<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 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} $
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