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??
 A: In 1907, A. Hurwitz considered 
$$ x_1^2 + x_2^2 + \cdots + x_n^2 = x \, x_1 x_2 \cdots x_n  $$
The conclusions, with all the $x_i \geq 0$ integers and $x$ an integer, included $x \leq n.$ The main thing, though is that, for a fixed pair $n,x,$ all solutions collected into a finite set of rooted trees. Travel within a tree is by "Vieta Jumping." He called a tree root a Grundlösung, I usually say fundamental solution. He gave enough inequalities to find all fundamental solutions for any given pair $n,x,$ including showing when there were no actual solutions using that pair.
making jpeg, just a minute

A: 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 $0$.
A: We notice that one of the solution is $ x=y=z=0 $. Now let's try to find other solutions for the equation. 
Suppose if none of the x,y,z is even. Then, $$x^2+y^2+z^2\equiv(1+1+1)\mod{4},2xyz=2\mod4 $$ If exactly one is even, $$x^2+y^2+z^2\equiv(0+1+1)\mod4 ; 2xyz \equiv 0 \mod4$$ If two of x,y,z are even and one is odd then, $$x^2+y^2+z^2\equiv(0+0+1)\mod4 ; 2xyz \equiv 0 \mod4$$  So, the only possibility is that all are even. 
Let x = 2X , y = 2Y ,z  = 2Z. Then, $$4X^2 +4Y^2 +4Z^2 = 16XYZ $$ $$\implies X^2 + Y^2 + Z^2 = 4XYZ$$ The same argument goes show that X,Y,Z are even. The process can be continued indefinitely. 
This is possible only when x = y = z = 0. So ,there exist no other solution over $\mathbb{N} $
A: $2xyz$ 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 $\Bbb{N}$
