Proving the Divisibility of (9^n - 4^n) by 5 Using Mathematical Induction

Mathematical induction is a powerful proof technique that can be used to prove statements about all natural numbers or integers. In this article, we will provide a detailed step-by-step proof to show that (9^n - 4^n) is always divisible by 5 for all integers (n geq 0). We will cover the principles of mathematical induction, explain each step, and provide a clear proof for the given statement.

Step 1: Base Case

First, we verify the base case, where (n 0).

[9^0 - 4^0 1 - 1 0]

Since 0 is divisible by 5, the base case holds true.

Step 2: Inductive Hypothesis

Assume that the statement is true for some integer (k). This means that:

[9^k - 4^k 5m]

where (m) is an integer. We will use this hypothesis to prove the statement for (n k 1).

Step 3: Inductive Step

Next, we need to show that the statement holds for (n k 1), i.e., we need to prove that:

[9^{k 1} - 4^{k 1} 5p]

for some integer (p).

Starting with the expression for (n k 1):

[9^{k 1} - 4^{k 1} 9 cdot 9^k - 4 cdot 4^k]

Using the inductive hypothesis, we can write:

[9^{k 1} - 4^{k 1} 9 cdot (9^k - 4^k 4^k) - 4 cdot 4^k]

[9^{k 1} - 4^{k 1} 9 cdot 9^k - 4 cdot 4^k 9 cdot 4^k - 4 cdot 4^k]

[9^{k 1} - 4^{k 1} 9 cdot 9^k - 9 cdot 4^k 5 cdot 4^k]

[9^{k 1} - 4^{k 1} 5 cdot 5m 5 cdot 4^k]

[9^{k 1} - 4^{k 1} 5(5m 4^k)]

This shows that (9^{k 1} - 4^{k 1}) is a multiple of 5, which completes the inductive step.

Conclusion

By the principle of mathematical induction, the statement (9^n - 4^n) is divisible by 5 for all integers (n geq 0).

While we have used mathematical induction to prove the statement, it's worth noting that the same result can be obtained using a simpler observation. Notice that:

[9^n - 4^n (9 - 4)(9^{n-1} 9^{n-2} cdot 4 cdots 4^{n-1}) 5(9^{n-1} 9^{n-2} cdot 4 cdots 4^{n-1})]

Since the expression inside the parentheses is an integer, (9^n - 4^n) is divisible by 5. This completes the proof without using mathematical induction.