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I'm trying to prove using induction that $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ whenever $n$ is a non-negative integer. So far, I have:

  1. Base case: $P(1) = (1) + (8) + (27) = 36, 36$ can be divided by $9$ so the base case is valid
  2. Inductive step: let $P(n)$ be the statement $9$ divides $n^3 + (n+1)^3 + (n+2)^3$. Assume $P(k)$ is true, so $9$ divides $k^3 + (k+1)^3 + (k+2)^3$.

And this is where I'm stuck. I'm not sure how to demonstrate that $9$ divides $P(n)$ when $n = k+1$. If someone could step me in the right direction that would be awesome. Thanks!

  • 0
    k does not equal k+1.2010-10-14
  • 1
    In the last paragraph, “when k = k+1” should be “when n = k+1.”2010-10-14
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    N.B. This question has absolutely nothing to do with induction.2010-10-14
  • 0
    @muad: Why do you say that? Because it has a trivial proof mod 9?2010-10-14
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    @Bill Dubuque, no. The question is about proving $P(k)\Rightarrow P(k+1)$. It happens to come from part of an induction proof but the question is not about any aspect of induction.2010-10-14
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    But induction questions often focus on the only nontrivial step.2010-10-14
  • 0
    If you're proving for "non-negative integers", that means your base case should be 0 instead of 1, since non-negative integers is 0,1,2,....2010-10-15
  • 0
    See also http://math.stackexchange.com/questions/859572/the-sum-of-three-consecutive-cubes-numbers-produces-9-multiple2016-10-04

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