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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?

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    $\begingroup$ 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. $\endgroup$
    – GH from MO
    May 24 at 3:33

1 Answer 1

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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.

How did this problem arise?

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

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