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

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?

• @HappyDay That was intentional, and was what was asked of them. Of course $n(n+2)$ is not divisible by $4$ for odd $n$. Commented Mar 3, 2023 at 14:13
• Oh, I'm sorry. I misread the question. Commented Mar 3, 2023 at 14:14
• @ OP, this is a correct proof, though as you pointed out this is not really a proof by induction but is rather a direct proof. I suppose if you insist on phrasing this by induction, you could try to fiddle with it all, writing as $(2k+2)(2k+2+2) = (2k)(2k+2)+4k+8+4k+4$... and using the induction hypothesis for the piece on the left and elementary number theory to say the items on the right are all multiples of four... Personally, I see this as a problem with the question itself and would not have asked this question in such a way as to imply it needed induction. Commented Mar 3, 2023 at 14:17
• I assume the intention of the problem was starting with $n=2$, and then showing the statement is true for $n=k+2$ if it is true for $n=k$. Commented Mar 3, 2023 at 14:18

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 $$4$$.

• Maybe the person who originally posed the question wanted us to notice that $(n+2)(n+4) = n(n+2) + 4(n+2)$, assert that $n(n+2)$ is a multiple of 4 by induction, etc, etc. Commented Mar 3, 2023 at 14:22
• @kimchilover My course instructor told us to prove by induction, so I will go with your suggestion, and discard my attempt. Commented Mar 3, 2023 at 14:25
• It’s even a multiple of 8
– lhf
Commented Mar 3, 2023 at 15:07

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,

• I'm nitpicking: basis case is slightly wrong. P(0) is not valid as the basis step since 0 is not positive. Commented Mar 3, 2023 at 14:28
• Ok, I extended the proof to all non-negative integers ;) Commented Mar 3, 2023 at 14:31