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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?

poetasis
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    Well, why not take a primitive triple and multiply each term by an odd square? – lulu Sep 25 '19 at 18:03
  • I know the subset contains odd square multiples of primitives. I don't need to find them. I'm looking for what has been studied about their properties. I have my own observations and I would like to compare them to what's been done. I even developed a formula that generates the entire subset but I'm sure it must have been done before. – poetasis Sep 25 '19 at 18:40
  • all pythagorean triples have certain properties... –  Sep 25 '19 at 19:52
  • All Pythagorean triples have been characterized (in particular, primitive triples). You can find this in any number theory book. – xxxxxxxxx Sep 27 '19 at 16:10
  • I have Principia Mathematica by Whithhead/Russell and History of the Theory of Numbers by Dickson but the closest I could find was the sum of two squares in volume 3 of the latter, no mention of any kind of subsets of Pythagorean triples. I'm looking for any reference to the distinct sets of triples that I found. – poetasis Sep 28 '19 at 10:08
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    I don't know, if Maor, Eli, 2007: The Pythagorean Theorem: a 4000-year History. Princeton Univ. Press. ISBN 9-780-691-14823-6 helps (I haven't read it), but generally it's useful to read books about the history of the Pythagorean triple. – user90369 Oct 11 '19 at 12:28
  • @user90369 Thanks for the suggestion. I've spent about a thousand dollars on books so far looking for a reference. There's a chance this could be the $one$. If you ever find a reference the these sets or my formula, let me know. The bounty expires in $56$ minutes as I write. I wonder what will happen to it. – poetasis Oct 11 '19 at 15:04
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    Your splitting into sets is very special, it's not clear for which purpose (although I've read the explanation for Nilotpal Kanti Sinha). Perhaps you are the first one who is splitting into sets. ;) If you don't really find what you are looking for, it might be better to broaden your topic (as long as the core of the topic is maintained) and allow more ideas. Then more readers can be reached and the probability of finding suitable literature increases. ;) --- For the bounty a grace period of 24 hours is left. :) – user90369 Oct 11 '19 at 16:14
  • @user90369 I want the topic to be this narrow. I want the focus to be on my formula and the sets it generates. I've asked in various forms but most responders haven't focused on the question. They've gone off on common knowledge things that are not related to the $question$ I ask. – poetasis Oct 11 '19 at 16:45
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    Yes, I have already noticed that here. But very special questions don't usually have many readers. If I find suitable literature, I write it of course, but the hope is unfortunately small. – user90369 Oct 11 '19 at 20:01
  • Neither of the two answers I received answered or even addressed the question which was about prior discovery of the sets I discovered. Only the comments by user90369 were somewhat useful. – poetasis Oct 12 '19 at 18:25

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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.

Somos
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  • It will take me a while to understand your post but it looks like a step in the right direction. None of the other posts were useful and I downvoted them both. I lost my bounty points but at least they did not get them. – poetasis Jul 18 '21 at 22:44
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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

Angela Pretorius
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  • I'm sorry, I did not see anything related to the sets I discovered or the formula I developed in the link. If someone has found these sets, surely the formula would follow. e.g. $F(1,1)=(3,4,5), F(1,2)=(5,12,13), F(1,3)=(7,24,25), F(1,4)=(9,40,41)$ $F(2,1)=(15,8,17), F(2,2)=(21,20,29), F(2,3)=(27,36,45)$ and so on. Can you, perhaps, find a reference to a variation of Euclid's formula where $A=((2m-1+n)^2-n^2)\quad B=(2(2m-1+n)n)\quad C=((2m-1+n)^2+n^2)$? It produces the sets shown in the example. – poetasis Oct 09 '19 at 18:22
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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.

Nilotpal Sinha
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  • I know my formula is a special case. It generates only and all triples where $GCD(A,B,C)=(2m-1)^2, m\in \mathbb{N}$ which includes all primitives. One advantage is that it eliminates the trivials, the doubles, the even squares and other non-odd-square multiples of primitives generated by Euclid's formula. What I'm looking for is a reference to any earlier discovery of the $sets$ shown in the example. My discovery appears to be original but I know that is vanity so I'm looking to give credit if prior art can be found. – poetasis Oct 11 '19 at 15:12