the discriminant of $Q$
with respect to each variable is a square. We choose $d$
(the others work the same way), and find
$$
{\rm disc}\_d(Q) = (2c)^2 (2a^4 - 4a^3b + 4ab^3 + 2b^4).
$$
Thus either $c=0$ or the second factor is a square.
The former is possible (with $d=0$ as well) but the problem statement
forbids it. The latter gives rise to an elliptic curve,
which turns out to be curve
[40.a3](https://www.lmfdb.org/EllipticCurve/Q/40/a/3)
with rank $0$ over the rational numbers;
its four torsion points correspond to $a=\pm b$
(each of which gives rise to two points on the curve).
Since the problem statement also forbids $a^2=b^2$ we're done.


How did this problem arise?


