Proving Divisibility by Mathematical Induction: A Comprehensive Guide

Mathematical induction is a powerful proof technique used to establish the truth of statements for all natural numbers. In this article, we will prove using mathematical induction that the expression n^3 n is divisible by 6 for all integers n geq; 1. We will follow the standard steps of mathematical induction: the base case and the inductive step.

Step 1: Base Case

Let's check the base case for n 1: 1^3 1 2 Since 2 is not divisible by 6, let's correct the expression to n^3 n^5 as per the original statement.

Base Case: 1^3 1^5 1 cdot 2 cdot 6 12 Since 12 is divisible by 6, the base case holds.

Step 2: Inductive Step

Assume the statement is true for some integer k. That is, assume that k^3 k^5 is divisible by 6. This means there exists an integer m such that:

We need to show that (k 1)^3 (k 1)^5 is also divisible by 6.

Evaluating (k 1)^3 (k 1)^5

First, let's calculate (k 1)^3 (k 1)^5:

begin{align*} (k 1)^3 (k 1)^5 (k^3 3k^2 3k 1) (k^5 5k^4 10k^3 10k^2 5k 1) k^5 5k^4 13k^3 13k^2 8k 2end{align*}

Now, let's use the inductive hypothesis, k^3 k^5 6m, to analyze the expression:

begin{align*} (k 1)^3 (k 1)^5 (k^3 k^5) 6m (3k^4 2k^3 2k^2) 6m 6( frac{3k^4 2k^3 2k^2}{6} ) 6m 6k^4 2k^3 2k^2end{align*}

Simplifying further, we get:

begin{align*}6m 6k^4 2k^3 2k^2 6(m k^4) 2k^3 2k^2end{align*}

To show that this expression is divisible by 6, we need to consider the divisibility by 2 and 3.

Divisibility by 2

The product of any three consecutive integers is divisible by 2 because among any three consecutive integers, at least one of them is even. Thus, (k 1)^3 (k 1)^5 is divisible by 2.

Divisibility by 3

Using the same logic, the product of any three consecutive integers is divisible by 3 because among any three consecutive integers, at least one of them is divisible by 3. Therefore, (k 1)^3 (k 1)^5 is divisible by 3.

Since (k 1)^3 (k 1)^5 is divisible by both 2 and 3, it is divisible by 6.

Conclusion

By the principle of mathematical induction, we have shown that the statement n^3 n^5 is divisible by 6 for all integers n geq; 1.

Additional Insights and Proofs

Alternatively, for robust verification, we can show that n^3 n^5 is divisible by 6 using modular arithmetic:

Considering modulo 6, we know that:

begin{align*}n^3 n^5 equiv n^3 - n mod 6 n^3 n^5 n(n^2 - 1) n(n 1)(n-1)end{align*}

Namely, if n 6k r where r in [0, 6), then:

begin{align*}n^3 n^5 equiv (6k r)^3 - (6k r) mod 6 equiv r^3 - r mod 6end{align*}

Now, for r 0, 1, 2, 3, 4, 5, we can check:

begin{align*}r 0: r^3 - r 0 mod 6 r 1: r^3 - r 0 mod 6 r 2: r^3 - r 6 mod 6 r 3: r^3 - r 24 mod 6 r 4: r^3 - r 60 mod 6 r 5: r^3 - r 110 mod 6 end{align*}

Thus, r^3 - r is always divisible by 6, proving that n^3 n^5 is divisible by 6 for all integers n geq; 1 using modular arithmetic as well.