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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