Let $k>1$ be an integer and let $x_1,x_2,y_1,y_2,z_1$ and $z_2$ be the unknowns.

How can I solve for the unknowns given the following equations?

  1. $[(2k-1)+x_1]^2+[(2k^2-2k)+y_1]^2=[(2k^2-2k+1)+z_1]^2+1$
  2. $[(2k-1)+x_2]^2+[(2k^2-2k)+y_2]^2=[(2k^2-2k+1)+z_2]^2+1$
  3. $x_1+x_2=2k-1$
  4. $y_1+y_2=2k^2-2k$
  5. $z_1+z_2=2k^2-2k+1$

Note: I know that $x_1=1, x_2=2k-2, y_1=2k-1, y_2=2k^2-4k+1,z_1=2k-1$ and $z_2=2k^2-4k+2$ is a solution for the system of equation. Unfortunately I cant show how to arrive on it. I think I am missing some steps on showing what I want to show.

My attempt is this:

First using 3,4, and 5 I can express $x_2,y_2$ and $z_2$ in terms of $x_1,y_1$ and $z_1$ and $k$.

Next, I will used the result in first step to change $x_2,y_2$ and $z_2$ in terms of $x_1,y_1$ and $z_1$ and $k$ in equation 2.

Then I will expand 1 and 2 yielding two equations involving $x_1,y_1$ and $z_1$. But after this, I don't know how to proceed and thus I need your help. Thanks a lot.

  • $\begingroup$ do you have an example for $$k$$? $\endgroup$ – Dr. Sonnhard Graubner Aug 24 '15 at 16:37
  • $\begingroup$ @Dr.SonnhardGraubner Thanks for the comment Dr. I am working on a general $k$, that is I want to solve the equation for all integer $k>1$. Since the solution I know is valid for all integer $k>1$. Is it advisable to consider a particular $k$? $\endgroup$ – Jr Antalan Aug 24 '15 at 16:43

For what it's worth, getting rid of $k$ gives you the two simultaneous equations:

$$(2x_1 + x_2)^2 + (2y_1 + y_2)^2 = (2z_1 + z_2)^2 + 1$$


$$(x_1 + 2x_2)^2 + (y_1 + 2y_2)^2 = (z_1 + 2z_2)^2 + 1$$

together with the constraint

$$y_1 + y_2 + 1 = z_1 + z_2.$$

Hint: Subtract the second equation from the first. Then square the third equation.

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