Show that $n(2n + 1)(7n + 1)$ is divisible by 6 for all integers $n$ and find all integers $n$ such that $n(2n + 1)(7n + 1)$ is divisible by 12. a) Show that $n(2n + 1)(7n + 1)$ is divisible by 6 for all integers $n$.
b) Find all integers $n$ such that $n(2n + 1)(7n + 1)$ is divisible by 12.
I tried expanding but it doesn't really help much.  Thanks in advance!

$$n(2n+1)(7n+1)\equiv n(2n+1)(n+1)\equiv 0\pmod{2},$$
because exactly one of $n$, $n+1$ is even.
$$n(2n+1)(7n+1)\equiv n(-n+1)(n+1)\pmod{3}$$
$$\equiv n(-1)(n-1)(n+1)\equiv 0\pmod{3},$$
because exactly one of $n$, $n-1$, $n+1$ is divisible by $3$.
Therefore $2\cdot 3\mid n(2n+1)(7n+1)$ for all $n\in\mathbb Z$.
The condition $2^2\cdot 3\mid n(2n+1)(7n+1)$ is therefore equivalent to $2^2\mid n(2n+1)(7n+1)$.
Since $2n+1$ is odd for all $n\in\mathbb Z$, this is equivalent to $2^2\mid n(7n+1)$ (by common sense, or you can see Generalized Euclid's lemma. We have $\gcd\left(2^2,2n+1\right)=1$ for all $n\in\mathbb Z$).
$$n(7n+1)\equiv n(-n+1)\equiv -n(n-1)\pmod{2^2}$$
is equivalent to $0$ if and only if either $2^2\mid n$ or $2^2\mid n-1$ (because $n$, $n-1$ are coprime).
The answer is: all $n\in\mathbb Z$ such that $n\equiv \{0,1\}\pmod{4}$.
