Problem 20 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We need to demonstrate that the expression \(3^{2n} - 1\) is divisible by 8 for any natural number \(n\). This means when we divide \(3^{2n} - 1\) by 8, the remainder should be 0.

Step 2: Simplify the Expression

First, rewrite \(3^{2n}\) in terms of small powers to observe a pattern. Start with small values of \(n\) and observe:* For \(n = 1\), \(3^{2\cdot1} = 9\), so \(3^{2\cdot1} - 1 = 8\).* For \(n = 2\), \(3^{2\cdot2} = 81\), so \(3^{2\cdot2} - 1 = 80\).* For \(n = 3\), \(3^{2\cdot3} = 729\), so \(3^{2\cdot3} - 1 = 728\).

Step 3: Identify the Pattern

Notice that each computed result of \(3^{2n} - 1\) is a multiple of 8. Thus, it suggests the expression is generally divisible by 8 for any positive integer \(n\).

Step 4: Prove by Induction

To prove by induction:\[P(n): 3^{2n} - 1\text{ is divisible by }8\]**Base Case**:For \(n = 1\), \(3^{2\cdot1} - 1 = 8\), which is divisible by 8.**Inductive Step**: Assume true for some \(n = k\), that is, \(3^{2k} - 1 = 8m\) for some integer \(m\).Now, prove it for \(n = k+1\):\[3^{2(k+1)} - 1 = 3^{2k+2} - 1 = (3^{2k}\cdot 9) - 1 \]Using the assumption:\[3^{2k} = 8m + 1 \]Substituting:\[((8m + 1) \cdot 9) - 1 = 72m + 9 - 1 = 72m + 8\]Clearly, \(72m + 8\) is divisible by 8, so \(P(k+1)\) is true. Thus, by induction, \(3^{2n} - 1\) is divisible by 8 for all natural numbers \(n\).

Step 5: Conclusion

The expression \(3^{2n} - 1\) is indeed divisible by 8 for all natural numbers \(n\), as proved by mathematical induction and confirmed through the pattern.

Key Concepts

Divisibility

Divisibility is a concept in mathematics where one integer can be divided by another without leaving a remainder. In this exercise, we need to show that \(3^{2n} - 1\) is divisible by 8. This implies that when \(3^{2n} - 1\) is divided by 8, the remainder is zero.
The remainder of a division is what is left over after the division. For instance, 17 divided by 5 leaves a remainder of 2, because 5 times 3 is 15, and 17 minus 15 is 2. However, in divisibility, we want this remainder to be zero.
To check divisibility, we start with small values of the variable involved and look for patterns. By verifying that the initial few values of \(3^{2n} - 1\) (like 8, 80, etc.) are divisible by 8, it suggests a pattern of divisibility, which we can then prove holds true for all natural numbers \(n\). When proving such patterns mathematically, tools like mathematical induction are often used.

Exponentiation

Exponentiation refers to the process of raising a number to a power. The power, often called the exponent, denotes how many times the number, which is the base, is multiplied by itself. For example, \(3^2 = 3 \cdot 3 = 9\).
In this exercise, we have the expression \(3^{2n}\). This means that the base 3 is raised to the power represented by \(2n\), essentially meaning \(3\) is multiplied by itself \(2n\) times. It's a powerful concept because it allows very large numbers to be expressed concisely, making the calculations needed in proofs more manageable.
When working with exponentiation in divisibility problems, observing patterns in smaller powers can be helpful. Here, we observe how numbers like \(3^2\), \(3^4\), and \(3^6\) behave concerning divisibility by 8, building confidence in a pattern before formally proving it through methods like induction.

Algebraic Proofs

Algebraic proofs, such as those using mathematical induction, are methods used to prove that a mathematical statement is true for all natural numbers. The goal is to show not just for a select few values, but for an infinite number by establishing a base case and then using it to prove further cases.
This specific problem is solved using mathematical induction, a popular technique for proving statements about natural numbers. The proof involves two main steps:

• Base Case: Verify the statement for the first number, typically \(n = 1\). In this case, we find \(3^{2 \cdot 1} - 1 = 8\), which is divisible by 8.
• Inductive Step: Assume the statement is true for some arbitrary natural number \(k\). This is our induction hypothesis.

Next, we demonstrate that if it holds for \(n = k\), it must also hold for \(n = k+1\). By substituting and simplifying, we show that under the assumption that \(3^{2k} - 1\) is divisible by 8, we can prove that \(3^{2(k+1)} - 1\) is as well.
Thus, by combining the base case and the inductive step, we conclude that \(3^{2n} - 1\) is divisible by 8 for all natural numbers \(n\). This approach effectively covers all possible cases, justifying the pattern we observed earlier in small examples.