Proving Divisibility Using Mathematical Induction and Modular Arithmetic

Understanding the divisibility of a number by another, such as demonstrating that 3^{2n^2} - 8n - 9 is divisible by 16 for all integers n geq 0, is a fundamental concept in number theory. This article explores this problem through both mathematical induction and modular arithmetic. Let's delve into the detailed steps.

Mathematical Induction

Mathematical induction is a method of mathematical proof that is used to establish the truth of a statement for an infinite sequence of cases. The general idea is to show that the statement is true for a base case and then prove that if it is true for any one case, it is also true for the next. Here’s how we prove the statement using this method:

Step 1: Base Case

First, we check the base case where n 0:

3^{2 cdot 0^2} - 8 cdot 0 - 9 3^0 - 0 - 9 1 - 9 -8

Since -8 is divisible by 16 (i.e., -8 equiv 8 equiv 0 mod 16), the base case holds.

Step 2: Inductive Step

Now, assume the statement holds for some integer n k:

3^{2k^2} - 8k - 9 equiv 0 mod 16

This implies:

3^{2k^2} - 8k - 9 16m text{ for some integer } m

Step 3: Inductive Hypothesis

We need to show that the statement holds for n k 1:

3^{2(k 1)^2} - 8(k 1) - 9

This simplifies to:

3^{2k^2 4k 2} - 8k - 8 - 9 3^{2(k^2 2k 1)} - 8k - 17 3^{2k^2 4k 2} - 8k - 17

Step 4: Express 3^{2k^2 4k 2} Using the Property of Exponents

Using the property of exponents, we can rewrite:

3^{2k^2 4k 2} 3^{2k^2} cdot 3^4 3^{2k^2} cdot 81

Substituting this back into our expression:

3^{2k^2 4k 2} - 8k - 17 81 cdot 3^{2k^2} - 8k - 17

Step 5: Replace 3^{2k^2} Using the Inductive Hypothesis

From our inductive hypothesis, we have:

3^{2k^2} 8k 9 16m

Substituting this into our expression gives:

81 cdot (8k 9 16m) - 8k - 17

Expanding this:

648k 729 1296m - 8k - 17

Combining like terms:

640k 712 1296m 64(10k 11 20m)

Since 64(10k 11 20m) is clearly divisible by 16, we have shown:

3^{2(k 1)^2} - 8(k 1) - 9 equiv 0 mod 16

Conclusion

By the principle of mathematical induction, 3^{2n^2} - 8n - 9 is divisible by 16 for all integers n geq 0.

Alternative Proof Using Modular Arithmetic

Alternatively, we can solve this problem using modular arithmetic. Let’s consider:

S_n 9^n - 9 - 8n

Given that 9 mod 16 -7 mod 16 and 9^2 mod 16 1 mod 16, we can rewrite:

9^{2k} 1 mod 16 for any k text{ (when } n 2k text{ is even})

9^{2k 1} 9 mod 16 text{ for any } k text{ (when } n 2k 1 text{ is odd})

When n text{ is even}:

S_n mod 16 9^{2k} - 9 - 16k mod 16 1 - 9 - 16k mod 16 -8 mod 16 0 mod 16

When n text{ is odd}:

S_n 9^{2k 1} - 9 - 16k - 8 mod 16 9 - 9 - 16k - 8 mod 16 -8 mod 16 0 mod 16

Thus S_n mod 16 0 mod 16 for all n. This concludes our alternative proof.

Both methods demonstrate the divisibility of 3^{2n^2} - 8n - 9 by 16 for all integers n geq 0. Such proofs are not only educational but also serve as foundational skills in number theory and computer science.

Keywords: Mathematical Induction, Divisibility, Modular Arithmetic