This has probably been known for several thosand years, but not by me. (Well, ok, I'm not that old, but...) I found out several weeks ago when I was really bored. (and I do mean bored!)

```(All variables are integers.)
Expand:      n = x^2 - x^2 + 2x - 1
Simplify:    n = 2x - 1
Solve for x: x = { (n-1)/2 , (n+1)/2 }
```

Ok... well as a mathematical proof, that probably sucks goldfish, but it works for me! Now, I'm sure someone would have a use for that...

Also, for all multiples of four: x = { (n-4)/4 , (n+4)/4 }

Not only is this true, but any odd number can be represented as the difference of two consecutive squares.

This is intimately related with basic calculus. The derivative (that is, rate of change) of a quadratic function like x^2 will change linearly as one moves along the function. That is, the rate of change at a given integer x-value along x^2 is two plus the rate of change at the previous integer x-value.

This trend is the same for higher powers of x. The rate of change of x^3 is proportional to x^2, and the rate of change of x^4 is proportional to x^3, etc.

BTW, I am one of those math nuts.

I think that what you have noticed is another way of saying that for any positive integer n, n2 is equal to the sum of the first n odd numbers, e.g. 32 is the sum of 1, 3, and 5, the first 3 odd numbers. While this fact seems odd when thought of this way, it makes perfect sense if you consider playing blocks with a kid.

Imagine, for a moment, that you have a huge pile of blocks and that you are making perfect squares out of them. You start with a 1x1 square.

```          X
```

Now, to make it a 2x2 square you need to add 3 blocks

```          XX
XX
```

3x3 is just the same, but it now requires 5 additional blocks.

```          XXX
XXX
XXX
```

In fact, to move from any nxn square to the (n+1)x(n+1) square requires the addition of 2n + 1 blocks. Coincidentally, this sequence of additional blocks happens to be all of the odd numbers.