# Three square sums solution structures

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!!

• 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... – individ Dec 15 '17 at 14:07
• 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...? – Rene Dec 15 '17 at 16:47

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

$$A^2+B^2+C^2=X^2+Y^2+Z^2=Q^2$$

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

We write as....

$$A=2ac$$

$$B=2bc$$

$$C=a^2+b^2-c^2$$

$$Q=a^2+b^2+c^2=x^2+y^2+z^2$$

$$X=2xz$$

$$Y=2yz$$

$$Z=x^2+y^2-z^2$$

And instead of solving the system, we solve it.

$$Q=a^2+b^2+c^2=x^2+y^2+z^2$$

You can write a simple solution:

$$a=q^2+s^2+k^2-p^2-t^2$$

$$b=q^2+s^2+k^2-p^2+t^2+2pt-2kt-2st-2qt$$

$$c=q^2+s^2+k^2+p^2-t^2+2tp-2kp-2sp-2qp$$

$$x=p^2+t^2+k^2-q^2-s^2-2kt-2pk+2sk+2qk$$

$$y=p^2+t^2-k^2-q^2+s^2-2ts-2ps+2ks+2qs$$

$$z=p^2+t^2-k^2+q^2-s^2-2tq-2pq+2kq+2sq$$

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

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

• 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) – Rene Dec 16 '17 at 18:49
• 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. – Rene Dec 18 '17 at 9:44
• Q can also be an irrational number ----??????? You can always keep the number of rational to integer and Vice versa. – individ Dec 18 '17 at 10:09
• 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. – Rene Dec 18 '17 at 10:56
• Substitute numbers are irrational. You have irrational values. I think you don't know what you want. – individ Dec 18 '17 at 11:02