A new year indeed

If there exist $n,m \in \mathbb{N}$ such that $f(n) = m^2$ , then (the inequalities are strict, as no integer can have a square root which is a half-integer) $\begin{array}{rccccl} m^2 - \tfrac12 & < & n + \sqrt{n} & < & m^2 +\tfrac12 \\ 4m^2 - 1 & < & (2\sqrt{n} + 1)^2 & < & 4m^2 + 3 \\ \tfrac12\big[\sqrt{4m^2 - 1} - 1\big] & < & \sqrt{n} & < & \tfrac12\big[\sqrt{4m^2 + 3} - 1\big] \\ \tfrac14\big[4m^2 - 2\sqrt{4m^2-1}\big] & < & n & < & \tfrac14\big[4m^2 + 4 - 2\sqrt{4m^2 + 3}\big] \\ m^2 - \tfrac12\sqrt{4m^2 - 1} & < & n & < & m^2 + 1 - \tfrac12\sqrt{4m^2 + 3} \end{array}$ But $\sqrt{4m^2 - 1} < 2m < \sqrt{4m^2+ 3}$ , and hence $m^2 - m \;<\; n \, < \; m^2 - m + 1$ which is not possible. Thus $f(n)$ is never the square of an integer for any positive integer $n$ . Thus the set of integers $1 \le k \le 2017$ such that $f^k(k)$ is a perfect square is emply. Adopting the standard convention, the sum of the empty set is $\boxed{0}$ .