A New formula For Generating Pythagorean Triples?

Question

(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.

Answer 1

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.

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

Answer 2

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.

Answer 3

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?