Proof Verification: $2|n^2-n,\ \forall n\ \in \mathbb Z$

Question

$$\text{Prove: } 2|n^2-n,\ \forall \ n \in \mathbb Z$$

My Proof is as follows. I am unsure as to how good it is and I presume there is a much better way of proving this that is simpler and clearer.

Proof: $2|n^2-n,\ \forall \ n \in \mathbb Z$ there are two cases, 1, when $n$ is odd, and 2, when $n$ is even.

Case 1: Let $n=2k+1$ $$2|(2k+1)^2 - (2k+1) = 4k^2 + 4k +1 - 2k - 1$$ $$2|4k^2 + 2k \ \text{ ...Which is true}$$

Case 2: Let $n=2k$ $$2|(2k)^2 - (2k) $$ $$2|4k^2 - 2k \ \text{ ...Which is also true}$$

Therefore as both Case 1 & 2 are true $2|n^2-n,\ \forall \ n \in \mathbb Z$

It is correct! You can also do this with $2 | n^2+n$, and see that $\frac{n^2+n}{2}=\frac{n(n+1)}{2}=1+2+3+\dots+n$. So $2 | n(n+1)$ and you can replace $n-1$ in $n$ and get the same result. — Tio Zuca, Nov 25 '23 at 16:18

score 6 · Accepted Answer · answered Nov 25 '23 at 16:16

6

Your proof is fine.

Here is how I prove it:

$n^2-n=n(n-1)$, it's a product of two consecutive integers, hence it is divisible by $2$.

answered Nov 25 '23 at 16:16

Siong Thye Goh

3

To add some detail, either $n$ or $n-1$ will be even…the claim follows – Andrew Nov 25 '23 at 16:23
Thank you, that’s a lot simpler than my proof, I will definitely keep this in mind for future proofs. – Ilikemath Nov 25 '23 at 16:33

score 0 · Answer 2 · answered Nov 25 '23 at 21:00

Alternatively notice that $n$ and $n^2$ have the same parity, so their difference is even.

You can prove it in a similar way:

This in facts opens the road to modular arithmetic (that you'll learn later in your studies), i.e.

$n^2\equiv n\pmod 2$ therefore $n^2-n\equiv n-n\equiv 0\pmod 2$

2 Answers2