$n^{3} + 2n$ is divisible by $3$. Is this induction proof correct?

Question

Question: Prove by means of the principle of induction that for every $n ∈ N$ the number $n^{3} + 2n$ is divisible by $3$.

Proof

Denote "$n^{3} + 2n$ is divisible by 3" by $P(n)$. Check $P(n)$ for an arbitrary $n$, for instance $n=1$. $1^{3}+2*1=3*1$ and thus divisible by three, therefore $P(1)$ holds.

Induction step: Assume $P(n)$ is is true, let $n ∈ N$. Then $(n+1)^{3}+2(n+1) =(n+1)(n^{2}+2n+1)+2(n+1)=(n+1)(n^{2}+2n+3)=n^{3}+2n^{2}+3n+n^{2}+2n+3=n^{3}+2n+3(n^{2}+n+1).$

We assumed $P(n)$ holds, thus the $n^{3}+2n$ part of the induction holds. We can obviously see that $3(n^{2}+n+1)$ is divisible by 3 and that concludes the proof.

Note also that $$n^3+2n=3n+n^3-n=3n+\underbrace{n(n-1)(n+1)}_{\text{product of 3 consecutive integers}}$$ is necessarily divisible by $3$. — Kim Jong Un, Oct 13 '14 at 12:25
As you have written it, $P(n)$ is not a statement, so it can't be true or false. (See Thomas's answer for how to fix this.) — Tara B, Oct 14 '14 at 15:20

score 3 · Answer 1 · edited Jun 12 '20 at 10:38

I suggest a slight rewording.

Proof

Denote the statement

$n^{3} + 2n$ is divisible by 3

by $P(n)$. We check $P(n)$ for an $n=1$: $1^{3}+2*1=3*1$. Thus $P(1)$ holds.

Induction step: Assume $P(n)$ is is true, let $n ∈ N$. Then $(n+1)^{3}+2(n+1) =(n+1)(n^{2}+2n+1)+2(n+1)=(n+1)(n^{2}+2n+3)=n^{3}+2n^{2}+3n+n^{2}+2n+3=n^{3}+2n+3(n^{2}+n+1).$

We assumed $P(n)$ holds, thus $n^{3}+2n$ is divisible by $3$. We can obviously see that $3(n^{2}+n+1)$ is divisible by 3 and that concludes the proof.

score 1 · Answer 2 · answered Oct 13 '14 at 12:24

1

Your proof is correct. Well done.

answered Oct 13 '14 at 12:24

amWhy

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$n^{3} + 2n$ is divisible by $3$. Is this induction proof correct?

Proof

2 Answers2

Proof