There and Back Again With Squares

TLDR; - There is no practical use for this, it's just a little pattern exploration to think about. Seriously, don't read it.

Have you ever noticed a pattern with how squares grow?

With every value of x, our results of x² is the result of (x-1)² added with (2x -1). Why is this? The pattern is more obvious if we see our squares visualized as a grid of blocks:

You can see how the results of x² can be broken up to other components including (x-1)² and (2x-1).
Cool. Is there another pattern we can toy with here? I've been
interested in breaking up these square grids in other patterns,
especially in patterns that deal with whatever x is. So consider selecting a diagonal in our square grids. I'm going to choose a few different sizes here:

In a 3x3 square, a diagonal is 3 squares. For our 5x5 square, it's 5 squares. And for our 9x9, it's 9. So for a grid of x², its diagonal is x squares (or units). If we wanted to find another way to express x², we could break it up and include it's diagonal added to the other components in the grid.

What are the other components of the grid? 2 triangles of same size leftover from the diagonal cutting the box in half. I want to express the area of these grids (these happen to be the same as x², BTW) in terms of these components, but I don't know the formula at this point. I just see the start of a pattern. So lets say x²=x+2(a). What's the 'x'? That's the diagonal. What's the 'a'? That's our undetermined area of 1 of our 2 triangles (that's why we are multiplying by 2). Now lets consider what our 'a' would be for each value of x and see if there's a pattern.

The most obvious pattern is that as x grows, so does the value of a. This is expected. Why? The triangles in our boxes get bigger as our overall square (or x) gets bigger. This means there is some kind of relationship between x and a. Let's see if we can figure out a way to express the relationship.

There are several ways to get where we are going, such as putting the area of a triangle formula in terms of whatever x is - we will get to that later. But instead I want to simply explore what happens if we break up our unknown triangle area into 2 components: x and a new unknown we will refer to as 'b'. In other words, we will let a=x+b. In this case, our above table can look like this:

So for x=1, b=-1. x=2, b=-1. For x=3, b=0. x=4, b=2.

If we throw these on a number graph we can see that we are looking at a quadratic formula with a (very) rough shape like this for our b values:

Because the graph intercepts the x axis at 0 and 3 (where f(x)=0) we know from a review of graphing quadratics that our function for b will look something like f(x)=x(x-3). However, this would produce a parabola that is too pronounced and the values are doubled (f(1)=-2, f(2)=-2, f(4)=4). So the quick fix and formula to put our b in terms of x is x(x-3)/2.

When we plug this back into our formula, we get an overall formula for x² that looks like:
x+2(x+(x(x-3)/2))

I'm a believer that definitions shouldn't include the term they are defining which means we are creating complexity we don't need. Obviously this is a more difficult definition of x² than the simple x*x that we know. But lets continue with the exercise anyways. Let's simplify this a little more:
x+2((x²-x)/2)

This is an interesting place to pause at.

Triangular Number Sequence

Remember that the part in parentheses (the (x²-x)/2 part) that we are multiplying by 2 describes the growth of our triangles that we've divided with a diagonal. This is almost like an already well known formula called the triangle number sequence (see more about the triangular number sequence): this formula is (x²+x)/2.

As soon as we recognized our equilateral triangles, we could have started using this function if we were aware of it. This would be us putting the area of the triangles in terms of x as I alluded to in the beginning. Consider this 5 level triangle of asterisks:

*
**
***
****
*****

Plugging 5 into our triangle sequence formula gives us
(5²+5)/2 = 15
(which is indeed how many *'s we have in the above example).
Why is our triangle number formula shifted to (x²-x)/2 in our squares formula? Because our triangles aren't being formed until x=2. When x=2, our triangle dimensions are 1.

Go back and look at the boxes with the yellow diagonals and you'll see that when x=5, the triangle formula would have x=4 (for our 2 triangles divided by the yellow diagonal). If we plug x-1 into our triangle sequence formula we get:
((x-1)²+(x-1))/2 = (x²-2x+1+x-1)/2 = (x²-x)/2

We are almost there.

When we last paused our formula discussion we were at:
x²=x+2((x²-x)/2)

and this quickly simplifies to
x²=x+x²-x

and finally, and of course,
x²=x²

And now you're back where you started. This was like taking a trip to the grocery store only to realize you forgot your wallet and you have to go back home. Maybe not as helpful as a proof, but hopefully a little more fun to read.

edit 1/16 - fixed equation typos