So I had to prove this in my calculus exam today
Prove that $n^3+5n$ is always divisible by 6.
I know that there have been other people around here having asked this already, but I chose a different approach and would really appreciate to hear whether my train of thought here was right or wrong. So I am especially not looking for the "right way" to prove it (i.e., I don't need a proper solution) I would just appreciate for somebody to assess my solution, if it's completely wrong or possibly not complete etc.
so, what I did:
Initial case prove that $n^3+5n \bmod 6 = 0$ for $n=1$.
$$1^3+5\cdot 1 \bmod 6 = 0 \Leftrightarrow 6 \bmod 6 = 0$$
which is trivially true.
Induction step Prove that for every $n$, if the statement holds for $n$, then it holds for $n + 1$.
$$((n+1)^3+5(n+1)) \bmod 6 =0 \\ \Leftrightarrow ((n^3+3n^2+3n+1)+5n+5) \bmod 6 =0 \\ \Leftrightarrow (n^3+5n \bmod 6) + (3n^2+3n+6 \bmod 6)=0 \\ \Leftrightarrow 0+(3n^2+3n+6 \bmod 6)=0 \\ \Leftrightarrow 3n^2+3n+6 \bmod 6 =0.$$
And this is true for all $n \in \mathbb{N}$, where from the third to the fourth line, we replaced with the induction hypothesis that $n^3+5n \bmod 6 =0$ . Take for instance $1, 2,3,…$ it is always divisible by 6 without leaving any rest!!
However, my other friends that took the exam said that this is in need of some extension to show that the last line actually holds true for whatever n you insert. But I cannot grasp what they mean; by the definition of the modulus, it should already trivially be given that the last line is right…
I'm really looking forward to your thoughts, do you think I would at least get partial points for this? 🙁 I really don't see how this can be wrong…
Thanks so much,
Lin
Best Answer
In view of the final step, $3n^2+3n+6\equiv 0\mod 6$ is equivalent to $3n^2+3n\equiv 0\mod 6$, i.e.,
$3(n^2+n)\equiv 0\mod 6$.
So it is sufficient to show that
$2\mid n^2+n$.
But $n^2+n = n(n+1)$ is a product of consecutive integers and one of which will be even. Thus $n^2+n$ is a multiple of $2$ and hence $3(n^2+n)$ is a multiple of $6$, as required.