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This is somewhat embarrassing, but I am having some trouble figuring out how to do a strong induction question. We have to prove that $n^3 - n$ is divisible by $6$ for all $n \in \mathbb{N}$. Now I can see that this must be true, since $n^3 - n = (n+1)n(n-1)$, i.e. the product of three consecutive integers. Therefore at least one of them will be even, and one will be a multiple of $3$. Hence the product will be divisible by $6$. However, I can't figure out how to do it by induction. It's probably quite easy to do once you see one of those trick moves, like adding and subtracting the same thing, etc, but I can't figure it out. Please help!

Rodrigo de Azevedo
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Gerrit Goosen
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  • Base case tells you if $n=1, 6 | 0.$. Now assume $6 | k^3-k$ And consider $n=k+1$ – Eleven-Eleven Aug 01 '16 at 13:24
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    Never liked this kind of questions... use induction to prove something that is trivial to prove without induction. (Besides, in the proof by induction, one needs to prove that $2 \mid n^2 + n$ for all $n$, and to stay within the spirit of the problem, that should ALSO be done by induction - otherwise it's no different from the original problem). –  Aug 01 '16 at 13:29
  • Yes you are correct. I was trying to come up with an algebraic trick to make more direct use of the induction hypothesis, instead of seeing that there is another induction layer here. Thanks. – Gerrit Goosen Aug 01 '16 at 13:38
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    @mathguy But it is not "trivial to prove without induction". If you are going to use the Lemma that a sequence of $k$ consecutive integers contains a multiple of $k$ then you need to prove it, and that either uses induction (or further Lemmas that use induction). But "uses induction" is a fuzzy concept at this informal level, and the buck has to stop somewhere (e.g. proofs of the associative. commutative, distributive laws of natural addition and multiplication do use induction, and every proof uses these). – Bill Dubuque Aug 01 '16 at 13:45
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    My point wasn't that the lemma isn't "easier" (to prove) than the result. I would have no problem with that if it wasn't of the same "nature" as the result itself. Namely: It is not OK to use the lemma that a sequence of $k$ consecutive integers contains a multiple of $k$ for $k$ = 3, but it IS OK to use it for $k=2$? Why? –  Aug 01 '16 at 14:08

2 Answers2

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Base case holds: $6|1-1$.

For induction step:

Assume that $6|n^3-n$.

We have $$(n+1)^3-(n+1)=(n^3-n)+3n^2+3n=(n^3-n)+3n(n+1).$$ By induction assumption $6|n^3-n$.

Also, $$2|n(n+1),$$ since product of two consecutive numbers is divisible by $2$.

(Induction proof of the previous fact: $2|1*2$, so induction base holds. Induction step: assume $2|n(n+1)$, write $(n+1)(n+2)=n(n+1)+2(n+2)$ and conclude from that: $2|(n+1)(n+2)$.)

Therefore, $$6|3n(n+1).$$ Summing those two gives $$6|(n+1)^3-(n+1).$$

alans
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  • "product of two consecutive numbers is divisible by 2" should be proved, and likely by induction (otherwise how is it different from "product of three consecutive numbers is divisible by 3" which is almost the same thing as the thing to prove in the first place?) –  Aug 01 '16 at 13:36
  • @mathguy I added induction proof of "product of two consecutive numbers is divisible by 2", as you suggested. – alans Aug 01 '16 at 13:44
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As $n^3$ and $n$ are both even or both odd, $n^3-n$ is always even. Hence it is enough to show $n^3-n$ is divisible by $3$.

This of course is just Lil' Fermat, since $3$ is prime. But it's easy to prove it by induction: $$(n+1)^3-(n+1)=(n^3-n)+3n^2+3n$$ from which it follows at once.

Bernard
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  • Immediat de-merit - the OP did not ask for "any" proof but specifically one by induction. Learning induction, not arbitrary facts like $6 \mid n^3 - n$. –  Aug 01 '16 at 13:35
  • I'm sorry, I also showed it by induction. And showed (or tried to) how one may simplify induction with preliminary considerations. – Bernard Aug 01 '16 at 13:39