<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] We use a classical infinite descent argument.
Note that the right-hand side is even, so the left-hand side must be. It follows that two of $x$, $y$, $z$ are odd and the third even, or all three are even.
But two odd and one even is impossible, for then the right-hand side is divisible by $4$ and the left-hand side is not.
Thus $x=2x_1$, $y=2y_1$, $z=2z_1$ for some integers $x_1$, $y_1$, $z_1$.
Substituting we get $x_1^2+y_1^2+z_1^2=4x_1y_1z_1$.
Repeat the argument. We find that $x_1=2x_2$, and so on, with $x_2^2+y_2^2+z_2^2=8x_2y_2z_2$.
Continue. We conclude that $x$, $y$, $z$ are each divisible by arbitrarily high powers of $2$, so are all $0$.
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