System!

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

#Algebra #SystemOfEquations #AlgebraicManipulation

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Nihar Mahajan - 6 years 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.

Ryan Tamburrino - 6 years 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.

Nihar Mahajan - 6 years ago

@Nihar Mahajan – Now that's Brilliant!

Ryan Tamburrino - 6 years ago

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

Nihar Mahajan - 6 years ago

@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 - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$