How to prove that $2x$ is integer too? Let  $x , y, z$ are rational and  $x^2+y^2+z ,\ y^2+z^2+x,\ x^2+z^2+y$ are integers. 
How to prove that $2x$ is integer too?
I know that $$(z-y)(z+y-1)=a,\ (y-x)(y+x-1) = b,\ (x-z)(x+z-1)=c$$
such that  $a,b,c \in \mathbb{Z}$.
 A: For simplicity, for a prime $p$, Let $\mbox{ord}_p(x)$ be an integer such that $p^{-\mbox{ord}_p(x)}\cdot x$ does not contain factor $p$ both in numerator and denominator. The question is equivalent to saying that $\mbox{ord}_p(x) \geq 0$ for any odd prime $p$ and $\mbox{ord}_2(x) \geq -1$ (the same for $y$ and $z$, by symmetry).
Suppose $p$ is an odd prime such that $\min\{\mbox{ord}_p(x),\mbox{ord}_p(y),\mbox{ord}_p(z)\} = -n < 0$. There exists a $d \in \mathbb Z$; $p\!\not | d$ such that $x' := dp^nx$, $y' := dp^ny$ and $z' := dp^nz$ are integers. Observe that $p|(x'^2 + y'^2)$, $p|(y'^2 + z'^2)$ and $p|(z'^2 + x'^2)$. This immediately implies $p|x'$, $p|y'$ and $p|z'$ which leads to a contradiction.
Similarly, if we consider the case $p = 2$, and suppose that $\min\{\mbox{ord}_2(x),\mbox{ord}_2(y),\mbox{ord}_2(z)\} = -n < -1$. There exists a $d \in \mathbb Z$; $2\!\not | d$ such that $x' := d2^nx$, $y' := d2^ny$ and $z' := d2^nz$ are integers. Then $4|(x'^2 + y'^2)$, $4|(y'^2 + z'^2)$ and $4|(z'^2 + x'^2)$. This immediately implies $2|x'$, $2|y'$ and $2|z'$ which leads to a contradiction.
