<s>[INST] Prove that: $x^2+y^2+z^2=2xyz$ has no answer over $\Bbb{N}$ 
Prove that: $x^2+y^2+z^2=2xyz$ has no answer over $\Bbb{N}$

$$LHS=(x+y+z)^2-2(xy+yz+xz)=2xyz \implies (x+y+z)^2=2(xy+yz+xz)+2xyz$$
now what??
 [/INST] $2xyz$ even $\rightarrow$ $x$ even and $y$,$z$$\in$ odd OR $x$,$y$,$z$$\in$ even


*

*$x=2k$ $y$,$z$$\in$odd:


$4k^2 + y^2 + z^2 = 4kyz$
$4k(k-yz) +y^2 + z^2 = 0$
We have this problem: $4|y^2 + z^2$ so $y$ or $z$ must be even but are odd.


*$x=2k, y=2l, z=2m$:


$4k^2 + 4l^2 + 4m^2 = 16klm$
$k^2 + l^2 + m^2 = 4klm$
And the problem is infinite (problem is the same like at the beginning - 4 instead of 2 nothing change) but $x,y,z$ should be finite.
There is no answer over $\Bbb{N}$
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