Integral part of `(7 + 5sqrt2)^(2n+1) ` is `(n in N)`

Options

A. an even number
B. an odd number
C. an even or an odd number depending upon the value of n
D. nothing can be said

Answer

To find the integral part of \( (7 + 5\sqrt{2})^{2n+1} \) for \( n \in \mathbb{N} \), we can follow these steps:

### Step 1: Define the expression
Let \( x = (7 + 5\sqrt{2})^{2n+1} \) and \( y = (7 - 5\sqrt{2})^{2n+1} \).

### Step 2: Analyze the second term
Since \( 7 - 5\sqrt{2} \) is a positive number less than 1 (approximately 0.071), \( y \) will be a very small positive number as \( n \) increases. Thus, \( 0 < y < 1 \).

### Step 3: Write the sum
Now, consider the sum:
\[ x + y = (7 + 5\sqrt{2})^{2n+1} + (7 - 5\sqrt{2})^{2n+1} \]
This sum is an integer because it is the sum of two conjugate expressions.

### Step 4: Identify the integral part
The integral part of \( x \) can be expressed as:
\[ \lfloor x \rfloor = x + y - y \]
Since \( 0 < y < 1 \), we have:
\[ \lfloor x \rfloor = (7 + 5\sqrt{2})^{2n+1} + (7 - 5\sqrt{2})^{2n+1} - y \]
This shows that the integral part of \( x \) is equal to the integer part of \( x + y \) minus a small fraction \( y \).

### Step 5: Conclude the result
Since \( x + y \) is an integer and \( y \) is a small positive number, the integral part of \( x \) is:
\[ \lfloor x \rfloor = (7 + 5\sqrt{2})^{2n+1} + (7 - 5\sqrt{2})^{2n+1} - 1 \]
Thus, the integral part of \( (7 + 5\sqrt{2})^{2n+1} \) is an odd integer.

### Summary of the steps:
1. Define \( x \) and \( y \).
2. Analyze \( y \) as a small positive number.
3. Write the sum \( x + y \) as an integer.
4. Identify the integral part of \( x \).
5. Conclude that the integral part is an odd integer.