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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.

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  • $\begingroup$ What is meant by $b_r$? How does your recursive formula relate to the sequences of triples you've written? $\endgroup$
    – kccu
    Feb 12, 2021 at 2:19
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    $\begingroup$ Thanks for the suggestion, I have fixed it $\endgroup$ Feb 12, 2021 at 2:27
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    $\begingroup$ Your triples are the upper and lower paths in the ternary tree of Pythagorean triples $\endgroup$ Feb 12, 2021 at 2:28
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    $\begingroup$ 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? $\endgroup$
    – David K
    Sep 26, 2021 at 2:11
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    $\begingroup$ 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). $\endgroup$
    – robjohn
    Sep 26, 2021 at 23:27

3 Answers 3

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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.

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  • $\begingroup$ 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. $\endgroup$
    – robjohn
    Sep 27, 2021 at 8:07
  • $\begingroup$ @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. $\endgroup$
    – poetasis
    Sep 27, 2021 at 9:30
  • $\begingroup$ (+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. $\endgroup$
    – robjohn
    Sep 27, 2021 at 17:47
  • $\begingroup$ @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. $\endgroup$
    – poetasis
    Sep 27, 2021 at 22:32
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    $\begingroup$ 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. $\endgroup$
    – robjohn
    Sep 28, 2021 at 4:50
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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 $\qquad k_x=k_{(x-1)}+(2n-1)$

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  • $\begingroup$ Although the Pythagorean triples my formula produces match this it does not show the proof of its recursiveness. $\endgroup$ Sep 28, 2021 at 22:11
  • $\begingroup$ @SpoonedBread What is reclusiveness? $\endgroup$
    – poetasis
    Sep 28, 2021 at 22:12
  • $\begingroup$ sorry look again $\endgroup$ Sep 28, 2021 at 22:14
  • $\begingroup$ @SpoonedBread Please review my update. $\endgroup$
    – poetasis
    Sep 28, 2021 at 23:20
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    $\begingroup$ @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. $\endgroup$
    – poetasis
    Sep 30, 2021 at 0:50
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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.

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