Answer to Question #349163 in Discrete Mathematics for Nopi

Question #349163

6 divides n^3 - n for n>=2


1
Expert's answer
2022-06-09T05:22:57-0400

Let "P(n)" be the proposition that for the positive integer "n\\ge 2: n^3-n" is divisible by 6.

Basis Step:

"P(2)"is true, because "2^3-2=6" is divisible by 6.

Inductive Step:

For the inductive hypothesis we assume that "P(k)" holds for an arbitrary

positive integer "k." That is, we assume that "k^3-k" is divisible by 6


"k^3-k=6m, m\\in \\Z, m\\ge 1"

Under this assumption, it must be shown that "P(k + 1)" is true, namely, that

"(k+1)^3-(k+1)" is divisible by 6, is also true.


"(k+1)^3-(k+1)=(k+1)(k^2+2k+1-1)"




"=k^3+2k^2+k^2+2k=k^3-k+3k(k+1)"


If "k" is even, then "3k(k+1)" is divisible by 6.

If "k" is odd, then "k+1" is even, and "3k(k+1)" is divisible by 6.



"(k+1)^3-(k+1)=(k^3-k)+3k(k+1)"

This last equation shows that "P(k + 1)" is true under the assumption that "P(k)" is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we know that "P(n)" is true for all integers "n\\ge 2." That is, we have proved that for the positive integer "n\\ge 2: n^3-n" is divisible by 6.


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