Is my proof by mathematical induction that $n(n+2)$ is divisble by 4 correct?

Question

Problem:

Prove that $n(n+2)$ is divisible by $4$ by using mathematical induction, if $n$ is any even positive integer.

My attempt:

$P(q):$ "$2q(2q+2)$ is divisible by $4$", where $q$ is a natural number. Note that we get the original expression if we substitute $2q$ with $n$.

Basis step:

$P(1)$ is true because $2(1)(2(1)+2)=8$ is divisible by $4$.

Inductive step:

Inductive hypothesis: Let $P(k)$ is true, where $k$ is any arbitrary natural number.

Now, we must show that $P(k+1)$ is also true under this assumption. That is, we have to show that

$$(2k+2)(2k+2+2)=(2k+2)(2k+4)=4(k+1)(k+2)$$ is divisible by $4$.

$4(k+1)(k+2)$ is divisible by $4$.

Both the basis and inductive steps have been proved. So, $P(q)$ is always true, where $q$ is a natural number. In other words, $n(n+2)$ is divisible by $4$.

My question:

Is my proof correct? We did not need to use the inductive hypothesis at all, which we usually do. Is that okay?

Answer 1

The proof looks ok, though, as you said, induction is not needed for it. One can simply notice that if $n$ is even, then $n=2k$ which means $$n(n+2) = 2k (2k + 2) = 2\cdot k \cdot 2 \cdot (k+1) = 4\cdot k(k+1)$$ and the final expression is obviously divisible by $4$.

Answer 2

Define $P(n): n(n+2) \space \text{is divisible by} \space 4$.

Clearly $P(0)$ is true ($0$ is divisible by $4$). Now, let's assume $P(k)$ is true. We are left with showing that $P(k+2)$ is true. In other words, we know $k(k+2)$ is divisible by $4$, and would like to show that $(k+2)(k+2+2)$ is also divisible by $4$.

$$(k+2)(k+2+2)=(k+2)(k+4)=k(k+2)+4(k+2)$$

Obviously $4(k+2)$ is divisible by $4$, and by our induction hypothesis so is $k(k+2)$. Thus, we have a sum of two terms that are divisible by $4$, which is also divisible by $4$. This is precisely what we needed to prove.

By the way, your proof is technically correct, though as you mentioned, it is not really induction,