# Pythagorean Triples Function

Question that’s intriguing me today (I know this may be a simple google search away on the internet, but I’m enjoying working it out myself):

There is a subset of Pythagorean triples such that b+1 = c. The familiar ones are

3, 4, 5

5, 12, 13

7, 24, 25

These can be generated for any odd a. For instance, for a = 13, 13^2= 169. The two consecutive numbers whose squares differ by 169 are 84 and 85 for the triple 13, 84, 85. These can be found quickly by taking (169-1)/2$latex i\hbar\frac{\partial}{\partial t}\left|\Psi(t)\right>=H\left|\Psi(t)\right>$

My question: what is the function to generate each b from an a? Continuing the triples looks like this:

9, 40, 41

11, 60, 61

13, 84, 85

15, 112, 113

17, 144, 145

Below is a graph in GeoGebra of these pairs (ignoring c). Haven’t been able to find the generating function. That’s the question of the night.

## 2 thoughts on “Pythagorean Triples Function”

1. Kristy

An intriguing question. You may have found your function already, or you may have forgotten about it altogether… but it’s worthwhile if you teach any right triangle trig! I may actually pose this very question with my High Level class next week.

nth term (a, b, c) a=(2n+1) b=an+n or b=2((n^2)+n)
n=1 (3, 4, 5) a=(2(1)+1) b=(3)(1)+(1) or b=2((1^2)+(1))
n=2 (5,12,13) a=(2(2)+1) b=(5)(2)+(2) or b=2((2^2)+(2))

This holds for all values of n. It would then follow that c=2((n^2)+n)+1

Thanks for the challenge [a year ago 🙂 ]
Happy mathing!

2. Kristy

nth term….. (a, b, c)…….. a=(2n+1)……. b=an+n……….. or…. b=2((n^2)+n)
n=1………… (3, 4, 5)…….. a=(2(1)+1)…. b=(3)(1)+(1)….. or…. b=2((1^2)+(1))
n=2………… (5,12,13)…… a=(2(2)+1)….. b=(5)(2)+(2)…. or…. b=2((2^2)+(2))

yay, formatting.