Croatian Mathematical Society Competitions

Note that $D(k)$ is odd if and only if $k$ is a perfect square. From $m^2<m^2+1<m^2+2m+1=(m+1{)}^2$ it follows that $m^2+1$ is not a perfect square for any positive integer $m$. Hence $b,c$ and $d$ of any green quadruple are not perfect squares, i.e. $D(b),D(c)$ and $D(d)$ are even. Therefore, $D(a)$ must be odd and $a$ is a perfect square. Since $10^6>d>c^2>b^4>a^8$, i.e. $2^{10}=1024>10^3>a^4$, we get $6^2>2^5>a^2$, i.e. $a<6$. Hence $a=1$ or $a=4$. A direct computation confirms that both options yield green quadruples: $(1,2,5,26)$ and $(4,17,290,84101)$. Therefore, the answer is 2.