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I was trying and could not, as it shows that $$8\mid n(n^{2}-1)(3n+2);\forall n \in \text{N}$$ Induction; looking eight consecutive numbers, what to do and how to do?$$$$Sorry, forgot to add a detail: I understand the statement without using modular arithmetic (content not yet studied).

Andrés E. Caicedo
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benjamin_ee
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  • Are you familiar with modular arithmetic? Just show $n(n^2-1)(3n+2) \equiv 0 \rm{mod} 8$ for every $n \in 0..7$. – Anthony Carapetis Aug 05 '13 at 13:21
  • Oh, I see, you are the OP of the original question. Is there a reason that you've asked it again here? – Thomas Andrews Aug 05 '13 at 13:28
  • @ThomasAndrews Did not understand, and got the other party, then posted just different part; – benjamin_ee Aug 05 '13 at 13:33
  • @AnthonyCarapetis: Yes the problem can be solved with a long (though not too long) and tedious calculation, but such a proof does not give much insight into the problem or in how to solve related problems when computational approaches are not feasible. Also, aesthetically, massive case analysis is icky and to be avoided whenever possible. – Aaron Aug 05 '13 at 13:38

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HINT: $n(n^2-1)=(n-1)n(n+1)$; show that if $n$ is odd, this is a multiple of $8$. Then show that if $n$ is even, one of $n$ and $3n+2$ is a multiple of $4$. There is no need to use induction.

Brian M. Scott
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