Let $D=\left\{\left( x^{2} +h\right)^{2} |\ x\in \mathbb{Z} ,\ h\in \{1,2\}\right\}$ Suppose that $a\in D$ and $a$ is written. So we have $a=\left( n^{2} +t\right)^{2}$ for some $n\in \mathbb{Z}$ and $t\in \{1,2\}$. If we create a new number $b$ from $a$, then $b\in \left\{\left( n^{2} +t\right)^{4} ,\left(\left( n^{2} +t\right)^{2} +1\right)^{2} ,\left(\left( n^{2} +t\right)^{2} +2\right)^{2}\right\}$. So we must have $b\in D$. Since $1\in D$, $D$ contains all written numbers. Assume that there exists a written number $z$ which is divisible by $2015$. So we have $z=\left( x^{2} +h\right)^{2}$ for some $x\in \mathbb{Z}$ and $h\in \{1,2\}$. Note that $2015$ is divisible by $31$. So we must have $31\mid x^{2} +h$. However, this contradicts that $\left(\frac{-1}{31}\right) =-1$ and $\left(\frac{-2}{31}\right) =-1$. $\blacksquare$ Note : $\left(\frac{.}{.}\right)$denotes the Legendre Symbol.