Solve the equation $x^{2}+y^{2}+z^{2}=4(x y+y z+z x)$ in integers.
The equation is homogeneous: if $\left(x_{0}, y_{0}, z_{0}\right)$ is a solution, then ( $t x_{0}, t y_{0}, t z_{0}$ ) is also a solution ( $t$ - any integer). Therefore, it is sufficient to find non-zero solutions where the unknowns are coprime, and the other solutions can be obtained by multiplying them by $t$. However, the equation has no such solutions.

Indeed, the right-hand side is even (even divisible by 4), so the left-hand side must have an even number of odd terms. But they must be (otherwise all unknowns are even, and we are considering the case without common divisors). Therefore, there are exactly 2 odd unknowns. But the sum of two odd and one even square is not divisible by 4.

Therefore, the equation has no non-zero solutions.

## Answer

$(0,0,0)$