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(The comments and answers, are no longer relevant to this post other than poestasis's answer, they were caused by misinterpretation)

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, $12r^2 + a_{r - 1}b_{r-1} = a_rb_r$, a recursive formula in terms of inradius($r$). 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 $r + r^2k$ ($k$ represents the $k$th term of a sequence) 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 $1$:

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

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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    The recursive formula is unclear, the start is for $r=1$? initiale terms? – Toni Mhax Feb 17 '21 at 16:39
  • sorry about that, the start for the recursive formula is r = 3, as $(8 ,15,17)$ has an inradius of $3$. – SpoonedBread Feb 17 '21 at 16:40
  • $$15=4^2-12,8=2\cdot4\cdot1,17=4^2+1^2$$ $$259=22^2-15^2,660=2\cdot22\cdot15,709=22^2+15^2$$ $$\cdots$$ All your triples are found by the classical formula, so what ? –  Feb 17 '21 at 16:47
  • @Yves Daoust I'm not sure what you mean by that, could you please expand? – SpoonedBread Feb 17 '21 at 16:53
  • Where is the "new set" ? –  Feb 17 '21 at 16:54
  • @Yves Daoust Sorry, is there a formula that produces these sets? I do not know much about Pythagorean triples could you provide a reference? – SpoonedBread Feb 17 '21 at 16:56
  • Web search is your friend. –  Feb 17 '21 at 16:56
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    All primitive Pythagorean triples can be found as $a=r^2-s^2$, $y=2rs$, $z=r^2+s^2$, where $r$ and $s$ are arbitrary integers of opposite parity, $r\gt s\gt 0$, $\gcd(r,s)=1$. This is well known. For example, Theorem 5.5 in Niven, Zuckerman, Montgomery's An Intorduction to the Theory of Numbers, 5th edition. There is no "new" set of Pythagorean triples. And this is just one known way of ennumerating all primitive Pythagorean triples. – Arturo Magidin Feb 17 '21 at 17:19
  • @poetasis, It is able to obtain all of k =1 and set 1 and set 2 if you include that the odd values of a will produce the inradius of triplets with an odd a starting with (5,12,13). – SpoonedBread Aug 10 '21 at 23:13
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    If you change the question so that the existing answers are no longer applicable, perhaps it would be better to post a new question. People don't enjoy having their efforts minimized. – robjohn Sep 25 '21 at 02:54
  • Furthermore, I don't see why you said that my answer was no longer relevant. I had shown why $ab-12r^2$ is the product of two legs for another Pythagorean triangle. In any case, I have added a bit to that and included a section on generating the sets. – robjohn Sep 25 '21 at 19:54

3 Answers3

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Generating the Sets

Given the sets in the question, we can match each triple to one generated by the classical formula: $\left(m^2-n^2,2mn,m^2+n^2\right)$ where $(m,n)=1$, $2\not\mid m-n\gt0$: $$ \begin{array}{c|c} \text{set}_0&3,4,5&5,12,13&7,24,25&9,40,41\\\hline m,n&2,1&3,2&4,3&5,4\\\hline\text{set}_1&15,8,17&33,56,65&51,140,149&69,260,269\\\hline m,n&4,1&7,4&10,7&13,10\\\hline \text{set}_2&35,12,37&85,132,157&135,352,377&185,672,697\\\hline m,n&6,1&11,6&16,11&21,16\\\hline \text{set}_3 &63,16,65&161,240,289&259,660,709&357,1276,1325\\\hline m,n&8,1&15,8&22,15&29,22\\\hline \text{set}_4&99,20,101&261,380,461&423,1064,1145&585,2072,2153\\\hline m,n&10,1&19,10&28,19&37,28\\\hline \end{array} $$ Noticing the pattern in the table above, we get that $$ \begin{align} \text{column $j$ of set}_k&=\left(m^2-n^2,2mn,m^2+n^2\right)\\ \text{where }(m,n)&=(1+(2k+1)(j+1),1+(2k+1)j) \end{align} $$ Column $0$ is the leftmost column of the table.

To see that $(m,n)=1$, note that $n(j+1)-mj=1$
To see that $2\not\mid m-n\gt0$, note that $m-n=2k+1$

I have added $\text{set}_0$ to the table, following the pattern in the subsequent sets.

These sets do not cover all of the Pythagorean Triples. For example, $(21,20,29)$ is not covered in any of these sets.

Original Answer: Why $\boldsymbol{ab-12r^2=a'b'}$

This answer, and probably many others, shows the following: all relatively prime Pythagorean triples can be written as $\left\{m^2-n^2,2mn,m^2+n^2\right\}$ where $(m,n)=1,\ 2\nmid m-n\gt0$.

The area of a triangle is the inradius times the semi-perimeter. Since a Pythagorean triangle is a right triangle, the area is half the product of the legs. Thus, the inradius is $$ \begin{align} \text{inradius} &=\frac{\text{area}}{\text{semi-perimeter}}\\ &=\frac{mn\left(m^2-n^2\right)}{m^2+mn}\\[9pt] &=n(m-n) \end{align} $$ Suppose that $\{a,b,c\}=\left\{m^2-n^2,2mn,m^2+n^2\right\}$ is a Pythagorean triple. Then $$ \begin{align} \overbrace{2mn\left(m^2-n^2\right)}^{\large ab}-12\overbrace{n^2(m-n)^2}^{\large r^2} &=2m(m+n)n(m-n)-12n(m-n)n(m-n)\\ &=2n(m-n)(m(m+n)-6n(m-n))\\ &=2n(m-n)\left(m^2-5mn+6n^2\right)\\ &=2n(m-n)(m-2n)(m-3n)\\ &=\underbrace{2n(m-2n)\vphantom{\left(n^2\right)}}_{\large e}\underbrace{\left((m-2n)^2-n^2\right)}_{\large d} \end{align} $$ where $\{d,e,f\}=\left\{(m-2n)^2-n^2,2n(m-2n),(m-2n)^2+n^2\right\}$ is another Pythagorean triple.

In fact: $$ \begin{bmatrix}d\\e\\f\end{bmatrix} =\begin{bmatrix} -1&-2&2\\ 2&1&-2\\ -2&-2&3 \end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix} $$ and, inversely, $$ \begin{bmatrix} -1&2&2\\ -2&1&2\\ -2&2&3 \end{bmatrix} \begin{bmatrix}d\\e\\f\end{bmatrix} =\begin{bmatrix}a\\b\\c\end{bmatrix} $$ Note that $d,e,f$ have the same parity as $a,b,c$, respectively.

Because $$ \begin{bmatrix} -1&-2&2\\ 2&1&-2\\ -2&-2&3 \end{bmatrix}^T \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&-1 \end{bmatrix} \begin{bmatrix} -1&-2&2\\ 2&1&-2\\ -2&-2&3 \end{bmatrix} =\begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&-1 \end{bmatrix} $$ we have $d^2+e^2-f^2=a^2+b^2-c^2$.

Furthermore, as mentioned above, $$ \begin{align} \text{inradius} &=\frac{\text{area}}{\text{semi-perimeter}}\\[6pt] &=\frac{ab}{a+b+\sqrt{a^2+b^2}}\\ &=\frac{a+b-\sqrt{a^2+b^2}}2 \end{align} $$

robjohn
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Connecting Pythagorean triples by inradius is well-known (though kudos to you for finding it on your own!). For example, see this document by Neville Robbins.

Kieren MacMillan
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Your formula is interesting but generates only about $1/3$ of the primitive triples that exist. More complete is the subset of triples where $GCD(A,B,C) =(2x-1)^2, x\in\mathbb{N}.\quad$ This subset includes no trivial triples, all primitives triples, and only about $1/3$ of the non-primitives as Euclid's formula. Here is a sample, showing how your formula generates only $Set_2$ and above and only columns $1,4,7,10,\cdots$:

\begin{array}{c|c|c|c|c|c|} n & 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 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 \\ \hline Set_{6} &43,24,145 &165,52,173 &187,84,205 &209,120,241 &231,160,281 \\ \hline \end{array}

These are generate by the formula:: \begin{align*} A=(2n-1)^2+ & 2(2n-1)k \\ B= \qquad\quad\quad & 2(2n-1)k+ 2k^2\\ C=(2n-1)^2+ & 2(2n-1)k+ 2k^2\\ \end{align*}

and is equivalent to Euclid's formula with the following substitutions: $A=(2n-1+k)^2-k^2\quad B=2(2n-1+k)k\quad C=(2n-1+k)^2+k^2$

Is there a way that your formula can generated the missing primitives?

poetasis
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