(i) Show that the product of any three consecutive integers is divisible by 3.
(ii) Show that the product of any three consecutive integers is divisible by 6.
(i) Let $n$ be the smallest of the three consecutive integers, Then we have
$ n(n+)(n+2) \equiv n^3 + 3n^2 + 2n \pmod 3$
There are two cases for $n$. It is either divisible by 3, in which case the product is, and we are done. If it is not, then by Corollary 4.2 we have $n^3 \equiv n \pmod 3$. And so
$ n(n+)(n+2) \equiv (n) + 3n^2 + 2n \equiv 3(n^2 +n) \equiv 0 \pmod 3$
We have shown the product of three consecutive integers is divisible by 3.
(ii) Let $n$ be the smallest of the three consecutive integers, Then we have
$ n(n+)(n+2) \equiv n^3 + 3n^2 + 2n \pmod 2$
There are two cases for $n$. It is either divisible by 2, in which case the product is, and we are done. If it is not, then by Corollary 4.2 we have $n^2 \equiv n \pmod 2$. And so
$ n(n+)(n+2) \equiv n(n) + 3n^2 + 2n \equiv 4n^2 + 2n \equiv 2(2n^2 +n) \equiv 0 \pmod 3$
We have shown the product of three consecutive integers is divisible by 2.
Since that product is also divisible by 3, and both 3 and 2 are co-prime, so the product is divisible by $2 \times 3 = 6$.