YLG_123 wrote: The numbers $1- \sqrt{2}$, $\sqrt{2}$ and $1+\sqrt{2}$ are written on a blackboard. Every minute, if $x, y, z$ are the numbers written, then they are erased and the numbers, $x^2 + xy + y^2$, $y^2 + yz + z^2$ and $z^2 + zx + x^2$ are written. Determine whether it is possible for all written numbers to be rational numbers after a finite number of minutes. Starting from $(1-\sqrt 2,+\sqrt 2,1+\sqrt 2)$, the three next steps are : $(3-\sqrt 2,7+3\sqrt 2,5)$ $(93+38\sqrt 2,127+57\sqrt 2,51-11\sqrt 2)$ $(50307+31673\sqrt 2,30693+14866\sqrt 2,18287+6861\sqrt 2)$ Here, all the $a+b\sqrt 2$ have $a,b\in\mathbb Z_{>0}$ and easy induction implies that this property will be true always in next steps. And so we'll never get any $b=0$ and so we'll never have rational numbers from there.

Solution: The answer is negative. Let's look to the set of the differences, initially it is $\{x-y,y-z,z-x\}$, after the operation it is $\{(x-y)(x+y+z),(y-z)(x+y+z),(z-x)(x+y+z)\}$, that is, at any point the numbers written on the board will be on the form $\{k(x-y),k(y-z),k(z-x)\}$ for some $k$ real number. At the beginning the set is $\{2\sqrt 2, 1, 2\sqrt 2-1\}$, so if all the numbers are rational, then $k=0$. But $x^2+xy+y^2=(x+\frac{y}{2})^2+\frac{3}{4}y^2\geq 0$ with equality only if $x=y=0$, so if in the $n$ step all the 3 numbers are $0$, in the $n-1$ step they must be $0$ too. Therefore all the 3 initial numbers are $0$. Contradiction