Show that for all $n ∈ Z$, $2n^3 + 3n^2 + n$ is divisible by both 2 and 3.

Question

I have to show that this is divisible by both 2 and 3. For the next, part I have to show that it's divisible by 6 which from what I understand is the same as showing it's divisible by 2 and 3.

Answer 1

Hint: $\;2n^3 + 3n^2 + n = 2n(n+1)(n+2) -3n(n+1)\,$, then use that the product of $\,2\,$ consecutive numbers is divisible by $\,2\,$ and the product of $\,3\,$ consecutive numbers is divisible by $\,6\,$.

Answer 2

Factorising $2n^3 + 3n^2 + 1$ we get $n(n+1)(2n+1)$ As we know that sum of squares of first n natural numbers $$1^2+2^2+...+n^2 = {n(n+1)(2n+1)\over 6}$$ And since $$1^2+2^2+ ... + n^2$$ is an positive integer we can say that $n(n+1)(2n+1)$ is divisible by 6 and hence $2n^3 + 3n^2 + 1$ is divisible by 2, 3 and 6.

Answer 3

By fermat theorem $n^2=n (mod2)$ then $2n^3+3n^2+n = 2n+3n+n = 6n =0 (mod2)$ $2n^3+3n^2+n =2n+3n^2+3n $