# Sides of a Right-angled Triangle

$$(2n + 1)^2 + (2n^2 + 2n)^2 = (2n^2 +2n +1)^2$$

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.

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.

Your formula notes that often Pythagorean Triples are of the form $$(t_1, t_2, t_2+1)$$. It attempts to see when this is the case and this formula generates that, we have: $$(2n^2+2n+1)^2-(2n^2+2n)^2=4n^2+4n+1=(2n+1)^2$$

Since the difference here is just a square, it's a valid triple. However, this misses many triples, particularly multiples of existing triples, but an example is $$(8,15,17)$$. I prefer the usage of:

$$(p^2-q^2)^2+(2pq)^2=(p^2+q^2)^2$$ This is most notable for its link to complex numbers, it is $$\Re(z^2)+\Im(z^2)=|z^2|$$ and this can be used to generate them. Set your calculator to Complex mode and use $$(1000\text{Ran#}+\text{1000Ran#}i)^2=$$ and every result will be the base and height of a Pythagoras triangle.

(Ran# is my calculators random number generator, it might be different to yours. Also it generates from $$0.001$$ to $$1$$, max $$3$$dp, hence my multiplying by $$1000$$)

Consider the Primitive Pythagorean triple $$(a, b, c)$$. Consider $$b=4T_n$$ where $$T_n$$ is the $$n^{\text{th}}$$ triangular number. Notice that if this is so, then the value of $$c$$ is $$4T_n+1$$.

Note that the formula for the $$n^{\text{th}}$$ triangular number is given by $$\frac{n(n+1)}{2}$$. All that we are left to do is solve for $$a$$ from the following equation: $$a^2+[4T_n]^2=[4T_n+1]^2$$. Solving this gives the value of $$a$$ as $$2n+1$$. Hence, $$(a, b, c)=(2n+1, 4T_n, 4T_n+1)$$ is a Pythagorean triple.