Answer to Question #349162 in Discrete Mathematics for Nopi

Question #349162

Prove by mathematical induction: 6 divides n^3 - n for n>=2



1
Expert's answer
2022-06-09T05:21:20-0400

Let P(n)P(n) be the proposition that for the  positive integer n2:n3nn\ge 2: n^3-n is divisible by 6.

Basis Step:

P(2)P(2)is true, because 232=62^3-2=6 is divisible by 6.

Inductive Step:

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

positive integer k.k. That is, we assume that k3kk^3-k is divisible by 6


k3k=6m,mZ,m1k^3-k=6m, m\in \Z, m\ge 1

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

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


(k+1)3(k+1)=(k+1)(k2+2k+11)(k+1)^3-(k+1)=(k+1)(k^2+2k+1-1)




=k3+2k2+k2+2k=k3k+3k(k+1)=k^3+2k^2+k^2+2k=k^3-k+3k(k+1)


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

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


(k+1)3(k+1)=(k3k)+3k(k+1)(k+1)^3-(k+1)=(k^3-k)+3k(k+1)

This last equation shows that P(k+1)P(k + 1) is true under the assumption that P(k)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)P(n) is true for all integers n2.n\ge 2. That is, we have proved that for the  positive integer n2:n3nn\ge 2: n^3-n is divisible by 6.


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