# Number of real solutions for the following set of equations? [closed]

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.

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## closed as off-topic by mau, Magdiragdag, Daryl, TMM, Lost1Jan 21 '14 at 11:02

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Add all the to get $x+y+z=0$ and go on –  Blah Jan 21 '14 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 '14 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 '14 at 9:36

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 $$(x,y,z)=(0,0,0),(1,-1,0),(-1,0, 1),(0,1,-1).$$

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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: $$(0,0,0),(1,-1,0),(-1,0,1),(0,1,-1).$$

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