Weird formula

If $N > 1$ then $g(N)> 1$ divides $N$ .

If $n > 1$ is such that $g(n^2-n+1)$ divides $n+1$ , then $g(n^2-n+1)$ must divide both $n^2-n+1$ and $n+1$ . Since $n^2-n+1 = (n-2)(n+1)+3$ , this means that $g(n^2-n+1)$ divides $3$ , and hence must be equal to $3$ . Thus we deduce that $n^2-n+1 = 3^K$ for some $K \ge 1$ . If $n \equiv 0 \pmod{3}$ or $n \equiv 1 \pmod{3}$ it is easy to see that $n^2 - n + 1 \equiv 1 \pmod{3}$ . Thus we must have $n \equiv 2 \pmod{3}$ . But if we write $n = 3m-1$ we see that $n^2-n+1 = 9m(m-1) + 3$ is divisible by $3$ but not by $9$ . Thus we deduce that $K=1$ and hence that $n^2-n+1= 3$ . Thus $(n-2)(n+1)=0$ and so (since $n > 1$ ), $n = 2$ . It is easy to see that $n=2$ is indeed a solution. Thus the only possible answer, and hence the sum of all possible answers, is $\boxed{2}$ .