Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to solve the following set of equations for real values of $x,y$ and $z$?

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

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

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

$(0,0,0)$ is an obvious one solution.

share|improve this question

closed as off-topic by mau, Magdiragdag, Daryl, TMM, Lost1 Jan 21 at 11:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – mau, Magdiragdag, Daryl, TMM, Lost1
If this question can be reworded to fit the rules in the help center, please edit the question.

1  
Add all the to get $x+y+z=0$ and go on –  Blah Jan 21 at 9:08
    
$(0,1,-1)$ is another solution, and any obivious permutation of this. Generally $x,y,z$ are related by $x+y+z=0,\;$ you can see this by adding the equations. –  gammatester Jan 21 at 9:09
    
What have you tried and where are you stuck? If you tell us, it will help us give an answer at the appropriate level of experience. Furthermore, this looks like homework; if it is, please add the (homework) tag. –  Magdiragdag Jan 21 at 9:36

2 Answers 2

up vote 2 down vote accepted

Since$$\begin{align}(x^2-y^2)+(y^2-z^2)+(z^2-x^2)=x+y+z&\Rightarrow x+y+z=0\\&\Rightarrow x+y=-z,\end{align}$$ you'll have $$\begin{align}(x-y)(x+y)=z&\Rightarrow (x-y)(-z)=z\\&\Rightarrow z(1+x-y)=0\\&\Rightarrow z=0\ \text{or}\ y=x+1.\end{align}$$

1) When $z=0$, $(x-y)(x+y)=0\Rightarrow y=\pm x.$

If $y=x$, then $x(x+1)=0\Rightarrow x=-1,0.$

If $y=-x$, then $x(x-1)=0\Rightarrow x=0,1.$

2) When $y=x+1$, $(x-y)(x+y)=z\Rightarrow z=-2x-1.$ So, $$(x+1)^2-(-2x-1)^2=x\Rightarrow -3x^2-3x=0\Rightarrow x=-1,0.$$

Hence, you'll get $$(x,y,z)=(0,0,0),(-1,-1,0),(1,-1,0),(-1,0, 1),(0,1,-1).$$ However, $(x,y,z)=(-1,-1,0)$ does not satisfy $y^2-z^2=x$.

So, the answer is $$(x,y,z)=(0,0,0),(1,-1,0),(-1,0, 1),(0,1,-1).$$

share|improve this answer

By adding all equations we obtain $x+y+z=0$. Thus $$ z=x^2-y^2=(x-y)(x+y)=(y-x)z$$ which implies $z=0\lor y=x+1$. By cyclic symmetry, we also have $x=0\lor z=y+1$ and $y=0\lor x=z+1$. So if at least one of $x,y,z$ is nonzero, by cyclic symmetry wlog. $x\ne0$, then $z=y+1$; then at least one of $y,z$ is also nonzero. If $y\ne 0$, we find $x=z+1=y+2$ and since then $y\ne x+1$ necessarily $z=0$; this gives us $(x,y,z)=(1,-1,0)$. If on the other hand $z\ne 0$, we find $y=x+1=z-1$, so $x\ne z+1$, hence $y=0$ and the solution $(-1,0,1)$. So considering cyclic symmetry again, we obtain the full list of solutions: $$(0,0,0),(1,-1,0),(-1,0,1),(0,1,-1). $$

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.