Problem 983

Notice that if $x, y$ are integers, then by parity analysis we know
$$x^{2}+y^{2} \equiv 0,1,2(\bmod 4) \text {. }$$

If $n \in S$, then from the above, $n \equiv 1(\bmod 4)$. Thus, we can set
$$\begin{array}{l} n-1=a^{2}+b^{2}, a \geqslant b, \\ n=c^{2}+d^{2}, c>d(c, d \text { cannot be equal }), \\ n+1=e^{2}+f^{2}, e \geqslant f, \end{array}$$

where $a, b, c, d, e, f$ are all positive integers.
Then $n^{2}+1=n^{2}+1^{2}, n^{2}=\left(c^{2}+d^{2}\right)^{2}=\left(c^{2}-d^{2}\right)^{2}+(2 c d)^{2}$,
$$n^{2}-1=\left(a^{2}+b^{2}\right)\left(e^{2}+f^{2}\right)=(a e-b f)^{2}+(a f+b e)^{2} \text {. }$$

Assume $b=a$, and $f=e$, then $n-1=2 a^{2}, n+1=2 e^{2}$, subtracting the two equations gives $e^{2}-a^{2}=1$, then $e-a \geqslant 1$, and $1=e^{2}-a^{2}=(e+a)(e-a)>1$, a contradiction!
Therefore, $b=a, f=e$ cannot both hold. So $a e-b f>0$, thus $n^{2} \in S$.