Proving n3 - n is Divisible by 6 Using Mathematical Induction and Consecutive Integers

To prove that n3 - n is divisible by 6 for all integers n, we can use mathematical induction. This method involves two main steps: the base case and the inductive step. Additionally, we can use the product of consecutive integers to provide an alternative approach.

Mathematical Induction

Base Case: First, we check the base case n 1:
n3 - n 13 - 1 1 - 1 0
Since 0 is divisible by 6, the base case holds.

Inductive Step: Now we assume that the statement is true for some integer k, i.e., we assume:

k3 - k is divisible by 6

This means there exists an integer m such that:

k3 - k 6m

We need to show that (k 1)3 - (k 1) is also divisible by 6:

(i) Expanding this expression:

(k 1)3 k3 3k2 3k 1
(ii) Thus,
(k 1)3 - (k 1) (k3 3k2 3k 1) - k - 1
(iii) Simplifying this, we get:

(k 1)3 - (k 1) k3 3k2 2k

(iv) Now we can rewrite this expression as:

k3 3k2 2k (k3 - k) 3k2 3k
(v) Using our induction hypothesis, since k3 - k 6m, we substitute:

(k 1)3 - (k 1) 6m 3k2 2k

(vi) Now we need to show that 3k2 2k is divisible by 6. We can factor this expression:

3k2 2k k(3k 2)

Divisibility by 2: If k is even, then k(3k 2) is even. If k is odd, 3k 2 is also even since 3 times odd 2 odd 2 even.
Thus, k(3k 2) is divisible by 2.

Divisibility by 3: Consider k mod 3: If k ≡ 0 mod 3, then k(3k 2) ≡ 0 mod 3. If k ≡ 1 mod 3, then 3k 2 ≡ 3*1 2 ≡ 5 ≡ 2 mod 3, which is not divisible by 3. If k ≡ 2 mod 3, then 3k 2 ≡ 3*2 2 ≡ 8 ≡ 2 mod 3, which is not divisible by 3.
From the above, we conclude that k(3k 2) is divisible by 3.

Since both k3 - k and 3k2 2k are divisible by 6, we have shown that:

(k 1)3 - (k 1) is also divisible by 6.

By the principle of mathematical induction, n3 - n is divisible by 6 for all integers n.

Alternative Approach Using Consecutive Integers

Let's consider the expression n3 - n. This can be rewritten as:

n3 - n n(n2 - 1) n(n - 1)(n 1)

This is the product of three consecutive integers n - 1, n, and n 1.

One of any set of three consecutive integers is a multiple of 3.
At least one of any set of three consecutive integers is a multiple of 2.

Since 2 and 3 are distinct prime numbers, and as n(n - 1)(n 1) is a multiple of both 2 and 3, it is a multiple of 2 * 3 6.

Therefore, n3 - n is divisible by 6.

Conclusion

In conclusion, we have provided two methods to prove that n3 - n is divisible by 6 for all integers n. The first method uses the principle of mathematical induction, while the second method leverages the properties of consecutive integers.