# Two diophantine equations with lots of unknowns

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?

$$A^2 + B^2=C^2 D^2$$ $$2 C^4 + 2 D^4 = E^2 + F^2$$

• 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
• @WillJagy Well, the OP would be satisfied with any nontrivla solution ... – Hagen von Eitzen Jun 28 '14 at 19:15
• 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

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 $2 \cdot 5^4 + 2 \cdot 13^4 = (13^2 - 5^2)^2 + (13^2 + 5^2)^2 = 144^2 + 194^2.$

• 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
• @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
• Ah 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

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}