Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$.

I've started by letting $P(n) = n^3+11n$

$P(1)=12$ (divisible by 6, so $P(1)$ is true.)

Assume $P(k)=k^3+11k$ is divisible by 6.


Since $P(k)$ is true, $(k^3+11k)$ is divisible by 6 but I can't show that $(3k^2+3k+12)$ is divisible by 6

space monkeys
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    Clearly $3k^2 + 3k + 12 = 3(k^2 + k + 6)$ is divisible by 3. Can you prove that it is divisible by $2$, too? – A.P. Mar 24 '15 at 13:57
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    $12$ is divisible by 6, so all that is left is $3k^2+3k=3k(k+1)$. If $k$ is odd, $3(k+1)$ is divisible by 6; if $k$ is even, $3k$ is. – Michael Grant Mar 24 '15 at 13:57

5 Answers5


For any positive integer $n$, let $S(n)$ denote the statement $$ S(n) : 6\mid (n^3+11n). $$

Base step: For $n=1, S(1)$ gives $1^3+11(1) = 12 = 2\cdot 6$. Thus, $S(1)$ holds.

Inductive step: Let $k\geq 1$ be fixed, and suppose that $S(k)$ holds; in particular, let $\ell$ be an integer with $6\ell = k^3+11k$. Then \begin{align} (k+1)^3 + 11(k+1) &= (k^3+3k^2+3k+1) + (11k+11)\tag{expand}\\[0.5em] &= \color{red}{k^3+11k}+3k^2+3k+12\tag{rearrange}\\[0.5em] &= \color{red}{6\ell} + 3k(k+1)+2\cdot 6.\tag{by ind. hyp.} \end{align} Since one of $k$ and $k+1$ is even, the term $3k(k+1)$ is divisible by $6$, and so the last expression above is divisible by $6$. This proves $S(k+1)$ and concludes the inductive step $S(k)\to S(k+1)$.

By mathematical induction, for each $n\geq 1$, the statement $S(n)$ is true. $\blacksquare$

Daniel W. Farlow
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$$3k^2 + 3k + 12=3(k^2 + k +4)= 3(k(k+1)+4)$$

Can you see why $k(k+1)$ and $4$ are each divisible by $2$?

At least one of $k, k+1$ is even, as is $4$, hence $2$ divides $(k(k+1)+4)$, and with three as a factor of $\color{blue}{3}(k(k+1)+ 4)$, we have $$2\cdot 3 = 6\mid (3k^2 + 3k + 12).$$

Jordan Glen
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First, show that this is true for $n=1$:


Second, assume that this is true for $n$:


Third, prove that this is true for $n+1$:






  • $2 |n\implies6|3n \implies6|3n(n+1)\implies3n(n+1)=6m$
  • $2\not|n\implies2|n+1\implies6|3n(n+1)\implies3n(n+1)=6m$



Please note that the assumption is used only in the part marked red.

Alternatively, consider the following options:

  • $n\equiv0\pmod6 \implies n^3+11n\equiv 0+ 0\equiv6\cdot 0\equiv0\pmod6$
  • $n\equiv1\pmod6 \implies n^3+11n\equiv 1+11\equiv6\cdot 2\equiv0\pmod6$
  • $n\equiv2\pmod6 \implies n^3+11n\equiv 8+22\equiv6\cdot 5\equiv0\pmod6$
  • $n\equiv3\pmod6 \implies n^3+11n\equiv 27+33\equiv6\cdot10\equiv0\pmod6$
  • $n\equiv4\pmod6 \implies n^3+11n\equiv 64+44\equiv6\cdot18\equiv0\pmod6$
  • $n\equiv5\pmod6 \implies n^3+11n\equiv125+55\equiv6\cdot30\equiv0\pmod6$
barak manos
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  • I see you and I use "templates" for our inductive proofs. :) – Daniel W. Farlow Mar 24 '15 at 14:33
  • @crash: Hahaha, have you been catching up on my induction answers? Indeed, I've been copy-pasting them from one question to another, changing only a few numbers :) I think that once you have a good and clear template, it actually serves the purpose quite well. – barak manos Mar 24 '15 at 14:40
  • Well I haven't been catching up on them, but I've seen them here and there (I'm a big induction fan if you've seen some of my [questions](http://math.stackexchange.com/questions/1196303/subtly-flawed-induction-proofs)/[answers](http://math.stackexchange.com/questions/1139579/why-is-mathematical-induction-a-valid-proof-technique/1139606#1139606)). Indeed, many inductive proofs are so similar as only in need of changing a few numbers. Did you create your template or adapt it from somewhere else? – Daniel W. Farlow Mar 24 '15 at 14:43
  • @crash: Created it. I think that if you follow up with older answers of mine, then you'll notice it slightly changed during time (though not much - basically just the red marking). – barak manos Mar 24 '15 at 14:54
  • Haha colors *do* add some spice. And they are certainly helpful for noobs wondering how the hell you pulled something out (i.e., applying the inductive hypothesis). I got mine from David Gunderson's tome *Handbook of Mathematical Induction*. I was thinking about asking a question for induction book references (there aren't too many in the literature it seems). – Daniel W. Farlow Mar 24 '15 at 14:57
  • @crash: Yeah, I adopted it during the process of learning LaTex (on the fly, that is). I think that "book" questions are usually not very appreciated by the community here, but you can give it a try... Personally, I wouldn't know of any. – barak manos Mar 24 '15 at 14:59
  • No, you're definitely right--book questions aren't usually welcome here which is why I haven't posted it haha. I'll stop commenting to reduce noise now, but here is an induction problem that may prove to be a fun little challenge for you: $(\forall n\in\mathbb{N})(19\mid 2^{2^n}+3^{2^n}+5^{2^n})$. Doubt you'll be able to get by with just swapping numbers there ;) +1 for the colors in your answer to this specific problem (and because it's a good answer too of course). :) – Daniel W. Farlow Mar 24 '15 at 15:04
  • @crash: Thanks :) To be honest, for divisibility questions, sometimes it is easier to go in the "modulo way" (as shown in the second part of my answer). With $19$ as the divisor, you only have $19$ cases to handle. – barak manos Mar 24 '15 at 15:22
  • Your comment just made me lol. I'll give you a more challenging problem then (in "the modulo way"): $(\forall n\in\mathbb{N})(576\mid (5^{2n+2}-24n-25))$. Have fun with all those cases! Sorry to be a douche :( haha – Daniel W. Farlow Mar 24 '15 at 15:28

Since $n^{3} + 11n = 6m$ for some integer $m,$ we have $$(n+1)^{3} + 11(n+1) = 6m + 3n^{2} + 3n + 12 = 6m + 12 + 3(n^{2} + n).$$ It suffices to prove that $3(n^{2} + n)$ is a multiple of $6$. But, since if $n$ is odd then $n^{2} + n = 2m'$ for some integer $m'$ and if $n$ is even then of course $n^{2} + n = 2m''$ for some integer $m'',$ it follows that $6$ is indeed a multiple of $3(n^{2}+n),$ qed.

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I want to give a direct proof:


Note: a product of three consecutive integers needs to divisible by 6.

N. F. Taussig
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vudu vucu
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