$S(n) \equiv n \pmod{9}$ Thus we have $n^2 - n + 7 \equiv 0 \mod{9}$ $\implies n \equiv \frac{1 \pm \sqrt{-27}}{2} \equiv 5 \pmod{9}$ Note that $S(n)$ must be "fairly large" for the equation to hold. Ending with a $9$ is a promising case because after squaring it turns into $1$. The first "fairly large" case is $n=59$, but that fails. The next is $\boxed{n=149}$, and it simply remains to check cases now. EDIT: oops, fixed.

It is all wrong. $S(n)\equiv n\pmod3$ If $n\equiv 0\pmod {3}$ then clearly false. If $n\equiv 1\pmod {3}$ then $n^2\equiv 1\pmod {3}$ so $S(n^2)-S(n)\equiv0\pmod3$ If $n\equiv 2\pmod {3}$ then $n^2\equiv 1\pmod {3}$ so $S(n^2)-S(n)\equiv1\pmod3$ So $n\equiv 2\pmod{3}$. and $49^2=2401$ so $S(49)-S(49^2)=13-7=6$

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