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?
- $[(2k-1)+x_1]^2+[(2k^2-2k)+y_1]^2=[(2k^2-2k+1)+z_1]^2+1$
- $[(2k-1)+x_2]^2+[(2k^2-2k)+y_2]^2=[(2k^2-2k+1)+z_2]^2+1$
- $x_1+x_2=2k-1$
- $y_1+y_2=2k^2-2k$
- $z_1+z_2=2k^2-2k+1$
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.
For what it's worth, getting rid of $k$ gives you the two simultaneous equations:
$$(2x_1 + x_2)^2 + (2y_1 + y_2)^2 = (2z_1 + z_2)^2 + 1$$
and
$$(x_1 + 2x_2)^2 + (y_1 + 2y_2)^2 = (z_1 + 2z_2)^2 + 1$$
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.