I stumbled upon this puzzle when working on a problem for fun on my spare time and I seemed to hit a wall. (Not gonna reveal what problem it is just yet, should be a pleasent surprise if it gets solved :) )

What I noticed is the following:

If you have 2 three square sum solutions to a number $r^{2}$ (r $\in \mathbb{R}$):

${x_1}^{2}+{y_1}^{2}+{z_1}^{2} = {x_2}^{2}+{y_2}^{2}+{z_2}^{2} = r^{2}$

the two following conditions cannnot be true at the same time:

$x_1 > x_2 \ge y_2 > y_1 \ge z_2 > z_1 $

$x_1 -x_2 > y_2 - y_1 > z_2 - z_1$

$x_1,x_2,y_2,y_1,z_2,z_1$ can all be real numbers as well.

Using non tradition spherical coordinates (lattitude and longitude):

$ x_1 = r\cdot cos \phi_1 \cdot cos \lambda_1 \\ y_1 = r\cdot cos \phi_1 \cdot sin \lambda_1 \\ z_1 = r\cdot sin \phi_1 $

$ x_2 = r\cdot cos \phi_2 \cdot cos \lambda_2 \\ y_2 = r\cdot cos \phi_2 \cdot sin \lambda_2 \\ z_2 = r\cdot sin \phi_2 $

I was able to show the following statements about the angles:

$\lambda_2> \lambda_1 > \phi_2 > \phi_1 $

$\lambda_2 - \lambda_1 > \phi_2 - \phi_1$

$sin \lambda_2 - sin \lambda_1 > sin \phi_2 - sin \phi_1$

And that all angles must be smaller than $\frac{\pi}{4}$.

As well as:

$\lambda_1 > \frac{\pi}{8} + \frac{\lambda_1 - \phi_2}{2}$

But still seeking the contradiction some new ideas are here needed! Or maybe this was already done and someone can give me a link to a paper? (I did search for it for a long time and found nothing). Thanks in advance!!

  • $\begingroup$ Maybe better just to find a parameterization of the solutions to this equation? And then to draw conclusions? Moreover, such an assertion is not always... $\endgroup$ – individ Dec 15 '17 at 14:07
  • $\begingroup$ That was one of the first things I tried and failed. And I havent found one example in the literature which fits my question.. How would I start looking for the parametrization? And what do you mean by assertation is not always...? $\endgroup$ – Rene Dec 15 '17 at 16:47

For this type of equations - which has something in common...


Record regarding $Q$ - A different equation. And replace solving a system of equations, the solution of the same equation.

We write as....








And instead of solving the system, we solve it.


You can write a simple solution:







$p,t,k,q,s$ - any integer asked us.

A similar method can be used for other type of equations.

  • $\begingroup$ Thanks for the suggestion I will look into it! What I dont get is the following... It seems that in your parametrization Q has to be a natural number which is not sufficient enough for my problem. For my interests Q can also be an irrational number ... Does that change things? (Q² can of course only be a natural number) $\endgroup$ – Rene Dec 16 '17 at 18:49
  • $\begingroup$ Unfortunately I cant use your parametrization due to the reason mentioned above( Q can also be an irrational number) . Focusing on the distances between squares I was able to show :$y_2 + z_2 > x_1$ , $x_2> y_1 +z_1$, $z_2 + y_1 > x_2$ . But a contradiction is still needed. $\endgroup$ – Rene Dec 18 '17 at 9:44
  • $\begingroup$ Q can also be an irrational number ----??????? You can always keep the number of rational to integer and Vice versa. $\endgroup$ – individ Dec 18 '17 at 10:09
  • $\begingroup$ I think we have a missunderstanding. I need Q to be able to be an irrational number. And your parametrization only allows for Q to be a natural number. e.g 10²+5²+2²= 129. In this example Q would be $\sqrt 129$ . But in your parametrization a,b,c are natural numbers(since p,t,k,q,s are natural numbers) and therefore Q also has to be a natural number. $\endgroup$ – Rene Dec 18 '17 at 10:56
  • $\begingroup$ Substitute numbers are irrational. You have irrational values. I think you don't know what you want. $\endgroup$ – individ Dec 18 '17 at 11:02

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