I have a question regarding the correctness of my proof (and the point of error in case of its incorectness), since the logic varies from the given answer proof.
Statement
Prove or disprove: Let k ∈ N. If $k^2 + 5k$ is odd, then $(k + 1)^2 + 5(k + 1)$ is odd.
Textbook proof
Assume that $k^2 + 5k$ is an odd integer, where k ∈ N. Then $k^2 + 5k = 2ℓ + 1$ for some integer ℓ. Hence, $(k + 1)^2 + 5(k + 1) = k^2 + 2k + 1 + 5k + 5 = (k^2 + 5k) + (2k + 6) = (2ℓ + 1) + (2k + 6) = (2ℓ + 2k + 6) + 1 = 2(ℓ + k + 3) + 1.$ Since ℓ + k + 3 is an integer, $(k + 1)^2 + 5(k + 1)$ is an odd integer.
My proof
Case 1: k is odd. Then k = 2a+1, a $\in$ $\mathbb{Z}$. Then
$k^2 + 5k = (2k+1)^2 +5(2k+1) = 4k^2+4k+1+10k+5 = 2(k^2+7k+3).$
Since $k^2+7k+3\in\mathbb{Z}$, $k^2+5k$ is even.
Case 2: k is even. Then k = 2b, $b\in\mathbb{Z}$. Then
$k^2+5k=(2b)^2+5(2b)=4b^2+10b=2(2b^2+5b)$
Since $2b^2+5b\in\mathbb{Z}$, $k^2+5k$ is even.
Hence, the statement is true vacuously.
Indeed this is a vacuous truth, your proof is correct. However, you could simplify it even more by using modular arithmetic:
Case 1: $k$ odd, so $k\equiv 1 \mod 2$, so we have $$ k^2+5k \equiv k^2+1\cdot k \equiv 1 + 1 \equiv 0 \mod 2 $$ and similarly for case 2 that $k$ is even, i.e. $k\equiv 0\mod 2$: $$ k^2+5k \equiv 0 + 5\cdot 0 \equiv 0 \mod 2 $$ Then you don't even need to multiply anything out.
Or in fact as @lulu mentioned, just noting that $k(k+5)$ is a product of two different parities resulting in an even number.