Looking for references to Pythagorean triple subsets

Question

I knew nothing about generating Pythagorean triples in 2009 so I looked for them in a spreadsheet. Millions of formulas later, I found a pattern of sets shown in the sample below. $$\begin{array}{c|c|c|c|c|} Set_n & Triple_1 & Triple_2 & Triple_3 & Triple_4 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41\\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65\\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137\\ \hline \end{array}$$

In each $Set_n$, $(C-B)=(2n-1)^2$, the increment between consecutive values of $A$ is $2(2n-1)k$ where $k$ is the member number or count within the set, and $A=(2n-1)^2+2(2n-1)k$. I solved the Pythagorean theorem for $B$ and $C$, substituted now-known the expressions for $A$ and $(C-B)$, and got $\quad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2$.

I have since learned the my formula is the equivalent of replacing $(m,n)$ in Euclid's formula with $((2n-1+k),k)$. I found ways of using either my formula or Euclid's to find triples given only sides, perimeters, ratios, and areas as well as polygons and pyramids constructed of dissimilar primitive triples.

I found that the first member of each set $(k=1)$ and all members of $Set_1 (n=1)$ are primitive. I found that, if $(2n-1)$ is prime, only primitives will be generated in $Set_n$ if $A=(2n-1)^2+2(2n-1)k+\bigl\lfloor\frac{k-1}{2n-2}\bigr\rfloor $ and I found that, if $(2n-1)$ is composite, I could obtain only primitives in $Set_n$ by generating and subtracting the set of [multiple] triples generated when $k$ is a $1$-or-more multiple of any factor of $(2n-1)$. The primitive count in the former is obtained directly; the count for the latter is obtained by combinatorics.

I'm trying to write a paper "On Finding Pythagorean Triples". Surely someone has discovered these sets in the $2300$ years since Euclid but I haven't found and reference to them or any subsets of Pythagorean triples online or in the books I've bought and read. So my question is: "Where have these distinct sets of triples been mentioned before?" I would like to cite the work if I can find it.

The bounty just expired and neither of the two answers has been helpful. I have not quite a day to award the bounty. Any takers? Where and when have these sets been discovered before?

Answer 1

In L. E. Dickson, History of the Theory of Numbers, Volume II, page 167

T. Fantet de Lagny$^{18}$ replaced $m$ by $d+n$ in $(1)$ and obtained $$ x = 2n(d+n),\;\; y=d(d+2n),\;\; z = x+d^2=y+2n^2. $$

The footnote 18 is briefly "Hist. Acad. Sc. Paris, 1729, 318."

Your formulas are

$$A\!=\!(2n\!-\!1)^2\!+\!2(2n\!-\!1)k,\\ B\!=\!2(2n\!-\!1)k\!+\!2k^2,\\ C\!=\!(2n\!-\!1)^2\!+\!2(2n\!-\!1)k\!+\!2k^2.$$

Get this from Lagny's formulas if $\,d\,$ is replaced by $\,2n-1\,$ and $\,n\,$ is replaced with $\,k.\,$

Thus, your formula is equivalent to de Lagny's except $\,2n-1=d\,$ is always odd, however, if $\,d\,$ is even, the triple has a common factor of $2$ and can not be primitive.

Answer 2

This paper defines the 'height' of a triple as $C-B$ and classifies Pythagorean triples in terms of their height and a parameter $k$.

Height and excess of Pythagorean triples, D McCullough - Mathematics Magazine, 2005 - Taylor & Francis, https://doi.org/10.1080/0025570X.2005.11953298

Answer 3

Let $A^2 + B^2 = C^2$ be a Pythagoream triplet. The formula that you have mentioned is a special case of the general formula which gives all Pythagoream triplets.

$$ A = n(r^2 - s^2), B = 2nrs), C = n(r^2+s^2) $$ where $n,r,s$ are some positive integers. In case you want to generate all primitive Pythagorean triplets where $a,b,c$ have no common factors then take $\gcd(r,s) = n = 1$.

Every other special type of triplets can be generated from this general formula so there is actually nothing left.