For how many natural numbers $n$ from 1 to 100 is $n^2+n+1$ divisible by 7?

Check residues mod 7: $n^2+n+1\equiv 0$ only for $n$ ≡ 2 $(4+2+1=7)$ and $n$ ≡ 4 $(16+4+1=21).n$ ≡ 2: 2, 9, …, 100 gives 15 numbers; $n$ ≡ 4: 4, 11, …, 95 gives 14. Total 29.