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Prove that $\forall n \in Z$, $n^3+3n$ is even.

Attempt: I am solving this problem using proof by cases. Case 1 is when $n$ is even, i.e. $n=2b$. This one is easy. However, in case 2 when $n$ is odd, i.e. ($n=2a+1$) I am having difficulties with showing that $n^3+3n$ is even. Namely, after plugging in and expanding I am stuck: $$n^3+3n=(2a+1)^3+3(2a+1)=8a^3+12a^2+6a+1+3a+3=\space...$$ I tried regrouping the terms, but my effors did not amount to anything. I guess I am making a mistake somewhere. Help appreiciated. Thank you.

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I don't know if you know about congrunces, but if $n=1\mod 2$ then $n^3+3n=1+3=4=0\mod 2$. – Pedro Tamaroff Oct 18 '13 at 4:36
@Koba: $n^3 + 3n = (2a+1)^3 + 3 (2a+1) = 8 a^3+12 a^2+12 a+4$. – Amzoti Oct 18 '13 at 4:40
@Amzoti: Ahhh god dammit! Thanks. – Koba Oct 18 '13 at 6:05
@Koba: Happens to us all, you only have $3(2a)$ as an issue, but the rest was spot on. Regards – Amzoti Oct 18 '13 at 6:07
if $n$ is odd then $n^3$ and $3n$ are also odd. So the 2 added together would be even. Also what does the $a$ mean? – EpicGuy Oct 18 '13 at 7:39
up vote 2 down vote accepted

You may write $n^3+3n=n(n^2+3)$. Now, if $n$ is even, then $2$ divides $n$, so $2$ divides $n(n^2+3)$. If $n$ is odd, then $n^2$ is odd, so $n^2+3$ is even, for $3$ is odd, so $2$ divides $n^2+3$, so it divides $n(n^2+3)$.

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Thats all good and intuitive, but I want to show it by plugging in $n=2a+1$. – Koba Oct 18 '13 at 6:04
@Koba Well, as I said $(2a+1)^2=2(2a^2+2a)+1=2k+1$ is odd so $2k+1+3=2(k+2)=2j$ is even. Mindlessly "plugging in" is quite unrewarding. – Pedro Tamaroff Oct 18 '13 at 6:10

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