Sure, let's explore why [tex]\( p^2 + 1 \)[/tex] is always an even number when [tex]\( p \)[/tex] is an odd number.
### Step-by-Step Solution:
1. Understanding Odd Numbers:
- An odd number can be represented in the form [tex]\( p = 2n + 1 \)[/tex], where [tex]\( n \)[/tex] is an integer. This representation ensures that [tex]\( p \)[/tex] is always odd.
2. Expanding [tex]\( p^2 \)[/tex]:
- Let's square the odd number [tex]\( p = 2n + 1 \)[/tex].
- [tex]\( p^2 = (2n + 1)^2 \)[/tex]
3. Applying the Binomial Theorem:
- Expand [tex]\( (2n + 1)^2 \)[/tex]:
[tex]\[
(2n + 1)^2 = (2n)^2 + 2 \cdot 2n \cdot 1 + 1^2
\][/tex]
- Simplifying it:
[tex]\[
(2n + 1)^2 = 4n^2 + 4n + 1
\][/tex]
4. Adding 1 to [tex]\( p^2 \)[/tex]:
- Now, add 1 to [tex]\( p^2 \)[/tex]:
[tex]\[
p^2 + 1 = 4n^2 + 4n + 1 + 1
\][/tex]
- Simplifying it further:
[tex]\[
p^2 + 1 = 4n^2 + 4n + 2
\][/tex]
5. Factoring Out Common Terms:
- Notice that [tex]\( 4n^2 + 4n + 2 \)[/tex] can be factored to pull out a 2:
[tex]\[
4n^2 + 4n + 2 = 2(2n^2 + 2n + 1)
\][/tex]
6. Conclusion:
- The expression [tex]\( 2(2n^2 + 2n + 1) \)[/tex] shows that [tex]\( p^2 + 1 \)[/tex] is clearly a multiple of 2.
- Hence, [tex]\( p^2 + 1 \)[/tex] must be an even number because any number that can be expressed as [tex]\( 2 \times \)[/tex] some integer is even.
Therefore, [tex]\( p^2 + 1 \)[/tex] is always an even number when [tex]\( p \)[/tex] is an odd number.
