Pythagoras' Theorem Part 2

In my post, Pythagoras' theorem, I
wrote about how following an
algorithm, you could generate a number
of Pythagorean triples.

I had put the process into a word-
algorithm, but I forgot to put it in an
algebraic form. I have put that in this post.

Let's say your original number was a,
which has to be odd.

Your first number becomes: a
Your second number becomes: ( a²-1 ) / 2
Your third number becomes: ( a²-1 ) / 2 + 1

These three numbers, if you call them a, b and c,
will satisfy the equation a² + b² = c².

You can prove this algebraically, and I'll show
you how in a later post.