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The problem in the image reads:

"**El número n es un número impar. Explique por qué \( n^2 + 2 \) siempre es un número impar.**"

### Explanation:

- \( n \) is an odd number.
- The square of an odd number, \( n^2 \), is always odd. Here's why:
  - Let \( n = 2k + 1 \), where \( k \) is an integer (this is the general form of any odd number).
  - Squaring \( n \): 
    \[
    n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1
    \]
    - This expression shows that \( n^2 \) is odd because it is of the form \( 2m + 1 \), where \( m \) is an integer.
  
- Now, \( n^2 + 2 \):
  - Since \( n^2 \) is odd, we add 2 (an even number) to it:
    \[
    n^2 + 2 = \text{odd} + \text{even} = \text{odd}.
    \]
  - Therefore, \( n^2 + 2 \) will always be odd because the sum of an odd and even number is always odd.

Thus, **\( n^2 + 2 \) is always odd** when \( n \) is odd.

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Do you need further details or have any questions about this? Here are five related questions to deepen understanding:

1. Why is the square of an even number always even?
2. What happens when you add two odd numbers? Is the result even or odd?
3. Can \( n^2 + 3 \) also be an odd number for an odd \( n \)? Why or why not?
4. How does the general form \( n = 2k + 1 \) help prove properties of odd numbers?
5. How can you prove algebraically that the sum of any odd number and an even number is always odd?

**Tip**: To explore number properties further, practicing with specific examples can make abstract concepts clearer!

El número n es un número impar. Explique por qué n^2 + 2 siempre es un número impar.

Explore why n^2 + 2 is always an odd number when n is odd. This problem uses key algebraic concepts and demonstrates the relationship between odd and even numbers, suitable for students in Grades 8-10.

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