A Hard Problem

Does anybody know how to solve this?

Find all triples $(x,y,z)$ of positive real numbers such that

$x^2 + y^2 + z^2 = xyz +4$ $xy + yz + zx = 2(x+y+z)$

#Algebra

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

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

It's impossible to solve because their are 3 unknown variables and just two equations.

Abhilakshay Pathania - 5 years, 9 months ago

Actually, there is a way to solve this. The key part of this problem is to use a special substitution. For example, if $x^2 + y^2 + z^2 - xyz = 4$ , you can make the substitution $x = a+ \frac{1}{a} , y = b + \frac{1}{b} , z = c + \frac{1}{c}$ such that $abc = 1$ .

Alan Yan - 5 years, 9 months ago

I didn't know about this method . I'll have to check again .Anyways thanks !!

Abhilakshay Pathania - 5 years, 9 months ago

Can u please post the complete solution using this method ,I am nt getting it ?

Abhilakshay Pathania - 5 years, 9 months ago

@Abhilakshay Pathania – Okay I will try to post it soon, I am still in school.

Alan Yan - 5 years, 9 months ago

@Abhilakshay Pathania – Here is the solution that my friend has:

Let $a=x+y+z>0$ so that we got : $x+y+z=a$ $xy+yz+zx=2a$ $xyz=(x+y+z)^2-2(xy+yz+zx)-4=a^2-4a-4$

So $x,y,z$ must be the three positive real roots of $X^3-aX^2+2aX-(a^2-4a-4)=0$

Setting $X=Y+\frac{a}{3}$ , this becomes $Y^3-Y\frac{a(a-6)}3=\frac{(a-6)(2a^2+21a+18)}{27}$

If $a<6$ , this cubic obviously can not have three real roots ( $LHS$ is increasing) If $a=6$ this gives the solution $x=y=z=2$ If $a>6$ , setting $Y=\frac 23\sqrt{a(a-6)}Z$ , the cubic becomes $4Z^3-3Z=\frac{2a^2+21a+18}{2a\sqrt{a(a-6)}}$ and it is easy to show that this cubic can not have three real roots ( $RHS>1$ )

So a unique solution $\boxed{(x,y,z)=(2,2,2)}$

which although is probably not the main solution but is a good one nonetheless.

Alan Yan - 5 years, 9 months ago

@Alan Yan – i don't see how you show the two cubics don't have real roots, can you please explain?

Edgar Wang - 5 years, 6 months 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}`	$\boxed{123}$