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.

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KristyAn 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!

Kristynth 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.