Problem 18 - Algebra and Trigonometry

Question

Show that \(3^{2 n}-1\) is divisible by 8 for all natural numbers \(n.\)

Step-by-Step Solution

Answer

By induction, \(3^{2n} - 1\) is divisible by 8 for all natural numbers \(n\).

Step 1: Base Case

Let's check if the statement is true for the smallest natural number, which is typically \(n = 1\).Substituting \(n = 1\) into the expression, we get:\[3^{2(1)} - 1 = 3^2 - 1 = 9 - 1 = 8\]Since 8 is divisible by 8, the base case holds true.

Step 2: Inductive Hypothesis

Assume that for some natural number \(k\), the expression \(3^{2k} - 1\) is divisible by 8. This means:\[3^{2k} - 1 = 8m\]for some integer \(m\). This is our inductive hypothesis.

Step 3: Inductive Step

We need to show that if the hypothesis holds for \(n = k\), it also holds for \(n = k + 1\).Consider the expression for \(n = k + 1\):\[3^{2(k+1)} - 1 = 3^{2k+2} - 1\]Rewrite it as:\[3^{2k+2} - 1 = (3^{2k})(3^2) - 1 = 9 imes 3^{2k} - 1\]Using our inductive hypothesis, substitute \(3^{2k} = 8m + 1\):\[9(3^{2k}) - 1 = 9(8m + 1) - 1 = 72m + 9 - 1 = 72m + 8\]This can be rewritten as:\[72m + 8 = 8(9m + 1)\]Thus, \(3^{2(k+1)} - 1 = 8(9m + 1)\) is divisible by 8.

Step 4: Conclusion by Mathematical Induction

Since the base case is true and the truth of the statement for \(n = k\) implies its truth for \(n = k + 1\), we conclude by the principle of mathematical induction that \(3^{2n} - 1\) is divisible by 8 for all natural numbers \(n\).

Key Concepts

Divisibility

Divisibility is a fundamental concept in mathematics. It refers to the ability of one number to be divided by another without leaving a remainder. In this exercise, we explore whether the expression \(3^{2n} - 1\) can be divided by 8 for every natural number \(n\). To prove this, we use a powerful technique called mathematical induction.

Here's a simple overview of divisibility:

• A number \(a\) is divisible by \(b\) if there is an integer \(q\) such that \(a = b \cdot q\).


• If \(a\) divided by \(b\) leaves no remainder, \(a\) is divisible by \(b\).


• Example: 16 is divisible by 4 because 16 divided by 4 equals 4 with no remainder.

Understanding divisibility helps establish foundational knowledge in number theory and algebra. It is crucial for proving properties of numbers, like the one in our problem.

Natural Numbers

Natural numbers are the set of positive integers beginning from 1, sometimes including 0. They are the building blocks for arithmetic and number theory.

Some key points about natural numbers include:

• The natural numbers set is usually represented as \( \mathbb{N} \).
• Commonly, \( \mathbb{N} \) contains \{1, 2, 3, ... \}.
• Natural numbers are infinite as they continue indefinitely.

In the exercise, we are tasked with proving a statement for all natural numbers \(n\). This means, once established using mathematical induction, our result holds true for every possible value of \(n\) within this infinite set.

Algebraic Proof

Algebraic proofs are methods of showing the truth of a mathematical statement through algebraic expressions and reasoning. They consist of logical steps, often employing variables to symbolize unknown values.

Here's how algebraic proof applies to our problem:

• We begin with a **base case**: demonstrating the statement is true for \(n = 1\).
• Next, we assume it's true for \(n = k\) (inductive hypothesis).
• Finally, the **inductive step**: proving it holds for \(n = k + 1\).

The goal is to illustrate that if one case holds true, the next also does, thereby covering an infinite sequence of possibilities. As step-by-step logic forms the core of algebraic proofs, it demands attention to detail and understanding of how to manipulate expressions algebraically to reach a conclusion.