# Identity for the sum of three squares equal to sum of three squares

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.

• 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. – abiessu Dec 9 '13 at 18:18
• There is no identity of the product of $2$ squares, or $4$ squares type. – André Nicolas Dec 9 '13 at 18:24
• Presumably, @abiessu means adding two equations of the form $A^2+B^2=F^2$ and $C^2=D^2+E^2$? – Thomas Andrews Dec 9 '13 at 18:25
• @ThomasAndrews: good clarification point, you are right in your interpretation of the intent of my comment. – abiessu Dec 9 '13 at 18:33
• 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. – Dietrich Burde Dec 9 '13 at 22:15

Diophantine equation:

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

Has a solution:

$A=-p^2+2(t+k-q-2a)ps+2(4t^2+4k^2-2a^2-5q^2+q(k+t)+7a(q-t-k)-tk)s^2$

$B=-p^2+2(t+a-q-2k)ps+2(4t^2+4a^2-2k^2-5q^2+q(t+a)+7k(q-a-t)-ta)s^2$

$C=-p^2+2(k+a-q-2t)ps+2(4k^2+4a^2-2t^2-5q^2+q(k+a)+$$7t(q-a-k)-ka)s^2$

$D=-p^2+2(t+k+a-q)ps+2(4t^2+4k^2+4a^2-5q^2+q(t+k+a)-tk-ta-ka)s^2$

$E=p^2+4(a+k+t-q)ps+2(2t^2+2k^2+2a^2-7q^2+5q(t+k+a)-5(tk+ta+ka))s^2$

$F=-p^2+2(a+k+t-4q)ps+2(4t^2+4k^2+4a^2+q^2-5q(t+k+a)-tk-ta-ka)s^2$

more:

$A=p^2+2(t+k+2a-q)ps+2(2a^2+q^2-q(t+k)+3a(t+k-q)+tk)s^2$

$B=p^2+2(t+2k+a-q)ps+2(2k^2+q^2-q(t+a)+3k(t+a-q)+ta)s^2$

$C=p^2+2(2t+k+a-q)ps+2(2t^2+q^2-q(k+a)+3t(k+a-q)+ka)s^2$

$D=p^2+2(t+k+a-q)ps+2(q^2+tk+ta+ka-q(t+k+a))s^2$

$E=p^2+4(t+k+a-q)ps+2(2t^2+2k^2+2a^2+q^2+3(tk+ta+ka)-3q(t+k+a))s^2$

$F=p^2+2(t+k+a)ps+2(tk+ta+ka-q^2+q(t+k+a))s^2$

$p,s,t,k,a,q$ - 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 $ab+cd+ef = 0$