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Is there any solution to the following system of diophantine equations? $$ \left\{\begin{array}{l} 2.a^2 = b^2+c^2+d^2 \\ a^2 = e^2+f^2+g^2 , & \mbox{with }((a,b,c,d,e,f,g)>2)\in N\mbox{ and differents among them} \end{array} \right. $$

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  • $\begingroup$ Why wouldn't there be infinite solutions? $\endgroup$
    – Don Larynx
    Commented Aug 26, 2013 at 3:55
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    $\begingroup$ First part. Of course, solutions exist: $13^2 = 3^2+4^2+12^2, \quad 2 \cdot 13^2 = 7^2+8^2+15^2$. $\endgroup$
    – Oleg567
    Commented Aug 26, 2013 at 4:12

4 Answers 4

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Yes there are solutions. For your first, note that if $a,b,c$ is a Pythagorean triangle with $a$ the hypotenuse and $a=d$ you have a solution. There are more. Since the equations are homogeneous (all the terms are squared) any multiple of a solution is another solution. The easy one is then $a=15,b=9,c=12,d=15,e=f=10,g=5$ Maybe the common values violate "differents among them" but I don't know how to read that. But then you can use Oleg567's solution and scale up.

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Let $c=d=e$, and let $f=g$. This gives us the following system of equations:

$2 \left( a^2-c^2\right)=b^2$

$a^2-c^2=2f^2$

If you further make the substitution $c=a-2$ and $b=2f$, we can combine both equations to yield one equation:

$2 \left( a-1\right)=f^2$

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In the system of equations:

$\left\{\begin{aligned}&X^2+Y^2+Z^2=2R^2\\&F^2+G^2+T^2=R^2\end{aligned}\right.$

Solutions have the form:

$X=4(c^2+f^2+k^2)k^2-(2c^2-2d^2+f^2)f^2-(d^2-c^2)^2$

$Y=4kd(2k^2+d^2-c^2-f^2)$

$Z=4(f^2+c^2-k^2)k^2+(f^2+2c^2-2d^2)f^2+(d^2-c^2)^2$

$F=4kc(2k^2+d^2-c^2-f^2)$

$G=4kf(2k^2+d^2-c^2-f^2)$

$T=(2k^2+d^2)^2+(c^2+f^2)^2-2(d^2+4k^2)(c^2+f^2)$

$R=4(k^2+d^2)k^2+(2c^2+f^2-2d^2)f^2+(d^2-c^2)^2$

$k,c,d,f$ - integers and sets us.

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set of equations:

$\left\{\begin{aligned}&X^2+Y^2+Z^2=2(R^2+W^2)\\&F^2+G^2+T^2=R^2+W^2\end{aligned}\right.$

has solutions

$X=4(t^2+q^2-a^2-v^2)a^2+(v^2+t^2+q^2-z^2)^2$

$Y=4(v^2-a^2-t^2-q^2)a^2+(v^2+t^2+q^2-z^2)^2$

$Z=4az(v^2+t^2+q^2-2a^2-z^2)$

$F=4(a^2+z^2-2t^2-2q^2)a^2+(v^2+t^2+q^2-z^2)^2$

$G=4aq(v^2+t^2+q^2-2a^2-z^2)$

$T=4at(v^2+t^2+q^2-2a^2-z^2)$

$R=4(z^2+a^2-2v^2)a^2+(v^2+t^2+q^2-z^2)^2$

$W=4av(v^2+t^2+q^2-2a^2-z^2)$

$a,v,t,q,z$ - integers and sets us.

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