# Searching for prior work in generating Pythagorean triples.

by poetasis   Last Updated September 11, 2019 19:20 PM

I've developed a Formula for generating 'only' the complete subset of Pythagorean triples where GCD(A,B,C) is an odd square. It also shows a distinct pattern of subsets within that subset. I haven't found anything resembling it in my online searches or in books I've bought.

Has this been done before?

$$\textbf{Update:}$$ People have asked about trivial triplets and set patterns. The Formula generates non-trivial Pythagorean triples for every pair of natural numbers $$n,k$$ where $$A=(2n-1)^2+2(2n-1)k\quad B=2(2n-1)k+2k^2\quad C=(2n-1)^2+2(2n-1)k+2k^2$$

Here is a sample of the sets it generates where $$n$$ is the set number and $$k$$ is shown as $$Triple_k$$.

$$\begin{array}{c|c|c|c|c|} \text{Set_n}& \text{Triple_1} & \text{Triple_2} & \text{Triple_3} & \text{Triple_4}\\ \hline \text{Set_1} & 3,4,5 & 5,12,13& 7,24,25& 9,40,41\\ \hline \text{Set_2} & 15,8,17 & 21,20,29 &27,36,45 &33,56,65\\ \hline \text{Set_3} & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 \\ \hline \text{Set_{25}} &2499,100,2501 &2597,204,2605 &2695,312,2713 &2793,424,2825\\ \hline \end{array}$$ We can see that $$C-B=(2n-1)^2$$ in each set where $$n$$ is the set number and the step between adjacent values of $$A$$ is $$2(2n-1)$$ so $$k$$ increments the values of $$A$$. The functions for $$B$$ and $$C$$ are derived by solving the Pythagorean Theorem for them, substituting the known expressions for $$A$$ and $$C-B$$, and collecting terms. We can see that $$GCD(A,B,C)=(2m-1)^2,m\in\mathbb{N}$$. Note that $$Set_1$$ contains only primitives as does the first member of each set but, for other sets, when $$k\gt1$$, non-primitives are generated, e.g. $$F(2,3)=(27,36,45)$$, whenever $$k$$ is equal to or a multiple of any factor of $$(2n-1)$$. I have addressed that in a paper I'm writing.

So, "Can my formula or something similar be found elsewhere?"

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