crossoverman9b

2022-06-22

How many solutions are there to this equation?

$\begin{array}{rl}{x}^{2}-{y}^{2}& =z\\ {y}^{2}-{z}^{2}& =x\\ {z}^{2}-{x}^{2}& =y\end{array}$

$\begin{array}{rl}{x}^{2}-{y}^{2}& =z\\ {y}^{2}-{z}^{2}& =x\\ {z}^{2}-{x}^{2}& =y\end{array}$

klemmepk

Beginner2022-06-23Added 16 answers

We have

$\begin{array}{rl}{x}^{2}-{y}^{2}& =z\\ {y}^{2}-{z}^{2}& =x\\ {z}^{2}-{x}^{2}& =y\end{array}$

Adding the first two equations, we get that

${x}^{2}-{z}^{2}=z+x\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}z=-x\text{or}x-z=1$

Similarly, we get that

$x=-y\text{or}y-x=1$

$y=-z\text{or}z-y=1$

Hence, there are $8$ possible choices of getting $3$ equations. But the choice

$x-z=1\text{and}y-x=1\text{and}z-y=1$

gives no solution. The rest of the seen choices give the following solutions

$(0,0,0);(-1,0,1);(0,1,-1);(1,-1,0)$

$\begin{array}{rl}{x}^{2}-{y}^{2}& =z\\ {y}^{2}-{z}^{2}& =x\\ {z}^{2}-{x}^{2}& =y\end{array}$

Adding the first two equations, we get that

${x}^{2}-{z}^{2}=z+x\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}z=-x\text{or}x-z=1$

Similarly, we get that

$x=-y\text{or}y-x=1$

$y=-z\text{or}z-y=1$

Hence, there are $8$ possible choices of getting $3$ equations. But the choice

$x-z=1\text{and}y-x=1\text{and}z-y=1$

gives no solution. The rest of the seen choices give the following solutions

$(0,0,0);(-1,0,1);(0,1,-1);(1,-1,0)$

Roland Waters

Beginner2022-06-24Added 6 answers

Sum both sides and get $x+y+z=0$, so you're on a plane.

Further hint:

$z=({x}^{2}-{y}^{2})=(x-y)(x+y)=-z(x-y)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\dots $

Further hint:

$z=({x}^{2}-{y}^{2})=(x-y)(x+y)=-z(x-y)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\dots $