, 2010

Remove the parentheses, collect the terms and divide both sides by $2$ to get $x^2 + y^2 + z^2 - 2xy - 2yz + xz = 0$. This equality can be written as $(x - y + z)^2 = xz$. Hence $xz$ is a square of an integer. If $x$ and $z$ have a common divisor $d$, then $xz$ is divisible by $d^2$, and by the previous equality $(x - y + z)^2$ is divisible by $d^2$, therefore $x - y + z$ is divisible by $d$. Since $x$ and $z$ are divisible by $d$, $y$ must be divisible by $d$, hence $d = 1$, i.e. $x$ and $z$ do not have common divisors. Since $xz$ is a square of an integer, it follows that both $x$ and $z$ are squares of integers.