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

  • $\begingroup$ Why wouldn't there be infinite solutions? $\endgroup$
    – Don Larynx
    Commented Aug 26, 2013 at 3:55
  • 2
    $\begingroup$ 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


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.


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

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


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

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


In the system of equations:


Solutions have the form:








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


set of equations:


has solutions









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


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