# System!

Find all solutions, in positive integers, to the system of equations $\begin{cases} x^2+y^2+z^2 = xyz+4 \\ xy+yz+xz = 2(x+y+z) \end{cases}$

Note by Ryan Tamburrino
3 years, 1 month ago

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I am getting $$x=y=z=2$$.

- 3 years, 1 month ago

That's what I thought, but I am having trouble showing that these are the only solutions. I showed that the first equation implies that $$x=a+\frac{1}{a}, y=b+\frac{1}{b}, z=c+\frac{1}{c}$$ with $$abc=1$$ (you can input these definitions to verify it works). Then I put those into the second equation but I got stuck.

- 3 years, 1 month ago

Well , I didn't use the 1st equation at all. See my method:

$xy+yz+xz=2(x+y+z) \\ \Rightarrow xy-2x+yz-2y+xz-2z=0 \\ \Rightarrow x(y-2)+y(z-2)+z(x-2) = 0$

For the above expression to be $$0$$ , we have two possible cases : $$x=y=z=0 \ or \ x=y=z=2$$ of which the case of $$x=y=z=2$$ is valid.

- 3 years, 1 month ago

Now that's Brilliant!

- 3 years, 1 month ago

But something is wrong since I didn't use the 1st equation at all!

- 3 years, 1 month ago

Well, you could use the definition $$x=a+\frac{1}{a}$$ which has a minimum of $$2$$ only when $$a=1$$.

- 3 years, 1 month ago