Find the solutions for the diophantine equation x^2 + y^2 + z^2 = 2xyz

Solution x = y =
z = 0 is obvious.

Let's prove that we
don't have other solutions, so we'll suppose the
opposite.

x^2 + y^2 + z^2 is an even number and at least
one of the numbers x, y, z is even. Considering the symmetry of the equation, then ca
x is even, if x = 2x1.
Then 4|y2 + z2, and this is only when
y and z are even. It's true if
y is even and z is odd, then 4 is not
divisible by y2 + z2. If both are odd then:

y^2 + z^2 = (2u
+ 1)^2 + (2v + 1)^2 =

=4(u^2 + v^2 + u + v) +
2

So 4 is not divisible by
y^2+z^2.

So that x =
2x1, y = 2y1,
z = 2z1 and considering the equation, we'll
determine:

x1^2 +
y1^2 + z1^2 =
2^2x1y1z1.

By
taking the same arguments from the equality above, we'll have 2|x1,
2|y1, 2|z1, so that
2^2|x, 2^2|y, 2^2|z. We
can show that 2^n|x,
2^n|y,
2^n|z for any n belongs
to N. It contradicts the
hypothesis.

So, the equation does have a single solution
and this is (0,0,0).