Use mathematical induction to prove that n3−n is divisible by 3 whenever n is a positive integer.

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pierretteA · Accepted Answer

In the inductive step we assumed that k3−k is divisible by 3 which means that k3−k=3M where is an integer. Now we can easily deduce P(k)→P(k+1). (k+1)3−(k+1)=k3+3k2+3k+1−k−1 =(k3−k)+3k2+3k =3M+3k2+3k =3(M+k2+k) =3Z where Z=M+k2+k. Since Z is an integer, it is clear (k+1)3−(k−1) is divisible by 3 if k3−k is divisible by 3. Thus concludes the proof. Now there is a much easier proof that does not require induction, factor n3−n as n(n−1)(n+1)=(n−1)n(n+1). Now n-1, n and n+2 are 3 consecutive integers, so one of them must be divisible by 3 since n mod 3 can take on 3 different values(0, 1 and 2) and the 3 consecutive integers each have a different value, so one of them will be 0 (mod 3). Since one of the integers in the product is divisible by 3, so is the whole product.

Use mathematical induction to prove that n^{3}− n is divisible

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