$$(2n + 1)^2 + (2n^2 + 2n)^2 = (2n^2 +2n +1)^2$$
It can be used to generate infinitely many sides of right-angled triangles with integer lengths by putting values of $n = 1, 2, 3, … $
I wanted to know that how we came to this equation. How do we know that putting n = 1, 2, 3, … into this we'll get all the sides of a right-angled triangle. I'm trying to find more about this on the internet, if you can help me what should I find about.
Best Answer
If you start with the sequence of square numbers, and take all the differences between consecutive terms, you get all the odd numbers in sequence. For instance, $$ 2^2 - 1^2 = 3\\ 3^2 - 2^2 = 5\\ 4^2-3^3 = 7 $$ Some times, that odd number happens to be a square itself. For instance, we have $$ 5^2 - 4^2 = 9 = 3^2\\ 13^2 - 12^2 = 25 = 5^2 $$ Rearranging these, we get Pythagorean triples: $$ 3^2 + 4^2 = 5^2\\ 5^2 + 12^2 = 13^2 $$ If we want to describe all the different Pythagorean triples that appear this way, we end up with exactly your expression.