The problem states that if $n$ is an odd integer, then $n^2 + 3n + 5$ is odd. We need to prove whether this statement is true or false.

1. Problem Description

The problem states that if $n$ is an odd integer, then $n^2 + 3n + 5$ is odd. We need to prove whether this statement is true or false.

2. Solution Steps

Since $n$ is an odd integer, we can write $n$ as $n = 2k + 1$, where $k$ is an integer.

Now, substitute $n = 2k + 1$ into the expression $n^2 + 3n + 5$:

$n^2 + 3n + 5 = (2k + 1)^2 + 3(2k + 1) + 5$

$= (4k^2 + 4k + 1) + (6k + 3) + 5$

$= 4k^2 + 4k + 1 + 6k + 3 + 5$

$= 4k^2 + 10k + 9$

$= 4k^2 + 10k + 8 + 1$

$= 2(2k^2 + 5k + 4) + 1$

Since $2k^2 + 5k + 4$ is an integer, let $m = 2k^2 + 5k + 4$. Then the expression becomes $2m + 1$, which is the form of an odd integer.

Therefore, if $n$ is an odd integer, then $n^2 + 3n + 5$ is odd.

3. Final Answer

True