Proof:
Case 1: n is an even integer
Let n = 2k, where k is an integer.
Then n^2 + 3n + 5 = (2k)^2 + 3(2k) + 5
= 4k^2 + 6k + 5
= 2(2k^2 + 3k) + 5
Since 2k^2 + 3k is an integer, let it be represented by m.
Therefore, n^2 + 3n + 5 = 2m + 5
Since 2m is even and adding 5 to an even number results in an odd number, n^2 + 3n + 5 is an odd integer when n is even.
Case 2: n is an odd integer
Let n = 2k + 1, where k is an integer.
Then n^2 + 3n + 5 = (2k + 1)^2 + 3(2k + 1) + 5
= 4k^2 + 4k + 1 + 6k + 3 + 5
= 4k^2 + 10k + 9
= 2(2k^2 + 5k) + 9
Since 2k^2 + 5k is an integer, let it be represented by m.
Therefore, n^2 + 3n + 5 = 2m + 9
Since 2m is even and adding 9 to an even number results in an odd number, n^2 + 3n + 5 is an odd integer when n is odd.
Therefore, by considering both cases, it is proven that if n is an integer, then n^2 + 3n + 5 is an odd integer.