Does anybody know how to solve this?

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

\[x^2 + y^2 + z^2 = xyz +4 \] \[xy + yz + zx = 2(x+y+z) \]

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## Comments

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

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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 \(abc = 1\).

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

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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.

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

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