I have been searching on Internet about a general solution to the Diophantine equation $A^2+B^2+C^2=D^2+E^2+F^2$. However, I haven't found anything. Can anyone help me? I'm looking for a identity or something like that. Very important is that must be a general solution, it must contain all the odds with all the possible numbers.

  • 1
    $\begingroup$ Consider the paired-Pythagorean solution: $$(A^2+B^2)+C^2=(D^2+E^2)+F^2$$ This does not necessarily capture the entire solution space, but it should be included in it. $\endgroup$ – abiessu Dec 9 '13 at 18:18
  • $\begingroup$ There is no identity of the product of $2$ squares, or $4$ squares type. $\endgroup$ – André Nicolas Dec 9 '13 at 18:24
  • $\begingroup$ Presumably, @abiessu means adding two equations of the form $A^2+B^2=F^2$ and $C^2=D^2+E^2$? $\endgroup$ – Thomas Andrews Dec 9 '13 at 18:25
  • $\begingroup$ @ThomasAndrews: good clarification point, you are right in your interpretation of the intent of my comment. $\endgroup$ – abiessu Dec 9 '13 at 18:33
  • $\begingroup$ There are references about triples of squares with equal sum, and triples of sixth powers with equal sum, e.g., Andrew Bremner: a geometric approach to equal sums of sixth powers. $\endgroup$ – Dietrich Burde Dec 9 '13 at 22:15

Diophantine equation:


Has a solution:














$p,s,t,k,a,q$ - integers asked us.


You should be happy with some of the results available at https://sites.google.com/site/tpiezas/004

$(a+b)^2 + (c+d)^2 + (e+f)^2 = (a-b)^2 + (c-d)^2 + (e-f)^2$ where $ab+cd+ef = 0$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.