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So I am trying to prove by induction that $3$ divides $n^3+2n$ whenever $n$ is positive.

So far I have:

$$\text{Assuming $P(n)$ is true, we'll prove $P(n+1)$ is also true.} \\ (n+1)^3 +2(n+1)\\ n^3+3n^2+3n+1+2n+2 \\ n^3+3n^2+3n+2n+3\\ \color{red}{(1)} \space (n^3+2n) + (3n^2+3n+3)\\ \color{red}{(2)} \space(n^3+2n)+3(n^2+n+3)$$ $(1)$ I am following an example in the text and they do this. By grouping $(n^3+2n)$, what's its significance?

$(2)$ I know factoring out a $3$ shows something about $3$ being divisable, but I don't understand why?

Also, side note, what does "$3|0$" mean?

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    $3 \mid 0$ means that the number $3$ divides the number $0$. – inavda Oct 09 '20 at 21:17
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    $3 \mid n^3+2n$ is your $P(n)$, right? You are assuming $P(n)$ is true. – Sávio Oct 09 '20 at 21:20
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    3|a means 3 divides a. or a is divisible by 3. In other words, a = 3k for some integer k. Note that if 3|a and 3|b, then 3|(a+b). This is easily proven. a =3k and b = 3m. So a+b = 3k+3m = 3(k+m). So the sum is also divisible by 3. – Ameet Sharma Oct 09 '20 at 21:22
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    @Savio yes I am! :) –  Oct 09 '20 at 21:22
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  • Grouping $(n^3 + 2n)$ means that you know that $3$ divides that value. And you know that $3$ divides $3(n^2 + n + 3)$ so $3$ must divide $(n^3 + 2n) + 3(n^2 + n+3)$. 2) factoring out $3$ out of a number and haven't the quotient turn by an integer means $3$ divides what you had. $3n^2 + 3n + 3 = 3(n^2 + n+1)$ and $n^2 + n+1$ is an integer. So that means $3$ divides $3n^2 + 3n + 3$ because we can factor $3$ out and have an iteger left over. Side not $3|0$ means $3$ divides the number $0$. This is true because $0 = 3k$ if $k = 0.
    • – fleablood Oct 09 '20 at 21:24
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    So you are assuming $3|n^3 + 2n$ (so there is an integer $K$ so that $n^3 + 2n = 3K$) and you showed that $(n+1)^3 + 2n = (n^3 + 2n) + 3(n^2 + n+1)$ and we assume $3|n^3 + 2n$ and we can see that $3| 3(n^2 + n+1)$ [because there is an integer $n^2 + n+1$ so that $3(n^2 + n+1) = 3(n^2+n+1)$] so $3|(n^3 + 2n) + 3(n^2+n+1)$ because there is an integer $K + n^2 + n+1$ so that $(n^3 + 2) + 3(n^2 + n + 1) = 3(K + n^2 + n+1)$.] – fleablood Oct 09 '20 at 21:28