# A Hard Problem

Does anybody know how to solve this?

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

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

Note by Alan Yan
2 years, 9 months ago

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

- 2 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 $$abc = 1$$.

- 2 years, 9 months ago

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

- 2 years, 9 months ago

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.

- 2 years, 9 months ago

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

- 2 years, 7 months ago

Okay I will try to post it soon, I am still in school.

- 2 years, 9 months ago