Take your time (2)

Since $(n-1)(n^2+n+1)= n^3-1$ , we want integers $n$ such that $n^3 \equiv 1 \pmod{13}$ but $n \not\equiv 1 \pmod{13}$ . The cube roots of $1$ modulo $13$ are $1,3,9$ (the generator of the nonzero integers modulo $13$ is $2$ , and $2^4 \equiv 3 \pmod{13}$ ), and so we want $n \equiv 3,9 \pmod{13}$ . Since $2017 = 155\times13+2$ , we deduce that there are $2 \times 155 = \boxed{310}$ positive integers with the desired property.

Let us define $f(n)=n^2+n+1$ .First trying to find the smallest solution say it is $n_1$ .

Our smallest solution is $n_1=3$ because $f(n_1=3)=3^2+3+1=13 \equiv 0 \pmod{13}$

Assume next solution is $n_2 >n_1$ so, $f(n_2)=n_2^{2} +n_2+1 \equiv 0 \pmod{13}$

So, $13\,\,|\,\, f(n_2)-f(n_1)=n_2^{2}+n_2-n_1^{2}-n_1$

Therefore, $13\,\,|\,\, (n_2-n_1)(n_2+n_1+1)$

Then either $13\,\,|\,\, (n_2-n_1)$ or $13\,\,|\,\,(n_2+n_1+1)$

Case(1): When $13\,\,|\,\, (n_2-n_1)$

In this case we have $n_2-n_1=13k$ where $k \in \mathbb{N}$

The solution set is generated by $n_2=13k+3$ .

The set looks like this $(3,16,29,.......,2005)$ .It is an A.P sequence .So,total number of terms is $155$ .

So, number of solutions are $155$

Case(1): When $13\,\,|\,\,(n_2+n_1+1)$

In this case we have $n_2+n_1+1=13k$ where $k \in \mathbb{N}$

The solution set is generated by $n_2=13k-3-1=13k-4$ .

The set looks like this $(9,22,35,.......,2011)$ .It is an A.P sequence .So,total number of terms is $155$ .

So, number of solutions are $155$

Therefore, total solutions are $=155+155=310$