Diophantine eqn, general solution?

Question

Here's the equation: $$ 4 \left( x^2+y^2-z^2 \right)=\left( 2k+1 \right) \left( x+y-z \right) $$ Is there a nontrivial solution for this in integers? If not, why not? If there is, can a general solution be created?

Answer 1

This is an old post, but anyway. A broad (but not complete) solution to,

$$x^2 + y^2 - z^2 = n(x + y - z)$$

can be given as,

$$n =p^4+p^2-1$$

$$x = p^2 - p - 1$$

$$y = p^2 + p - 1$$

$$z = 2p^2$$

Answer 2

Let $x + y - z = u$ and $\eta = 2k+1$, we have

$$\begin{align} & 4(x^2 + y^2 - z^2 ) = \eta(x+y-z)\\ \iff & 4(x^2 + y^2 - (x+y - u)^2) = \eta u\\ \iff & 4( -2xy + 2u(x+y) - u^2 ) = \eta u\\ \iff & 8(x-u)(y-u) = (4u-\eta) u\\ \end{align} $$ As long as RHS of last expression is divisible by $8$, the last equation has solution.

Since $\eta$ is an odd number, we find $u$ itself is divisible by $8$. Let $u = 8s$, the equation reduces to

$$(x - 8s)(y - 8s) = (32s - \eta)s$$

This means the general solution of the equation at hand can be parametrized as $$ \begin{cases} x &= 8s + p\\ y &= 8s + \frac{\Delta}{p}\\ z &= x + y - 8s = 8s + p + \frac{\Delta}{p}\\ \end{cases} $$ where $s$ is any integer, $\Delta = s(32s - 2k - 1)$ and $p$ is any factor of $\Delta$.

For example, we can take $p = \pm s$ and obtain following two family of solutions

$$\begin{array}{lcrrr} (x,y,z) &=& ( 9s, & 40s-2k-1,& 41s-2k-1 )\\ (x,y,z) &=& ( 7s, &-24s+2k+ 1,& -25s +2k+1 ) \end{array} $$ which are available for any $k$.

Answer 3

If there was a need for a simpler entry to the equation:

$$4(x^2+y^2-z^2)=q(x+y-z)$$

Then write a simple solution using any two integers - $k,t$.

$$x=(8k-1)t$$

$$y=qk-8k(4k-1)t$$

$$z=qk-(32k^2-8k+1)t$$

In this form it is easier and for all coefficients.