Question

If $n\epsilon N$, the $n(n^2-1)$ is divisible by?

Answer 1

(1) $n$ is a prime number --> if $n=2$, then the answer is NO but if $n=5$, then the answer is YES. Not sufficient.

(2) $n$ is greater than$191$. Clearly insufficient (consider$n=24$ for a NO answer and $n=17$ for an YES answer).

(1)+(2) Given that n is a prime number greater than 191 so n is odd and not a multiple of $3$.$n^2-1=(n-1)(n+1)$--> out of three consecutive integers $(n-1)$, $n$ and $n+1$ one must be divisible by $3$, since it's not n then it must be either $(n-1)$ or $(n+1)$, so $(n-1)(n+1)$is divisible by $3$. Next, since n is odd then$(n-1)$ and$(n+1)$are consecutive even numbers, which means that one of them must be a multiple of $4$, so$(n-1)(n+1)$ is divisible by$2\ast 4=8$. We have that $(n-1)(n+1)$ is divisible by both $3$and$8$ so$(n-1)(n+1)$is divisible by $3\ast 8=24$. Sufficient.