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Prove by induction that for all $n\in\mathbb N$, $3\mid n^3+3n^2+2n$.

$$P(1) = (1)^3+3(1)^2+2(1) = 6$$

Which is clearly divisble by $3$. Therefore, $P(1)$ is true.

Assume $P(1),\ldots,P(n)$ and show $P(n+1)$ is true.

$$(n+1)^3+ 3(n+1)^2+ 2(n+1) = n^3+6n^2+11n+6$$

This can be rewritten: $(n^3+3n^2+2n)+3(n^2+3n+2)$.

First term is divisible by $3$ because $P(n)$ is true. The second term is divisible by $3$ because it is a multiple of $3$. Thus, their sum is divisible by $3$.

Is there anything I am missing?

Michael Hardy
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Vincent
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    Looks good. $ $ $ $ $ $ $ $ – William Chang May 10 '14 at 23:20
  • @CharlesNosbig : Your MathJax code style is atrocious. Please look at my edits. – Michael Hardy May 10 '14 at 23:27
  • I can see your corrections, but I don't quite understand why it is necessary. I am reasonably new here so I likely just haven't realized the importance just yet. What I did and what you corrected appear exactly the same, why does it matter if the MathJax looks a little different? – Vincent May 10 '14 at 23:30
  • Your proof is good, but I would suggest not writing $P(1) = (1)^3+3(1)^2+2(1) = 6$, because $P(1)$ is a proposition, not a number. – Steve Kass May 10 '14 at 23:43

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Your proof looks solid, but I might suggest an easier proof:

For all $n \in \mathbb{N}$, we have $n \equiv 0, 1, 2 \pmod{3}$. If $n \equiv 0 \pmod{3}$, we are done.

If $n \equiv 1 \pmod{3}$, then we have $n^3 + 3n^2 + 2n \equiv (1+2) \equiv 0 \pmod{3}$.

If $n \equiv 2 \pmod{3}$, then we have $n^3 + 3n^2 + 2n \equiv (2 + 1) \equiv 0 \pmod{3}$.

We conclude that $3|(n^3 + 3n^2 + 2n)$ for all natural numbers $n$.

Kaj Hansen
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