## n²-n is always even: a visual proof

start by drawing a **n** by **n** grid of points

```
╭─────── n ───────╮
╭ o o o o o o o
│ o o o o o o o
│ o o o o o o o
n o o o o o o o
│ o o o o o o o
│ o o o o o o o
╰ o o o o o o o
```

take the dots on the main diagonal and move them over to the side

```
╭─────── n ───────╮ 1
╭ o o o o o o • ╮
│ o o o o o o • │
│ o o o o o o • │
n o o o o o o ==> • n
│ o o o o o o • │
│ o o o o o o • │
╰ o o o o o o • ╯
```

we can see the points in the diagonal can be arranged into a line of **1** by **n** points, thus the main square without the diagonal is **n²-n**

we can also see that the remaining dots are divided up into two mirror images of each other, mirrored along the main diagonal. Since it can be evenly divided into two, it the remainder must be even

```
╭─────── n ───────╮
╭ • • • • • •
│ o • • • • •
│ o o • • • •
n o o o • • •
│ o o o o • •
│ o o o o o •
╰ o o o o o o
```

We can also slide all the point of one halve towards the other, creating a rectangle

```
╭───── n-1 ────╮
╭ • • • • • •
│ o • • • • •
│ o o • • • •
n o o o • • •
│ o o o o • •
│ o o o o o •
╰ o o o o o o
```

This gives us the factored form of the expression n (n-1)

From this expression we can see that it must always produce an even result as it takes the product of two consecutive numbers, one of them must be even, and the product of two numbers is even if either number is even