Let $ S$ be a set of all numbers written on the board. Let $ ( a,b)$ be an arbitrary pair of distinct elements of $ S$. We want show that $ b/a$ is a rational number. Let $ c$ be an arbitrary element of $ S$ with $ c\notin \{a,b\}$. Let $ d$ be an arbitrary element of $ S$ with $ d\notin \{a,b,c\}$. We can take $ x\in \mathbb{N}$ such that $ a^{2} +bc=x\ \cdots ( 1)$. We can take $ y\in \mathbb{N}$ such that $ b^{2} +ac=y\cdots ( 2)$. We can take $ x\in \mathbb{N}$ such that $ a^{2} +bd=z\cdots ( 3)$. We can take $ w\in \mathbb{N}$ such that $ b^{2} +ad=w\cdots ( 4)$. From $ ( 1)$ and $ ( 2)$, we have $ x-by/a=a^{2} -b^{3} /a\ \cdots ( 5)$. From $ ( 3)$ and $ ( 4)$, we have $ z-bw/a=a^{2} -b^{3} /a\cdots ( 6)$. From $ ( 5)$ and $ ( 6)$, we have $ ( y-w)\frac{b}{a} =x-z\ \cdots ( 7)$. Since $ c\neq d$, we have $ y\neq w\ \cdots ( 8)$. From $ ( 7)$ and $ ( 8)$, we have $ \frac{b}{a} =\frac{x-z}{y-w}$, which is a rational number. $ \blacksquare $