Proof by Induction - Wrong common factor I'm trying to use mathematical induction to prove that $n^3+5n$ is divisible by $6$ for all $n \in \mathbb{Z}^+$.
I can only seem to show that it is divisible by $3$, and not by $6$. This is what I have done:
Let $f(n) = n^3+5n$.
Basis Step: When $n=1$, $f(n)= 6$ and clearly $6$ divides $6$.
Assumption Step: Assume that $6$ divides $f(n)$ when $n=k$.
Inductive Step: Consider $f(k+1)-f(k)$:
$$f(k+1)-f(k) = [(k+1)^3+5(k+1)]-[k^3+5k]$$
$$=3k^2+3k+6$$
$$=3(k^2+k+2)$$
It follows that $f(k+1) = f(k)+3(k^2+k+2)$. I really wanted a common factor of $6$.

From your proof...
Note that if k is odd, k^2 + k + 2 is even and hence divisible by 2, and that the same is true if k is even.  Therefore k^2 + k + 2 is divisible by 2 for all k.  this gives you the extra factor of 2 you need to get a factor of 6.
