The number of subsets

$\displaystyle \frac { { n }^{ 2017 }+2 }{ n+1 } =\frac { { n }^{ 2017 }+1+1 }{ n+1 } \\ \displaystyle \frac { { n }^{ 2017 }+2 }{ n+1 } =\frac { { n }^{ 2017 }+1 }{ n+1 } +\frac { 1 }{ n+1 } \in \mathbb{N}$

$\frac { 1 }{ n+1 } \in \mathbb{N}$ , Hence $n + 1 \le n ...(*)$

The sign of the similarity of the inequality $(*)$ will apply if $n = 0$ . Though we know that $n \in \mathbb{N}$ . So there is no value $n$ that satisfies the statement. Then we find that $S = \{\}$ . So that many subsets of $S$ exist $1$ .