Using principle of mathematical induction, prove that `7^(4^(n)) -1` is divisible by `2^(2n+3)` for any natural number n.

Options

A. `2^(2n)`
B. `2^(2k+5)`
C. `3^(2n)`
D. `2^(n+3)`

Explanation

To prove that \( 7^{4^n} - 1 \) is divisible by \( 2^{2n + 3} \) for any natural number \( n \) using the principle of mathematical induction, we follow these steps:

### Step 1: Base Case
We start by checking the base case \( n = 1 \).

\[ 7^{4^1} - 1 = 7^4 - 1 \]

Calculating \( 7^4 \):

\[ 7^4 = 2401 \]

Thus,

\[ 7^4 - 1 = 2401 - 1 = 2400 \]

Now, we check if \( 2400 \) is divisible by \( 2^{2(1) + 3} = 2^5 = 32 \).

Calculating \( 2400 \div 32 \):

\[ 2400 \div 32 = 75 \]

Since \( 75 \) is an integer, the base case holds true.

### Step 2: Inductive Hypothesis
Assume that the statement is true for \( n = k \), i.e., assume that:

\[ 7^{4^k} - 1 \text{ is divisible by } 2^{2k + 3} \]

This means there exists an integer \( m \) such that:

\[ 7^{4^k} - 1 = m \cdot 2^{2k + 3} \]

### Step 3: Inductive Step
We need to prove that the statement is true for \( n = k + 1 \):

\[ 7^{4^{k+1}} - 1 \]

We can rewrite \( 7^{4^{k+1}} \) as:

\[ 7^{4^{k+1}} = 7^{4 \cdot 4^k} = (7^{4^k})^4 \]

Thus,

\[ 7^{4^{k+1}} - 1 = (7^{4^k})^4 - 1 \]

Using the difference of squares, we can factor this:

\[ (7^{4^k} - 1)((7^{4^k})^3 + (7^{4^k})^2 + 7^{4^k} + 1) \]

From our inductive hypothesis, we know that \( 7^{4^k} - 1 \) is divisible by \( 2^{2k + 3} \). Therefore, we can write:

\[ 7^{4^k} - 1 = m \cdot 2^{2k + 3} \]

Now, we need to analyze the second factor:

\[ (7^{4^k})^3 + (7^{4^k})^2 + 7^{4^k} + 1 \]

Let \( x = 7^{4^k} \). The expression becomes:

\[ x^3 + x^2 + x + 1 \]

Now, we will check the parity of \( x \):

Since \( 7 \equiv 1 \mod 2 \), we have \( 7^{4^k} \equiv 1 \mod 2 \). Thus, \( x \equiv 1 \mod 2 \).

Calculating \( x^3 + x^2 + x + 1 \mod 2 \):

\[ 1^3 + 1^2 + 1 + 1 \equiv 1 + 1 + 1 + 1 \equiv 0 \mod 2 \]

This shows that \( x^3 + x^2 + x + 1 \) is even.

### Step 4: Combine Results
Since \( 7^{4^k} - 1 \) is divisible by \( 2^{2k + 3} \) and \( x^3 + x^2 + x + 1 \) is even, we can conclude that:

\[ 7^{4^{k+1}} - 1 = (7^{4^k} - 1)((7^{4^k})^3 + (7^{4^k})^2 + 7^{4^k} + 1) \]

is divisible by:

\[ 2^{2k + 3 + 1} = 2^{2(k + 1) + 3} \]

Thus, the statement holds for \( n = k + 1 \).

### Conclusion
By the principle of mathematical induction, we conclude that \( 7^{4^n} - 1 \) is divisible by \( 2^{2n + 3} \) for all natural numbers \( n \).

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