# Non-trivial solutions for $-(c^2-d^2)(a^2-b^2)=2(ad-bc)(bd+ac)$?

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?

• Nice question. In the $\neq 0$ condition, you can omit the factor $(ad-bc)(bd+ac)$, since this is automatically nonzero if the equation holds and the other factors are nonzero. May 24 at 3:33

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 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.
• Not a very profound reason: I was looking for real solutions to $l a^2+ m c^2=l b^2+ m d^2=2(lab+mcd)$. May 25 at 3:49