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Using a right triangle with side lengths $(a,b,c)$ where $a , b < c$, I was thinking about how the area of a Pythagorean triple can be found using the Pythagorean triple right before it and I came across something that worked for a large number of Pythagorean triples, $12r^2 + a_{k- 1}b_{k - 1} = a_kb_k$, a recursive formula where $k$ represents the $k$th term in a sequence. This seemingly generates a sequence of Pythagorean triples that I could not find used in any other formula. Its important to note that $12r^2$ is twice the area of Pythagorean triples that stem from side lengths $(3,4,5)$. Using this formula we can find the $1st$ term of sets where the inradius of each Pythagorean triple is $r + r^2k$ and the relationship between the side lengths are still defined by our recursive formula.

These $1st$ terms are triplets with an even value of $a$ where $r$ increases by $2$:

$(8,15,17),(12,35,37),(16,63,65)...$

Note: We find this using $(8,15,17)$ as we have a recursive formula as well as the knowledge that $r = \frac{a + b - \sqrt{a^2 + b^2}}2$, which lets us find the side lengths of each Pythagorean triple.

Here is a sample of what they generate: $$\begin{array}{c|c|c|c} set_1&15,8,17&33,56,65&51,140,149&69,260,269 \\ \hline set_2&35,12,37&85,132,157&135,352,377&185,672,697 \\ \hline set_3 &63,16,65&161,240,289&259,660,709&357,1276,1325& \\ \hline set_4&99,20,101&261,380,461&423,1064,1145&585,2072,2153 \\ \hline \end{array}$$

I couldn't seem to find any similar formulas to this one or any method of generating Pythagorean triples that follow this sequence, I am looking for a proof.

SpoonedBread
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  • What is meant by $b_r$? How does your recursive formula relate to the sequences of triples you've written? – kccu Feb 12 '21 at 02:19
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    Thanks for the suggestion, I have fixed it – SpoonedBread Feb 12 '21 at 02:27
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    Your triples are the upper and lower paths in the ternary tree of Pythagorean triples – Bill Dubuque Feb 12 '21 at 02:28
  • To me, it looks like your formula is not for generating Pythagorean triples but rather for finding a multiplier given area. Also, for primitive triples, given Euclid's formula, et al, side-A is always odd, side-B is always even, and for half of all triples, $A>B$. There are ways of finding triples given an area if you are interested. – poetasis Feb 12 '21 at 15:53
  • @poetasis note that the value of r is known for the Pythagorean triple in the sequence and the area, which allows us to find the side lengths, also as before, make the condition true for a < b < c. The only two knowns required are (3,4,5), and (8,15,17) as we have a recursive formula – SpoonedBread 9 mins ago – SpoonedBread Feb 12 '21 at 16:31
  • https://mathoverflow.net/questions/225781/fricke-klein-method-for-isotropic-ternary-quadratic-forms/225995#225995 – individ Sep 25 '21 at 06:19
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    I don't see any method described here. Starting with $(3,4,5),$ step by step, exactly what calculations do you perform to "generate" another triple? – David K Sep 26 '21 at 02:11
  • @David K, (8,15,17) has inradius 3 so to generate the next triple which is the 1st term of a set, r increases by 2 so using inradius 5 and the product of the legs of the triple 8,15,17 (120) we substitute them into the formula and get 12(5)^2 + 120 = 420, since we are given the area and inradius we can find the side lengths fairly easily. Since this is the 1st term of a set, each consecutive terms inradius of that set is $5 + 25k$ where $k$ represents the $k$th term in the set. So applying the same method we can determine the rest of the triples in the set. – SpoonedBread Sep 26 '21 at 16:39
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    Is this significantly different from A New formula For Generating Pythagorean Triples?? There is an answer to that question which gives a formula for generating the Pythagorean Triples you have listed here (and there). – robjohn Sep 26 '21 at 23:27
  • @robjohn You misunderstand my question, I am asking for a proof of my formula, not for a known formula, also I changed that question because I wanted to put that one for a bounty, but changed my mind due to the already provided answers and instead made a new one. I would like to delete my previous post but it refuses to let me. – SpoonedBread Sep 27 '21 at 01:46
  • Proof of what formula? If you're talking about $12r^2+a_{k- 1}b_{k - 1}=a_kb_k$, that is answered to the other question. If you're talking about $\frac{a+b-\sqrt{a^2+b^2}}2$, that is also answered to the other question. I see no other formula mentioned in either question. You mention "the Pythagorean triple right before" another; how does one determine the triple "right before" another? – robjohn Sep 27 '21 at 08:00
  • -1 This is an incoherent mess. Voting to close as soon as the bounty expires. – Servaes Sep 29 '21 at 11:23
  • The bounty will be awarded by default in 1 hours. I will be the default but will only get half of it unless you award it to the answer you chose as "correct". – poetasis Oct 02 '21 at 23:23

3 Answers3

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Your formula does generate Pythagorean triples but misses most of them and appears to require seeds to work.

I'm not sure what you are generating. You do generate triples where $C-A=2$ in the first column but that can be generated more easily by $\quad A=4n^2-1\quad B=4n\quad C=4n^2+1.\quad $ The rest of the table shows no pattern that I can see, like a consistent side difference within a set or consistent increment of side values within a set. The following formula generates all primitives and a few that are not but there is a consistent $C-B=(2n-1)^2\quad$ and $\quad A_{n+1}-A_{n}=2(2n-1).\quad$ It is the formula derive when $A=(2n-1+k)^2-k^2,\space B=2(2n-1+k)k,\space C=(2n-1+k)^2+k^2$

\begin{align*} A=(2n-1)^2+ \quad &2(2n-1)k\\ B=\hspace{55pt} &2(2n-1)k\quad+2k^2\\ C=(2n-1)^2+ \quad &2(2n-1)k\quad +2k^2 \end{align*} Here is a sample of what it generates $$\begin{array}{c|c|c|c|c|} n & k=1 & k=2 & k=3 & k=4 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 \\ \hline \end{array}$$ Your formula generates the first column but nothing with a pattern I can see in the other cells. If you do want to work with areas, there is a list of them here. If you can figure out how to generate this sequence, I can show you how to find all of the $1,\space 2, \text{ or } 3\space $ triples that correspond to each area.

poetasis
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  • In this answer, I was able to generate the table given in these questions, but I am at a loss to decipher what formula is being used in these questions to generate this table. – robjohn Sep 27 '21 at 08:07
  • @robjohn I can see that you seem to have replicated the table in the OP and added a set zero to it. My answer, given before the bounty, simply says OP was incomplete as you indicated with $(21,20,29)$. As for “this” table, I gave the formula just above the table. It is the subset where $GCD(A,B,C)=(2m-1)^2, m\in\mathbb{N}$ which includes “all” primitives where $GCD(A,B,C)=1.$ The numbers $n,k$ are natural numbers and no trivials are generated. – poetasis Sep 27 '21 at 09:30
  • (+1) Your table is complete. You have filled in the sets in the question to cover all Pythagorean triples. Each set, in both your answer and the question, has a constant "hypotenuse minus even leg" (this difference being the square of an odd integer). As you can see from my answer, only a fraction of all the triples are covered in sets $\ge1$ (set $0$ does cover all triples that have hypotenuse minus even leg equal to $1$). The higher the set, the smaller the coverage in the sets from the question. – robjohn Sep 27 '21 at 17:47
  • @robjohn $(C-B)$ is always an odd square for primitives but for half of all triples, $A>B.\quad$ The OP table may be incomplete but I think my second answer addresses the question. – poetasis Sep 27 '21 at 22:32
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    I am not saying anything is wrong with your answer. Your answer covers all Pythagorean triples. Yes, the hypotenuse minus the even leg is always an odd square. It is the same odd square for all the triples in each $\text{set}_k$, and increases by $2$ for each successive $\text{set}_k$. The other part of my comment was about the percentage coverage by each successive $\text{set}_k$ in the question (not your answer); the percentage coverage generally gets smaller as $k$ increases. – robjohn Sep 28 '21 at 04:50
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This formula will generate your table \begin{align*} A=(2n-1)^2+ & 2(2n-1)k \\ B= \qquad & 2(2n-1)k+ 2k^2 \\ C=(2n-1)^2+ & 2(2n-1)k+ 2k^2 \end{align*}

if provided with the following $(n,k)$ values $$\begin{array}{c|c|c|c|c|c|} X & Term_1 & Term_2 &Term_3 & Term_4 & Term_5 \\ \hline Set_1 & (1,1) & (1,2) & (1,3) & (1,4 & (1,5) \\ \hline Set_2 & (2,1) & (2,4) & (2,7) &( 2,10) & (2,13) \\ \hline Set_3 & (3,1) & (3,6) & (3,11) & (3,16) & (3,21) \\ \hline Set_4 & (4,1) & (4,8) & (4,15) & (4,22) & (4,29) \\ \hline Set_5 & (5.1) & (5,10) & (5,19) & (5,28) & (5,37) \\ \hline Set_6 & (6,1) & (6,12) & (6,23) & (6,34) & (6,45) \\ \hline \end{array}$$`

\begin{align*} \text{To achieve this we replace k by}\qquad \bigg((2n-1)(k-1)+1\bigg) \end{align*}

\begin{align*} A=(2n-1)^2+ & 2(2n-1) \bigg((2n-1)(k-1)+1\bigg) \\ B= \qquad\qquad\quad & 2(2n-1) \bigg((2n-1)(k-1)+1\bigg)+ 2 \bigg((2n+-)(k-1)+1\bigg)^2. \\ C=(2n-1)^2+ & 2(2n-1) \bigg((2n-1)(k-1)+1\bigg)+ 2 \bigg((2n-1)(k-1)+1\bigg)^2 \end{align*}

and the result is

$$\begin{array}{c|c|c|c|c|c|} X & k=1 & k=2 & k=3 & k=4 & k=5 \\ \hline Set_1 & (3,4,5) & (5,12,13) & (7,24,25) & (9,40,41 ) & (11,60,61) \\ \hline Set_2 & (15,8,17) & (33,56,65) & (51,140,149) &( 69,260,269) & (87,416,425) \\ \hline Set_3 & (35,12,37) & (85,132,157) & (135,352,377) & (185,672,697) & (235,1092,1117) \\ \hline Set_4 & (63,16,65) & (161,240,289) & (259,660,709) & (357,1276,1325) & (413,1716,1765) \\ \hline Set_5 & (99,20,101) & (261,380,461) & (423,1064,1145) & (585,2072,2153) & (747,3404,3485) \\ \hline Set_6 & (143,24,145 ) & (385,552,673) & (627,1564,1685) & (869,3060,3181) & (1111,5040,5161) \\ \hline \end{array}$$`

$\textbf{Update}\qquad $. Strictly speaking, this formula is not recursive in that no triple is dependent on any other. It can be viewed as recursive, however, because $\quad\large{k_x = k_{x-1} + (2n-1)}$.

e.g. For $Set_3,\space$ where $(2n-1)=5,\space$ for column one, $\quad k_0=1+(1-1)(5)=1,\quad$ for column two $\quad k_1=k_0+(2-1)(5)=k_0+5=6,\quad$ for column three $\space k_2=k_0+(3-1)(5)=1+(5+5)=(1+5)+5=k_1+5=11,\space $ etc.

To implement this recursiveness, we let the first triple $T_1$ in each set be $$A=4n^2-1\qquad B=4n\qquad C=4n^2+1$$ and let all other $T_x$ be \begin{align*} A=(2n-1)^2+ & 2(2n-1)k_x \\ B= \qquad & 2(2n-1)k_x + 2k_x ^2 \\ C=(2n-1)^2+ & 2(2n-1)k_x + 2k_x ^2 \end{align*} where $\qquad k_x=k_{(x-1)}+(2n-1)$

poetasis
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  • Although the Pythagorean triples my formula produces match this it does not show the proof of its recursiveness. – SpoonedBread Sep 28 '21 at 22:11
  • @SpoonedBread What is reclusiveness? – poetasis Sep 28 '21 at 22:12
  • sorry look again – SpoonedBread Sep 28 '21 at 22:14
  • @SpoonedBread Please review my update. – poetasis Sep 28 '21 at 23:20
  • @SpoonedBread Thank you for accepting my answer. Do you also choose to award the bounty or to wait 2 more days to see if someone posts an even better answer? My post was adequate but hardly as rigorous as I imagine it could be. – poetasis Sep 29 '21 at 20:47
  • I will wait for 2 more days, but for now your answer is the most favorable. – SpoonedBread Sep 29 '21 at 21:47
  • @ SpoonedBread. Good thinking. I would wait too to see if a more rigorous treatment came along. – poetasis Sep 29 '21 at 22:43
  • @ SpoonedBread. I am still curious to know why you want these particular "sets" of triples since all but the first is missing most primitives with the "skips" between values of "k" that would otherwise increment by one instead of $(2n-1).$ – poetasis Sep 29 '21 at 22:51
  • I am hoping to find a geometric proof that can in some way explain this pattern. (Although I would prefer finding it on my own). – SpoonedBread Sep 29 '21 at 23:52
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    @SpoonedBread Challenging problem.. Perhaps one of these images might give you an idea for an approach. My email is in my profile if you would like to talk outside the forum. – poetasis Sep 30 '21 at 00:50
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We need to write generally speaking the more General equation:

$$aX^2+bXY+cY^2=jZ^2$$

Although I formula solutions recorded, but I see it is of interest expression solutions using any one of the known solution.

If we know what any one solution: $(x,y,z)$ - then you can write a formula for the solutions of this equation.

$$X=jxt^2-cxk^2+2(cyk-jzt)s+(by+ax)s^2$$

$$Y=jyt^2-2jztk+(cy+bx)k^2+2axks-ays^2$$

$$Z=jzt^2-(bx+2cy)kt+czk^2+(bzk-(2ax+by)t)s+azs^2$$

$k,t,s$ - any integer asked us.

individ
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