Is it possible (tractable) to determine if the following system of equations has any nontrivial solutions (ie, none of the unknowns are zero) in the domain of integers?
$$A^2 + B^2=C^2 D^2$$ $$2 C^4 + 2 D^4 = E^2 + F^2$$
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both have infinitely many nontrivial. with a little effort, you can force $C,D > 1,$ infinitely often – Will Jagy Jun 28 '14 at 19:12
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@WillJagy Well, the OP would be satisfied with any nontrivla solution ... – Hagen von Eitzen Jun 28 '14 at 19:15
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Although each many have infinitely many nontrivial solutions, do any of these non-trivial solutions use the same values for the same unknowns? – Mark Jun 28 '14 at 19:20
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for the second one, take $C > D > 0,$ then $$ E = C^2 - D^2, \; \; \; F = C^2 + D^2 $$
If you wanted a system, take any $C,D \equiv 1 \pmod 4$ distinct primes, such as $5,13.$ We get the Pythagorean triple $16^2 + 63^2 = 65^2 = 5^2 13^2.$ Then $2 \cdot 5^4 + 2 \cdot 13^4 = (13^2 - 5^2)^2 + (13^2 + 5^2)^2 = 144^2 + 194^2.$
Will Jagy
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The question is about one Diophantine problem with a system of equations, not two separate problems. – Hagen von Eitzen Jun 28 '14 at 19:25
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@HagenvonEitzen, I guess you're right. Does not change anything much, you can do the system in two stages. – Will Jagy Jun 28 '14 at 19:27
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1Ah yes, now I get it. For example, $A=3$, $B=4$, $CD=5$, so let $C=5$, $D=1$ and then $E=24$, $F=26$. – Hagen von Eitzen Jun 28 '14 at 19:29
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To solve,
$$A^2+B^2=C^2 D^2\\ (2C)^4+(2D)^4=E^2+F^2$$
Choose,
$$\begin{aligned} A&=2(ac-bd)(ad+bc)\\ B&=(ac-bd)^2-(ad+bc)^2\\ C&=a^2+b^2\\ D&=c^2+d^2\\ E&=(a^2+b^2 )^2-(c^2+d^2 )^2\\ F&=(a^2+b^2 )^2+(c^2+d^2 )^2\\ \end{aligned}$$
Tito Piezas III
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Bumblebee
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