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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
2 years, 6 months ago

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I am getting \(x=y=z=2\).

Nihar Mahajan - 2 years, 6 months ago

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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.

Ryan Tamburrino - 2 years, 6 months ago

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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.

Nihar Mahajan - 2 years, 6 months ago

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@Nihar Mahajan Now that's Brilliant!

Ryan Tamburrino - 2 years, 6 months ago

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

Nihar Mahajan - 2 years, 6 months ago

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

Ryan Tamburrino - 2 years, 6 months ago

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